use std::io;
use std::io::Read;
use std::io::Write;
use std::convert::TryInto;

/* What is this file?
 *
 * This is a tutorial that will show you how to bootstrap an interactive
 * programming environment from a small amount of code.
 *
 * First we will design a virtual computer.
 *
 * Then we will design software to run on that computer, to enable REPL-style
 * interactive programming.
 *
 * A REPL is a "Read, Evaluate, Print loop".  A REPL lets you type code at
 * the keyboard and immediately get a result back.  You can also define
 * functions, including functions that change how the environment works in
 * fundamental ways.
 */

/* What is Forth?
 *
 * Forth is the programming language we will use with our computer.
 *
 * Forth was invented by Chuck Moore in the 1960s as a tool for quickly
 * coming to grips with new computer systems.
 *
 * "Let us imagine a situation in which you have access to
 * your computer. I mean sole user sitting at the board with
 * all the lights, for some hours at a time. This is
 * admittedly an atypical situation, but one that can
 * always be arranged if you are competent, press hard, and
 * will work odd hours. Can you and the computer write a
 * program? Can you write a program that didn't descend from
 * a pre-existing program? You can learn a bit and have a
 * lot of fun trying."
 * -- Chuck Moore, "Programming a Problem-Oriented Language", 1970
 * https://colorforth.github.io/POL.htm
 *
 * As you will see, it does not take much work to get Forth running on a
 * new machine, including a machine with a completely unfamiliar instruction
 * set.
 *
 * But before we can do any of that we will need a machine.  Let's make one.
 */

/* ---------------------------------------------------------------------------
 *                           Part 1 - The Computer
 *  ------------------------------------------------------------------------ */

/* This computer will have a 16-bit CPU.  It will be able to address
 * 2^16 (65536) memory locations, numbered 0 to 65535.
 */
const ADDRESS_SPACE: usize = 65536;

/* The job of a CPU is to load numbers from memory, do math or logic on them,
 * then write the resulting number back into memory.
 *
 * The CPU needs a temporary place to hold numbers while it is working with
 * them.
 *
 * In most CPUs, this place is called a "register".  Registers work like
 * variables in a programming language but there are only a few of them
 * (most CPUs have between 1 and 32).
 *
 * On 64-bit ARM the registers are named  r0, r1, ..., r15.
 * On 64-bit Intel they are instead named rax, rbx, ....
 * Just in case those names ring any bells.
 *
 * Having immediate access to dozens of registers is quite handy, but it means
 * many choices are available to the programmer, or more likely, to the
 * compiler.  And making good choices is Hard.  A lot of work goes into
 * deciding what variable to store in what register ("register allocation") and
 * when to dump register contents back into memory ("spilling").
 *
 * Our CPU avoids these problems by not having registers; instead we store
 * numbers in a stack.  The CPU can only access the value that was most
 * recently pushed onto the stack.  This may seem like a big limitation
 * right now but you will see ways of dealing with it.
 *
 * This makes our CPU a "stack machine" as opposed to a "register machine".
 */

#[derive(Debug)]
struct Stack<const N: usize> {
    mem: [u16; N],
    tos: usize  /* top-of-stack */
}

impl<const N: usize> Stack<N> {
    /* Add a number to the stack. */
    fn push(&mut self, val: u16) {
        self.tos = (self.tos.wrapping_add(1)) & (N - 1);
        /* This stack is fixed-size and can hold N values.
         *
         * When a fixed-size stack fills up, there is a failure case
         * (stack overflow) that must be handled somehow.
         *
         * In this stack, the bitwise & makes it a circular stack, meaning, if
         * it ever fills up, it will discard the oldest entry instead of
         * signaling an error.  The lack of error handling makes the CPU
         * simpler.
         */
        self.mem[self.tos] = val;
    }

    /* Return the most recently pushed number. */
    fn pop(&mut self) -> u16 {
        let val = self.mem[self.tos];
        self.mem[self.tos] = 0;
        /* You don't have to set the value back to zero.  I am only doing
         * this because it makes makes the stack look nicer when dumped
         * out with print!().
         */
        self.tos = (self.tos.wrapping_sub(1)) & (N - 1);
        return val;
    }
}

/* Now that we have a stack let's use one!  Or two?
 *
 * Why two stacks?
 *
 * The first stack will be called the "data stack" and is used instead of
 * registers as already described.
 *
 * The second stack will be called the "return stack".  This one holds
 * subroutine return addresses.  Don't worry if you don't know what that
 * means; we'll get to it later when we talk about the instruction set.
 *
 * In addition to stacks we are going to give the CPU a couple more things:
 *
 * 1. An "instruction pointer", which holds the memory address of the next
 *    instruction that the CPU will execute.
 *
 * 2. To make life simpler we put main memory straight on "the CPU" even
 *    though in a real computer, RAM would be off-chip and accessed through a
 *    data bus.
 */

struct Core {
    ram: [u8; ADDRESS_SPACE],
    /* In our memory, each of the 65536 possible memory addresses will store
     * one 8-bit byte (u8 data type in Rust).  This makes it a 65536 byte
     * (64 KB) memory.
     *
     * We could have chosen to make each memory address store 16-bits instead.
     * That would make this a "word-addressed memory".
     *
     * Instead we are going with the "byte-addressed memory" that is more
     * conventional in today's computers.  This choice is arbitrary.
     */
    ip: u16,  /* instruction pointer */
    dstack: Stack<16>, /* data stack */
    rstack: Stack<32>  /* return stack */
}

/* Helper to initialize the cpu.
 * There is probably a better idiom for this but I am bad at rust */
fn new_core() -> Core {
    let c = Core {
        ram: [0; ADDRESS_SPACE],
        ip: 0,
        dstack: Stack {tos: 15, mem: [0; 16]},
        rstack: Stack {tos: 31, mem: [0; 32]}};
    /* Because these are circular stacks it doesn't matter where top-of-stack
     * starts off pointing.  I arbitrarily set it to the highest index so
     * the first value pushed will wind up at index 0, again because this
     * makes the stack look nicer when printed out.
     */
    return c;
}

