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building a computer in a programming language for kids

… or in other words, the dumbest way to get into embedded systems.

Watch it in action here!

The Goal

The goal was simple. Write some code in C or C++, and be able to execute in Scratch. Honestly, I just found the idea pretty funny: one of the fastest programming languages in one of the slowest. I had a feeling it was possible, but I wasn’t quite sure how. In the process, I learned way more about assembly languages, process memory, and executable files than I expected, and I hope you learn something new as well as I recount my journey.

Step 0: Making a Plan

My first idea was to take the code I wrote in C, break it up into pieces, and then put those pieces back together using Scratch. For example, a while loop in C might become a repeat until block in Scratch:

A while loop in C next to an equivalent “repeat until” loop in Scratch

In order for a C compiler to understand code, it first needs to generate an AST (abstract syntax tree), which is a tree representation of each important symbol in the source code. For example, an opening parenthesis, the name of a variable, or the return keyword might each be converted into distinct nodes. However, after looking at the AST for a simple Fibonacci number program…

Step 0b: Making a (Better) Plan

Okay, so that was out of the question. But what if instead of trying to re-compile the source code, we went one step down the ladder: assembly? In order for a program to run, it first needs to be compiled down into assembly. On my computer, that’s x86-64asm. Since assembly doesn’t have any complicated nesting structures, classes, or even variables, trying to parse a list of assembly instructions should (in theory) be easier than trying to parse the spaghetti monster of an AST, such as the one above. Here’s the same Fibonacci program but in x86 assembly.

The Fibonacci program above in x86–64asm.

Oh, brother. Okay, maybe it’s not that bad. How many total instructions are there?

That’s not a good sign…

Oh no…

Oh my God, it got worse…

Step 0c: The Complete Plan

Thankfully, x86 isn’t the only assembly language out there. As part of a college class, I learned about MIPS, a type of assembly language (over-simplifying) used in some video game consoles and supercomputers from the ’90s to early 2000s, that still sees some use today. Switching from x86 to MIPS brings the instruction count down from *unknown *to around 50.

The same Fibonacci program in MIPS I assembly.

Using a 32-bit version of MIPS, this assembly code can then be converted into machine code, where each instruction is converted into a 32-bit integer that the processor can understand, based on guidelines set by the processor’s architecture. There’s a book on the MIPS instruction set architecture available online, so if I take the machine code, and then emulate exactly what a MIPS processor would do, then I should be able to run my C code in Scratch!

The MIPS instruction  raw `addi $t0, $t1, -4` endraw  and its binary representation as machine code.

Now that that’s out of the way, we can get started.

Step 1: We Cannot Yet Get Started

Well, there’s already a problem. Usually if you have an integer and you want to pull out a series of bits from it, you calculate num & mask, where mask is an integer in which each important bit is 1, and each unimportant bit is 0.

   001000 01001 01000 1111111111111100
 & 000000 00000 00000 1111111111111111
--------------------------------------
   000000 0000 000000 1111111111111100
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The problem? There is no & operator in Scratch.

Now, I *could *simply go through both numbers bit by bit and check each of the four possible combinations of two bits, but that would be wastefully slow; after all, this will need to be done multiple times for *each *instruction. Instead, I came up with a better plan.

First, I wrote a quick Python script to calculate x & y for every x and every y between 0 and 255.

for x in range(256):
    for y in range(256):
        print(x & y)

0      (0 & 0 == 0)
0      (0 & 1 == 0)
0      (0 & 2 == 0)
...
0      (0 & 255 == 0)
0      (1 & 0 == 0)
1      (1 & 1 == 1)
0      (1 & 2 == 0)
...
254    (255 & 254 == 254)
255    (255 & 255 == 255)
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Now, for example, in order to calculate x & y for two 32-bit integers, we can do the following:

  1. Split x and y into four 8-bit integers (or bytes).

  2. Check what first_byte_in_x & first_byte_in_y is by looking in the table generated from the Python script.

  3. Similarly, look up what second_byte_in_x & second_byte_in_y is, and the third bytes, and the fourth bytes.

  4. Take the results of each of these calculations, and put them together to get the result of x & y .

The calculation  raw `0x2128fffc & 0x0000ffff` endraw  is turned into four calculations: (0x21 & 0x00), (0x28 & 0x00), (0xff & 0xff), (0xfc & 0xff). We can look up the value of each operation in our list, and then join the outputs together to get the final result.

However, once a MIPS instruction has been cut up into four bytes, we’ll only & the bytes we need. For example, if we only need data from the first byte, we won’t even look at the bottom three. But how do we know which bytes we need? Based on the opcode (i.e. the “type”) of an instruction, MIPS will try to split up the bits of an instruction in one of three ways.

The three ways a MIPS instruction can be decoded: as a register-type instruction, as an immediate-type instruction, or as a jump-type instruction.

Putting everything together, below is the Scratch code to extract opcode, $rs, $rt, $rd, shamt, funct, and immediate for any instruction.