/* Now we have a CPU sitting there but it does nothing.
 *
 * A working CPU would execute a list of instructions.  An instruction is
 * a number that is a command for the CPU.  For example:
 *
 * 65522 might mean "add the top two values on the data stack".
 * 65524 might mean "invert the bits of the top value on the data stack".
 *
 * The map of instruction-to-behavior comes from the CPU's
 * "instruction set" i.e. the set of all possible instructions and their
 * behaviors.
 *
 * Normally you program a CPU by putting instructions into memory and then
 * telling the CPU the memory address where it can find the first instruction.
 *
 * The CPU will:
 * 1. Fetch the instruction (load it from memory)
 * 2. Decode the instruction (look it up in the instruction set)
 * 3. Execute that instruction (do the thing the instruction set said to do)
 * 4. Move on to the next instruction and repeat.
 *
 * So now we will make the CPU do those things.
 * We'll start off by teaching it how to access memory, and then we will
 * define the instruction set.
 */

impl Core {
    /* Helper to read a number from the specified memory address. */
    fn load(&self, addr: u16) -> u16 {
        let a = addr as usize;
        /* We immediately run into trouble because we are using byte-addressed
         * memory as mentioned earlier.
         *
         * Each memory location stores 8 bits (a byte)
         *
         * Our CPU operates on 16 bit values and we want each memory operation
         * to take 16 bits at a time for efficiency reasons.
         *
         * What do we do?
         *
         * This CPU chooses to do the following:
         * - Read the low  byte of the 16-bit number from address a
         * - Read the high byte of the 16-bit number from address a+1
         *
         * 16 bit number in CPU: [00000000 00000001]        = 1
         *                        |        |
         *                        |        memory address a = 1
         *                        |
         *                        memory address a+1        = 0
         *
         * This is called "little endian" because the low byte comes first.
         *
         * We could have just as easily done the opposite:
         * - Read the high byte of the 16-bit number from address a
         * - Read the low  byte of the 16-bit number from address a+1
         *
         * 16 bit number in CPU: [00000000 00000001]          = 1
         *                        |        |
         *                        |        memory address a+1 = 1
         *                        |
         *                        memory address a            = 0
         *
         * This is called "big endian" because the high byte comes first.
         */
        return u16::from_le_bytes(self.ram[a..=a+1].try_into().unwrap());
        /* The le in this function call stands for little-endian. */
    }

    /* Helper to write a number to the specified memory address. */
    fn store(&mut self, addr: u16, val: u16) {
        let a = addr as usize;
        self.ram[a..=a+1].copy_from_slice(&val.to_le_bytes());
    }

    /* With that taken care of, we can get around to defining the CPU's
     * instruction set.
     *
     * Each instruction on this CPU will be the same size, 16 bits, for
     * the following reasons:
     *
     * 1. Instruction fetch always completes in 1 read.  You never have to
     *    go back and fetch more bytes.
     *
     * 2. If you put the first instruction at an even numbered address then
     *    you know all the rest of the instructions will also be at even
     *    numbered addresses.  I will take advantage of this later.
     *
     * 3. A variable length encoding would save space but 2 bytes per
     *    instruction is already pretty small so it doesn't matter very much.
     *
     * Here are the instructions I picked.
     *
     *  CALL
     *  ------------------------------------------------------------+----
     *  | n   n   n   n   n   n   n   n   n   n   n   n   n   n   n | 0 |
     *  ------------------------------------------------------------+----
     *
     * What CALL does:
     * ---------------
     *  - Push instruction pointer onto the return stack.
     *  - Set instruction pointer to address nnnnnnnnnnnnnnn0.
     *
     * This lets you call a subroutine at any even numbered address
     * from 0 to 65534.
     *
     * Why this is useful:
     * -------------------
     * Together with the return stack, CALL lets you call subroutines.
     *
     * A subroutine is a list of instructions that does something
     * useful and then returns control to the caller.
     *
     * For example:
     *
     * Address   Instruction   Meaning
     * 100 ->           200    Call 200
     * 102 ->           ???    Add the top two values on the data stack.
     * ...
     * 200 ->           ???    Push the value 3 onto the data stack
     * 202 ->           ???    Push the value 4 onto the data stack 
     * 204 ->           ???    Return to caller
     *
     * Don't worry about the other instructions I am using here.  I will
     * define them later.
     *
     * I mostly want to point out the three instructions that I put
     * at address 200 because they are a subroutine,
     * a small self contained piece of code (6 bytes) that 
     * performs a specific task.
     *
     * Do you think it's cool that you can count exactly how many bytes it
     * took?  I think it's cool.
     *
     * Here is what happens when the CPU begins execution at address 100.
     *
     * Address   Data stack   Return stack
     * 100       []           []    <--- About to call subroutine...
     * 200       []           [102]
     * 202       [3]          [102]
     * 204       [3 4]        [102] <--- About to return from subroutine...
     * 102       [3 4]        []
     * 104       [5]          []
     *
     * The return stack is there to make sure that returning from a subroutine
     * picks up where the caller left off.  We will talk more about the return
     * stack later when we talk about the RET instruction.
     *
     * Limitations of CALL:
     * --------------------
     * This CPU cannot call an instruction that starts at an odd address.
     * a.k.a. "unaligned call" is impossible.
     *
     * At first this seems like a limitation, but it really isn't.
     * If you put the first instruction at an even numbered address then
     * all the rest of the instructions will also be at even numbered
     * addresses.  So this works fine.
     *
     * Of course if you intersperse instructions and data in memory...
     *            _________
     *  ________ |_________| _____________
     * |________|    Data   |_____________|
     * Instructions         More instructions
     *
     * ...then you will have to be careful to make sure the second block
     * of instructions also starts at an even numbered address.
     * You might need to include an extra byte of data as "padding".
     *
     *  Data processing instructions
     *  --------------------------------------------+---------------+----
     *  | 1   1   1   1   1   1   1   1   1   1   1 | x   x   x   x | 0 |
     *  --------------------------------------------+---------------+----
     * Sixteen of the even numbers are reserved for additional instructions
     * that will be be described later.
     *
     * The even numbers 1111111111100000 to 1111111111111110 (65504 to 65534)
     * are reserved for these instructions.  This means that CALL 65504 through
     * CALL 65534 are not possible.  Put another way, it is not possible to
     * call a subroutine living in the top 32 bytes of memory.  This is not a
     * very severe limitation.
     *
     *  LITERAL
     *  ------------------------------------------------------------+----
     *  | n   n   n   n   n   n   n   n   n   n   n   n   n   n   n | 1 |
     *  ------------------------------------------------------------+----
     *
     * What LITERAL does
     * -----------------
     * - Place the value 0nnnnnnnnnnnnnnn on the data stack.
     *
     * Why this is useful:
     * -------------------
     * Program will often need to deal with constant numbers.
     * For example, you might want to add 2 to a memory address (to move
     * on to the next even-numbered address) or add 32 to a character code
     * (to convert it to lower-case).  These constants have to come from
     * somewhere.
     *
     * Limitations of LITERAL:
     * -----------------------
     * To differentiate it from a call, this instruction is always an
     * odd number.  The trailing 1 is discarded before placing the number on
     * the data stack.  This missing bit means that only 2^15 values can be
     * represented (0 to 32767).  32768 on up cannot be stored directly.
     * You would need to do some follow-up math to get these numbers.
     * The most direct way is to use the INV instruction, described later.
     */

     /* Now that the instruction set is generally described
      * let's look at the code that implements it */
    fn step(&mut self) {
        /* 1. Fetch the instruction.
         * Also advance ip to point at the next instruction for next time. */
        let opcode = self.load(self.ip);
        self.ip = self.ip.wrapping_add(2);

        /* 2. Decode and execute the instruction */
        if (opcode >= 0xffe0) && (opcode & 1 == 0) {
            /* Data processing instruction */
            PRIMITIVES[((opcode - 0xffe0) >> 1) as usize](self);
            /* These instructions get looked up in a table.  The bit
             * math converts the instruction code into an index in the
             * table as follows:
             *
             * 0xffe0 --> 0
             * 0xffe2 --> 1
             * ...
             * 0xfffe --> 15
             *
             * The table will be described below, and these instructions
             * explained.
             */
        }
        else if (opcode & 1) == 1 {
            /* Literal */
            self.dstack.push(opcode >> 1);
        }
        else {
            /* Call */
            self.rstack.push(self.ip);
            self.ip = opcode;
        }
    }
}

/* The names of the 16 remaining CPU instructions */
enum Op {
    RET = 0xffe0, TOR = 0xffe2, RTO = 0xffe4, LD  = 0xffe6,
    ST  = 0xffe8, DUP = 0xffea, SWP = 0xffec, DRP = 0xffee,
    Q   = 0xfff0, ADD = 0xfff2, SFT = 0xfff4, OR  = 0xfff6,
    AND = 0xfff8, INV = 0xfffa, GEQ = 0xfffc, IO  = 0xfffe,
}

type Primitive = fn(&mut Core);

/* A table of functions for each of the 16 remaining CPU instructions */
const PRIMITIVES: [Primitive; 16] = [
    /* Return-stack instructions */
    | x | {
        /* RET - Return from subroutine */
        x.ip = x.rstack.pop()
    },
    | x | {
        /* TOR - Transfer number from data stack to return stack */
        x.rstack.push(x.dstack.pop())
    },
    | x | {
        /* RTO - Transfer number from return stack to data stack */
        x.dstack.push(x.rstack.pop())
    },
    /* Memory instructions */
    | x | {
        /* LD - Load number from memory address specified on the data stack */
        let a = x.dstack.pop();
        x.dstack.push(x.load(a));
    },
    | x | {
        /* ST - Store number to memory address specified on the data stack */
        let a = x.dstack.pop();
        let v = x.dstack.pop();
        x.store(a, v);
    },
    /* Stack shuffling instructions
     *
     * Remember the problem of "register allocation" mentioned earlier,
     * and how stack machines are supposed to avoid that problem?  Well,
     * nothing comes for free.  Stack machines can only process the top
     * value(s) on the stack.  So sometimes you will have to do some work
     * to "unbury" a crucial value and move it to the top of the stack.
     * That's what these instructions are for.
     *
     * Their use will become more obvious when we start programming the
     * machine, soon.
     */
    | x | {
        // DUP - Duplicate the top number on the data stack
        let v = x.dstack.pop();
        x.dstack.push(v);
        x.dstack.push(v);
    },
    | x | {
        // SWP - Exchange the top two numbers on the data stack
        let v1 = x.dstack.pop();
        let v2 = x.dstack.pop();
        x.dstack.push(v1);
        x.dstack.push(v2);
    },
    | x | {
        // DRP - Discard the top number on the data stack
        let _ = x.dstack.pop();
    },
    /* Conditional branch instruction */
    | x | {
        /* Q - If the top number on the data stack is zero, skip the next
         * instruction.
         *
         * Note Q is the only "decision-making" instruction that our CPU
         * has.  This means that all "if-then" logic, counted loops, etc.
         * will be built using Q.
         */
        let f = x.dstack.pop();
        if f == 0 {
            x.ip = x.ip.wrapping_add(2)
            /* Because all of our instructions are two bytes, adding two
             * to the instruction pointer skips the next instruction. */
        };
    },
    /* Arithmetic and logic */
    | x | {
        // ADD - Sum the top two numbers on the data stack.
        let v1 = x.dstack.pop();
        let v2 = x.dstack.pop();
        x.dstack.push(v1.wrapping_add(v2));
    },
    | x | {
        /* SFT - Bit shift number left or right by the specified amount.
         * A positive shift amount will shift left, negative will shift right.
         */
        let amt = x.dstack.pop();
        let val = x.dstack.pop();
        x.dstack.push(
            if amt <= 0xf {
                val << amt
            } else if amt >= 0xfff0 {
                val >> (0xffff - amt + 1)
            } else {
                0
            }
        );
    },
    | x | { // OR - Bitwise-or the top two numbers on the data stack.
        let v1 = x.dstack.pop();
        let v2 = x.dstack.pop();
        x.dstack.push(v1 | v2);
    },
    | x | { // AND - Bitwise-and the top two numbers on the data stack.
        let v1 = x.dstack.pop();
        let v2 = x.dstack.pop();
        x.dstack.push(v1 & v2);
    },
    | x | { // INV - Bitwise-invert the top number on the data stack.
        let v1 = x.dstack.pop();
        x.dstack.push(!v1);
        /* You can use the INV instruction to compensate for the LITERAL
         * instruction's inability to encode constants 32768 to 65535.
         * Use two instructions instead:
         * - LITERAL the complement of your desired constant
         * - INV
         *
         * For example, LITERAL(0) INV yields 65535 (signed -1)
         * For example, LITERAL(1) INV yields 65534 (signed -2)
         * etc.
         */
    },
    | x | { // GEQ - Unsigned-compare the top two items on the data stack.
        let v2 = x.dstack.pop();
        let v1 = x.dstack.pop();
        x.dstack.push(if v1 >= v2 { 0xffff } else { 0 });
    },
    /* Input/output.
     *
     * The CPU needs some way to communicate with the outside world.
     *
     * Some machines use memory mapped IO where certain memory addresses are
     * routed to hardware devices instead of main memory.  This machine already
     * has the full 64K of memory connected so no address space is readily
     * available for hardware devices.
     *
     * Instead we define a separate input-output space of 65536 possible
     * locations.  Each of these possible locations is called an IO "port".
     *
     * For a real CPU you could hook up hardware such as a serial
     * transmitter that sends data to a computer terminal, or just an
     * output pin controller that is wired to a light bulb.
     *
     * This is a fake software CPU so I am going to hook it up to
     * stdin and stdout.
     */
    | x | { // IO - Write/read a number from/to input/output port.
        let port = x.dstack.pop();
        /* I'm loosely following a convention where even ports are inputs
         * and odd ports are outputs.  But each port acts different.
         * In a hardware CPU this would not be suitable but it is fine for
         * a software emulation.
         */
        match port {
            0 => {
                /* Push a character from stdin onto the data stack */
                let mut buf: [u8; 1] = [0];
                let _ = io::stdin().read(&mut buf);
                x.dstack.push(buf[0] as u16);
                /* You are welcome to make your own computer that supports
                 * utf-8, but this one does not. */
            }
            1 => {
                /* Pop a character from the data stack to stdout */
                let val  = x.dstack.pop();
                print!("{}", ((val & 0xff) as u8) as char);
                let _ = io::stdout().flush();
            }
            2 => {
                /* Dump CPU status.
                 * Like the front panel on a 1960s-1970s minicomputer. */
                println!("{:?} {:?}", x.dstack, x.rstack);
                let _ = io::stdout().flush();
            }
            _ => {}
        }
    }
];

/* ---------------------------------------------------------------------------
 *                            Part 2 - The Program
 *  ------------------------------------------------------------------------ */

/* "In a sense we're building a tree. We've now reached a
 * point where we can start making the roots. For a while
 * everything will be concealed but we'll eventually reach
 * daylight and start on branches."
 *
 * -- Chuck Moore, "Programming a Problem-Oriented Language", 1970
 * https://colorforth.github.io/POL.htm
 *
 * You now have an unfamiliar computer with no software.  Can you and the
 * computer write a program?
 *
 * The first program is the hardest to write because you don't have any tools to
 * help write it.  The computer itself is going to be no help.  Without any
 * program it will sit there doing nothing.
 *
 * What should the first program be?
 * A natural choice would be a tool that helps you program more easily.
 *
 * An interactive programming environment needs to let you do 2 things:
 *
 * 1. Call subroutines by typing their name at the keyboard
 * 2. Define new subroutines in terms of existing ones
 *
 * Begin with step 1:
 * Call subroutines by typing their name at the keyboard
 *
 * This is where we will meet Forth.
 *
 * The below is a small Forth for bootstrapping this computer.  If you want to
 * learn how to implement a full featured Forth, please read Jonesforth, and
 * Brad Rodriguez' series of articles "Moving Forth".  The simple Forth I write
 * below will probably help you understand those Forths a little better.
 *
 * Forth organizes all the computer's memory as a "dictionary" of subroutines.
 * The point of the dictionary is to give each subroutine a name so you
 * can run a subroutine by typing its name.  The computer will look up its
 * address for you and call it.
 *
 * The dictionary starts at a low address and grows towards high addresses.
 * It is organized like a linked-list, like this:
 *
 * [Link field][Name][Code .......... ]
 *  ^
 *  |
 * [Link field][Name][Code ...... ]
 *  ^
 *  |
 * [Link field][Name][Code ............... ]
 *
 * The reason it is a linked list is to allow each list entry to be a
 * different length.
 *
 * Each dictionary entry contains three things:
 *
 * - "Link field": The address of the previous dictionary entry.
 *                 For the first dictionary entry this field is 0.
 *
 * - "Name": A few letters to name this dictionary entry.
 *           Later you will type this name at the keyboard to call up
 *           this dictionary entry.
 *
 * - "Code": A subroutine to execute when you call up this dictionary
 *           entry.  This is a list of CPU instructions.  Note that one
 *           of the CPU instructions is "call".  So you can have a subroutine
 *           that call other subroutines, or calls itself.
 *
 *           This code must end with a return (RET) instruction.
 *
 *           Example subroutine:
 *
 *           Number Instruction  Meaning
 *           ------ -----------  -------
 *           7      Literal(3)   Push the value 3 onto the data stack
 *           9      Literal(4)   Push the value 4 onto the data stack 
 *           65504  RET          Return to caller
 *
 * A linked list is not a very fast data structure but this doesn't really
 * matter because dictionary lookup doesn't need to be fast.  Lookups are
 * for converting text you typed at the keyboard to subroutine addresses.
 * You can't type very fast compared to a computer so this lookup doesn't
 * need to be fast.
 *
 * In addition to the linked list itself, you will need a couple of
 * variables to keep track of where the dictionary is in memory:
 *
 * - Dictionary pointer:  The address of the newest dictionary entry.
 * - Here:                The address of the first unused memory location,
 *                        which comes just after the newest dictionary entry.
 *
 * [Link field][Name][Code .......... ]
 *  ^
 *  |
 * [Link field][Name][Code ...... ]
 *  ^
 *  |
 * [Link field][Name][Code ............... ]
 *  ^                                       ^
 *  |                                       |
 * [Dictionary pointer]                    [Here]
 *
 * Got all that?
 *
 * To create our Forth interactive programmming environment, we will start
 * by defining subroutines that:
 * - read names from the keyboard
 * - look up and execute dictionary entries by name
 *
 * We will put these subroutines themselves in the dictionary so they are
 * available for use once our interactive environment is up and running!
 *
 * If you were sitting in front of a microcomputer in 196x you would need
 * to create the dictionary with pencil and paper, but in 20xx we will
 * write a Rust program to help create the dictionary.
 *
 * First we need to keep track of where the dictionary is:
 */

struct Dict<'a> {
    dp: u16,   // The dictionary pointer
    here: u16, // The "here" variable
    c: &'a mut Core  // The dictionary lives in memory.  We are going to
                     // hang on to a mutable reference to the core to give
                     // us easy access to the memory.
}

/* Helpers to help put new routines in the dictionary */

enum Item {
    Literal(u16),
    Call(u16),
    Opcode(Op)
}
impl From<u16> for Item { fn from(a: u16) -> Self { Item::Call(a) } }
impl From<Op>  for Item { fn from(o: Op)  -> Self { Item::Opcode(o) } }

impl Dict<'_> {
    /* Helper to reserve space in the dictionary by advancing the "here"
     * pointer */
    fn allot(&mut self, n: u16) {
        self.here = self.here.wrapping_add(n);
    }

    /* Helper to append a 16 bit integer to the dictionary */
    fn comma(&mut self, val: u16) {
        self.c.store(self.here, val);
        self.allot(2);
    }

    /* Helper to append a CPU instruction to the dictionary */
    fn emit<T: Into<Item>>(&mut self, val: T) {
        match val.into() {
            Item::Call(val)    => { self.comma(val) }
            Item::Opcode(val)  => { self.comma(val as u16) }
            Item::Literal(val) => { assert!(val <= 0x7fff);
                                    self.comma((val << 1) | 1) }
        }
    }

    /* Helper to append a "name" field to the dictionary.  To save space and
     * to make each dictionary header a consistent size, I am choosing to not
     * store every letter of the name.  Instead I am storing only the length of
     * the name and then the first three letters of the name.
     *
     * That means these two names will compare equal:
     * - ALLOW (-> 5ALL)
     * - ALLOT (-> 5ALL)
     *
     * Even though their first three letters are the same, these two names
     * will compare unequal because they are different lengths:
     * - FORTH (-> 5FOR)
     * - FORGET (-> 6FOR)
     *
     * If a name is shorter than 3 letters it is padded out with spaces.
     * - X (-> 1X  )
     *
     * You can see that the name field is always four bytes regardless
     * of how many letters are in the name, and the link field is two bytes.
     * This means a dictionary header in this Forth is always six bytes.
     */
    fn name(&mut self, n: u8, val: [u8; 3]) {
        /* Store the length and the first character */
        self.comma(n as u16 | ((val[0] as u16) << 8));
        /* Store the next two characters */
        self.comma(val[1] as u16 | ((val[2] as u16) << 8));
    }

    /* Helper to append a new link field to the dictionary and update the
     * dictionary pointer appropriately. */
    fn entry(&mut self) {
        let here = self.here;
        self.comma(self.dp);
        self.dp = here;
    }
}

/* Now we can start building the dictionary. */
fn build_dictionary(c: &mut Core) {
    use Op::*;
    use Item::*;

    let mut d = Dict {
        dp: 0,  /* Nothing in the dictionary yet */
        here: 2,  /* Reserve address 0 as an "entry point", i.e. where the
                     CPU will jump to start running Forth.  We don't have a
                     Forth interpreter yet so we'll leave address 0 alone for
                     now and start the dictionary at address 2 instead. */
        c: c
    };

    /* Consider the following facts:
     * - The CPU knows how to execute a bunch of instructions strung together.
     * - Forth consists of a bunch of subroutine calls strung together.
     * - Subroutine CALL is a valid instruction of our CPU.
     *
     * This means that we can immediately begin programming our machine in
     * a language resembling Forth, just by writing a list of subroutine
     * calls into the dictionary.
     *
     * The line between "machine code program" and "Forth program" is
     * very blurry.  To illustrate:
     *
     * Here is a subroutine consisting of a few instructions strung together.
     *
     *       Instruction Number  Meaning
     *       ----------- ------  -------
     *       Literal(3)  7       Push the value 3 onto the data stack
     *       Literal(4)  9       Push the value 4 onto the data stack 
     *       RET         65504   Return to caller
     *
     * Here is a Forth subroutine consisting of a few subroutine calls strung
     * together.
     *       Call        Number  Meaning
     *       ----------- ------  -------
     *       S1          1230    Call subroutine S1 which happens to live
     *                           at address 1230
     *       S2          1250    Call subroutine S2 which happens to live
     *                           at address 1250
     *       RET         65504   Return to caller
     *
     * This duality between CPU instructions and Forth code comes from
     * an idea called "subroutine threading".  It is a refinement of an
     * idea called "threaded code".  This has no relation to the kind of
     * threading that lets you run programs in parallel.  You can read more
     * about threaded code on Wikipedia or in the other Forth resources I
     * mentioned earlier (Jonesforth, and Moving Forth by Brad Rodriguez).
     *
     * Our new language starts out with the sixteen (well, eighteen)
     * instructions built into the CPU.  We can string those instructions
     * together into a new subroutine.  This subroutine adds to the pool
     * of functions we have available for making new subroutines.
     *
     * Repeat until you have built what you wanted to build, via
     * function composition.  This is the idea behind Forth.
     */

    /*
     * We are going to be writing many series of instructions so let's
     * start out by making a Rust macro that makes them easier to type
     * and lets us specify a CPU instruction vs. a subroutine call with
     * equal ease.
     *
     * The macro below will convert:
     *
     *     forth!(Literal(2), ADD, RET)
     *
     * to:
     *
     *     d.emit(Literal(2));
     *     d.emit(ADD);
     *     d.emit(RET);
     *
     * which you probably recognize as code that will add a new subroutine
     * to the dictionary.
     */
    macro_rules! forth {
        ($x:expr) => (d.emit($x));
        ($x:expr, $($y:expr),+) => (d.emit($x); forth!($($y),+))
    }

    /* Now we can add the first subroutine to the dictionary!
     *
     * key: Reads a character from the keyboard and places its character
     * code on the stack.
     *
     * There is a tradition of writing stack comments for Forth subroutines
     * to describe the stack effect of executing the subroutine.
     * They look like this: key ( -- n )
     *
     * Read as: key does not take any parameters off the stack, and leaves
     * one new number pushed onto the stack.
     *
     * Also remember that a dictionary entry looks like this:
     * [Link field][Name][Code .......... ]
     */

    // key ( -- n )
    d.entry();           /* Compile the link field into the dictionary */
    d.name(3, *b"key");  /* Compile the name field into the dictionary */
    let key = d.here;    /* (Save off the start address of the code so we
                            can call it later) */
    forth!(
        Literal(0),      /* Compile a LITERAL instruction that pushes
                            0 to the stack */

        IO,              /* Compile an IO instruction.
                          *
                          * Remember from the CPU code that IO takes a
                          * parameter on the stack to specify which port
                          * to use.
                          *
                          * Also remember that IO port 0 reads
                          * a character from standard input.
                          */

        RET              /* Compile a RET instruction */
    );
    /* We have now compiled the "key" subroutine into the dictionary.
     * [Link field][Name][Code .......... ]
     *        0000  3key  1, 65534, 65504
     *
     * The next subroutine we will make is "emit".  This is a companion
     * to "key" that works in the opposite direction.
     *
     * key ( -- n ) reads a character from stdin and pushes it to the stack.
     * emit ( n -- ) pops a character from the stack and writes it to stdout.
     */

    // emit ( n -- )
    d.entry(); d.name(4, *b"emi");  let emit = d.here;
    forth!(Literal(1), IO, RET);

    /* I am tired of saying "subroutine" so many times, so I am going to
     * introduce a new term.  Remember the goal our language is working
     * towards -- we want to be able to type a word at the keyboard, and
     * let the computer look it up in the dictionary and execute the
     * appropriate code.
     *
     * So far we have two named items in the dictionary, call and emit.
     *
     * We are going to term a named dictionary item a "word".
     * This is a Forth tradition.
     *
     * So call and emit are "words", or "dictionary words" if you want to be
     * precise about it.  So far these are the only words we've defined.
     *
     * Let's define some more words.
     */

    /* Our CPU does not have subtraction so let's make subtraction by adding
     * the two's complement.
     *
     * To get the two's complement, do a bitwise invert and add 1.
     *
     * This will be the most complicated Forth that we've written so far
     * so let's walk through step by step. */

    // - ( a b -- a-b )
    d.entry(); d.name(1, *b"-  ");  let sub = d.here;
    forth!(         /* Stack contents:  a b, to start off with.
                     * We want to compute a minus b */

        INV,        /* Bitwise invert the top item on the stack.
                     * Stack contents: a ~b */

        Literal(1), /* Push 1 onto the stack.
                     * Stack contents: a ~b 1 */

        ADD,        /* Add the top two items on the stack.
                     * Stack contents: a ~b+1
                     * Note that ~b+1 is the two's complement of b. */

        ADD,        /* Add the top two items on the stack.
                     * Stack contents: n
                     * Note that n = (a + ~b+1) = a - b */

        RET         /* Done, return to caller, leaving n on the data stack. */
    );
    /* Writing it out like that takes a lot of space.  Normally Forth code
     * is written on a single line, like this:
     *
     * INV 1 ADD ADD RET
     *
     * Looking at it this way, it's easy to see the new word we just
     * created (-) is made from 5 instructions.  It's pretty typical for
     * a Forth word to be made of 2-7 of them.  Beyond that length, things
     * get successively harder to understand, and it becomes a good idea
     * to split some work off into helper words.
     *
     * We will see an example of this below.
     */

    /* Our next word will be useful for Boolean logic.
     *
     * 0= ( n -- f )
     *
     * In a stack comment, "f" means "flag", a.k.a. Boolean value.
     * By Forth convention, zero is false and any nonzero value is true.
     * However the "best" value to use for a true flag is 65535 (all ones)
     * so the bitwise logical operations can double as Boolean logical
     * operations.
     *
     * So what 0= does is:
     * - if n=0,    leave on the stack f=65535
     * - otherwise, leave on the stack f=0
     *
     * It is like C's ! operator.
     *
     * In Rust this could be implemented as:
     *
     * fn zero_eq(n: u16) {
     *     if (n == 0) {
     *         return 65535;
     *     } else {
     *         return 0;
     *     }
     * }
     *
     * Rust has an if-then and block scope, so this is easy to write.
     *
     * The literal translation to a typical register-machine assembly
     * language would look something like this:
     *
     * zero_eq:     compare r0, 0
     *              jump_eq is_zero
     *              move    r0, 0
     *              ret
     * is_zero:     move    r0, 65535
     *              ret
     *
     * It looks simple but I want to point out a couple things about it
     * that are not so simple.
     *
     * The conditional jump instruction, jump_eq.
     * ------------------------------------------
     * Our CPU doesn't have this.  Q is the only "decision-making"
     * instruction that our CPU has.
     *
     * Q - If the top number on the data stack is zero, skip the next
     * instruction.
     *
     * The forward reference
     * ---------------------
     * This is another problem.  Think of the job of an assembler which is
     * converting an assembly language program to machine code.  We are
     * currently writing our code in a tiny assembler that we made in Rust!  It
     * is very simple but so far it has worked for us.  The assembler of our
     * hypothetical register-machine below has a rather nasty problem to solve.
     *
     * zero_eq:     compare r0, 0
     *              jump_eq is_zero  <----- On this line.
     *              move    r0, 0
     *              ret
     * is_zero:     move    r0, 65535
     *              ret
     * 
     * It wants to jump to is_zero but that symbol has not been seen yet and is
     * unrecognized.  On top of that, the assembler also doesn't yet know what
     * address is_zero will have, so doesn't know what jump target to emit.
     * To successfully assemble that kind of program you would need an
     * assembler smarter than the assembler we made for ourselves in Rust.
     *
     * There are ways to solve this but let's NOT solve it.
     *
     * Our CPU has no jump instruction (only call) and our assembler only lets
     * us call things we already defined.  Instead of removing these
     * constraints, find a way to write 0= within the constraints.
     *
     * Here is a start at solving the problem
     * 
     * is_nonzero ( -- 0 )
     *     Literal(0)
     *     RET
     *
     * 0= ( n -- f )
     *     Q            <-- pop n, if n=0 skip next instruction
     *     is_nonzero   <-- f=0 is now pushed to stack
     *     Literal(0)
     *     INV          <-- f=65535 is now pushed to stack
     *     RET          <-- Return
     *
     * We got rid of the forward reference by defining is_nonzero before it
     * was used.
     *
     * We got rid of the jump instruction by using a subroutine call instead.
     *
     * This code is close to working but it doesn't quite work.  The problem
     * is that is_nonzero gives control back to 0= when done, just like
     * a subroutine call normally does, and then 0= runs as normal until it
     * hits the return instruction at the end.
     * So we wind up executing both the f=0 branch and the f=65535 branch,
     * instead of just executing the f=0 branch like we wanted in this case.
     *
     * It is possible to fix this last problem by adding the instructions
     * RTO DRP to is_nonzero.
     *
     * is_nonzero ( -- 0 )
     *     RTO          <-- Pop the return address, push to data stack
     *     DRP          <-- Discard it
     *     Literal(0)   <-- Put 0 on the data stack
     *     RET          <-- Return
     * 
     * Because we popped off and discarded one item from the return stack, the
     * final RET instruction will not return to 0= any more.  Instead it will
     * skip one level and return to whoever called 0=.  This has the result of
     * ending 0= early, which is what we wanted to do.
     * 
     * I call this pattern "return-from-caller".  It is used occasionally in
     * real Forth systems.  My dialect of Forth will use it extensively to work
     * around my CPU's lack of conditional branch.
     *
     * Now we've explained how 0= is going to work, let's make it.
     */

    /* First we define the helper.  It won't be reused, so I am not going
     * to bother giving it a dictionary header and name for easy lookup later.
     * Think of it as a private function. */
    let zero = d.here;
    forth!(Literal(0), RTO, DRP, RET);

    /* Now define 0= using the helper. */
    // 0= ( n -- f )
    d.entry(); d.name(2, *b"0= ");  let zero_eq = d.here;
    forth!(Q, zero, Literal(0), INV, RET);

    /* Next let's make a = equality comparison operator, using 0= and subtract.
     * I call it an "operator" because that's what other languages would
     * call it, but Forth has no special idea of an "operator".  Everything
     * is just words. */
    // = ( a b -- a=b )
    d.entry(); d.name(1, *b"=  ");  let eq = d.here;
    forth!(sub, zero_eq, RET);
    /* Note that 0= and subtract are both words, not CPU instructions.
     * This makes = the first "pure" Forth word we have defined, with no
     * direct dependency on the machine's instruction set.
     * We could define = as - 0= on a real standards-compliant Forth system
     * and it would still work.  So Forth gets you to the point of writing
     * "portable" code really quickly.  Often you can reuse routines early in
     * bootstrapping even though they were written and tested on a different
     * machine.  Many languages offer portability but few offer it so quickly.
     */

    /* Now that we've got some basics in place let's go back to solving
     * the real problem of getting our language to read words from the
     * keyboard.  The first problem we have is that we need some way to
     * separate words from each other so we know where one word ends and the
     * next begins.  This problem is called "lexing".  Forth has about the
     * simplest lexer ever, it just splits on whitespace.  Anything with
     * character code <=32 is considered whitespace.  Words are delimited by
     * whitespace.  And that is all the syntax Forth has.
     *
     * To read a word from the keyboard you will need to:
     * - Advance past any leading whitespace
     * - Read characters into a buffer until whitespace is seen again.
     */

    /* Let's start with the "advance past whitespace" part */
    let skip_helper = d.here;
    forth!(RTO, DRP, key, DUP, Literal(33), GEQ, Q, RET, DRP, skip_helper);

    // skipws ( -- c )
    d.entry(); d.name(6, *b"ski");  let skipws = d.here;
    forth!(skip_helper);

    // over ( a b -- a b a )
    d.entry(); d.name(4, *b"ove");  let over = d.here;
    forth!(TOR, DUP, RTO, SWP, RET);

    // 2dup ( a b -- a b a b )
    d.entry(); d.name(4, *b"2du");  let twodup = d.here;
    forth!(over, over, RET);

    // Buffer for parsing an input word, formatted as Nabcde.
    let word_buf = d.here;
    d.allot(6);

    // min ( a b -- n )
    d.entry(); d.name(3, *b"min");  let min = d.here;
    forth!(twodup, GEQ, Q, SWP, DRP, RET);

    // c@ ( a -- n )
    d.entry(); d.name(2, *b"c@ ");  let cld = d.here;
    forth!(LD, Literal(0xff), AND, RET);

    // c! ( n a -- )
    d.entry(); d.name(2, *b"c! ");  let cst = d.here;
    forth!(DUP, LD, Literal(0xff), INV, AND, SWP, TOR, OR, RTO, ST, RET);

    /* Load 1 letter into buffer. */
    let stchar = d.here;
    forth!(Literal(word_buf), cld, Literal(1), ADD, DUP, Literal(word_buf), cst,
           Literal(5), min, Literal(word_buf), ADD, cst, RET);

    /* Load letters into buffer until whitespace is hit again.
     * Return the whitespace character that was seen. */
    let getcs_helper = d.here;
    forth!(RTO, DRP, stchar, key, DUP, Literal(32), SWP, GEQ, Q, RET, getcs_helper);

    d.entry(); d.name(5, *b"get");  let getcs = d.here;
    forth!(getcs_helper, RET);

    // word ( -- )
    // Not quite standard.
    d.entry(); d.name(4, *b"wor");  let word = d.here;
    forth!(Literal(word_buf), DUP, Literal(2), ADD,
           Literal(0x2020), SWP, ST, Literal(0x2000), SWP, ST,
           skipws, getcs, DRP, RET);

    // latest ( -- a )
    /* Address of "latest" variable.  This variable stores the address of
       the latest word in the dictionary. */
    let latest_ptr = d.here; d.allot(2);
    d.entry(); d.name(6, *b"lat");  let latest = d.here;
    forth!(Literal(latest_ptr), RET);

    let matches = d.here;
    forth!(Literal(2), ADD, TOR,
           Literal(word_buf), DUP, Literal(2), ADD, LD, SWP, LD,
           RTO, DUP, TOR,
           LD, Literal(0x0080), INV, AND, eq,
           SWP, RTO, Literal(2), ADD, LD, eq, AND, RET);

    let matched = d.here;
    forth!(Literal(6), ADD, RTO, DRP, RET);

    let find_helper = d.here;
    forth!(RTO, DRP,
           DUP, Literal(0), eq, Q, RET,
           DUP, matches, Q, matched,
           LD, find_helper);

    // find ( -- xt|0 )
    d.entry(); d.name(4, *b"fin");  let find = d.here;
    forth!(latest, LD, find_helper);

    // ' ( -- xt|0 )
    d.entry(); d.name(1, *b"'  ");
    forth!(word, find, RET);

    /* --- The outer interpreter --- */

    // x10 ( n -- n*10 )
    d.entry(); d.name(3, *b"x10");  let x10 = d.here;
    forth!(DUP, DUP, Literal(3), SFT, ADD, ADD, RET);

    // here ( -- a )
    /* Address of "here" variable.  This variable stores the address of
       the first free space in the dictionary */
    let here_ptr = d.here; d.allot(2);
    d.entry(); d.name(4, *b"her");  let here = d.here;
    forth!(Literal(here_ptr), RET);

    // state ( -- a )
    /* Address of "state" variable.  This variable stores -1 if
     * interpreting or 0 if compiling. */
    let state_ptr = d.here; d.allot(2);
    d.entry(); d.name(5, *b"sta");  let state = d.here;
    forth!(Literal(state_ptr), RET);

    let word_addr = d.here;
    forth!(Literal(latest_ptr), LD, Literal(2), ADD, RET);

    // immediate ( -- )
    d.entry(); d.name(9 | 0x80, *b"imm");
    forth!(word_addr, DUP, LD, Literal(0x0080), OR, SWP, ST, RET);

    // smudge ( -- )
    d.entry(); d.name(6 | 0x80, *b"smu");  let smudge = d.here;
    forth!(word_addr, DUP, LD, Literal(0x0040), OR, SWP, ST, RET);

    // unsmudge ( -- )
    d.entry(); d.name(8 | 0x80, *b"uns");  let unsmudge = d.here;
    forth!(word_addr, DUP, LD, Literal(0x0040), INV, AND, SWP, ST, RET);

    // [ ( -- )
    d.entry(); d.name(1 | 0x80, *b"[  ");  let lbracket = d.here;
    forth!(Literal(0), INV, state, ST, RET);

    // ] ( -- )
    d.entry(); d.name(1 | 0x80, *b"]  ");  let rbracket = d.here;
    forth!(Literal(0), state, ST, RET);

    // , ( n -- )
    d.entry(); d.name(1, *b",  "); let comma = d.here;
    forth!(here, LD, ST,
           here, LD, Literal(2), ADD, here, ST, RET);

    let compile_call = d.here;
    forth!(DUP, Literal(4), sub, LD, Literal(0x0080), AND, state, LD, OR, Q, RET,
           comma, RTO, DRP, RET);

    let compile_lit = d.here;
    forth!(state, LD, Q, RET,
           DUP, ADD, Literal(1), ADD, comma, RTO, DRP, RET);

    let end_num = d.here;
    forth!(DRP, RTO, DRP, RET);

    let bad_num = d.here;
    forth!(DRP, DRP, DRP, Literal(0), INV, RTO, DRP, RET);

    let number_helper = d.here;
    forth!(RTO, DRP, DUP, Literal(word_buf), ADD, cld,
           Literal(48), sub, DUP, Literal(10), GEQ, Q, bad_num,
           SWP, TOR, SWP, x10, ADD, RTO,
           DUP, Literal(word_buf), cld, GEQ, Q, end_num,
           Literal(1), ADD, number_helper);

    // number ( -- n|-1 )
    d.entry(); d.name(6, *b"num");  let number = d.here;
    forth!(Literal(0), Literal(1), number_helper);

    // execute ( xt -- )
    d.entry(); d.name(7, *b"exe");  let execute = d.here;
    forth!(TOR, RET);

    let doit = d.here;
    forth!(RTO, DRP, compile_call, execute, RET);

    let bad = d.here;
    forth!(DRP, Literal(63), emit, RTO, DRP, RET);

    // dispatch ( xt -- )
    d.entry(); d.name(9, *b"int");  let dispatch = d.here;
    forth!(DUP, Q, doit,
           DRP, number, DUP, Literal(1), ADD, zero_eq, Q, bad,
           compile_lit, RET);

    // quit ( -- )
    d.entry(); d.name(4, *b"qui");  let quit = d.here;
    forth!(word, find, dispatch, quit);

    // create ( -- )
    d.entry(); d.name(6, *b"cre");  let create = d.here;
    forth!(word,
           here, LD, latest, LD, comma, latest, ST,
           Literal(word_buf), DUP, LD, comma, Literal(2), ADD, LD, comma, RET);

    // : ( -- )
    d.entry(); d.name(1, *b":  ");
    forth!(create, smudge, rbracket, RET);

    // ; ( -- )
    d.entry(); d.name(1 | 0x80, *b";  ");
    forth!(Literal(!(RET as u16)), INV, comma, lbracket, unsmudge, RET);

    /* Finally put the primitives in the dictionary so they can be
     * called interactively. */
    d.entry(); d.name(3, *b"ret"); forth!(RTO, DRP, RET);
    d.entry(); d.name(2, *b">r "); forth!(RTO, SWP, TOR, TOR, RET);
    d.entry(); d.name(2, *b"r> "); forth!(RTO, RTO, SWP, TOR, RET);
    d.entry(); d.name(1, *b"@  "); forth!(LD, RET);
    d.entry(); d.name(1, *b"!  "); forth!(ST, RET);
    d.entry(); d.name(3, *b"dup"); forth!(DUP, RET);
    d.entry(); d.name(4, *b"swa"); forth!(SWP, RET);
    d.entry(); d.name(4, *b"dro"); forth!(DRP, RET);

    d.entry(); d.name(1 | 0x80, *b"?  "); // This one only works in-line.
    forth!(Literal(!(Q as u16)), INV, comma, RET);

    d.entry(); d.name(1, *b"+  "); forth!(ADD, RET);
    d.entry(); d.name(5, *b"shi"); forth!(SFT, RET);
    d.entry(); d.name(2, *b"or "); forth!(OR, RET);
    d.entry(); d.name(3, *b"and"); forth!(AND, RET);
    d.entry(); d.name(3, *b"inv"); forth!(INV, RET);
    d.entry(); d.name(3, *b"u>="); forth!(GEQ, RET);
    d.entry(); d.name(2, *b"io "); let io = d.here; forth!(IO, RET);

    d.c.store(latest_ptr, io-6);
    d.c.store(here_ptr, d.here);
    d.c.store(state_ptr, 0xffff);
    d.c.store(0, quit);
}

fn main() {
    let mut c = new_core();
    build_dictionary(&mut c);
    c.ip = 0;
    loop {
        c.step();
    }
}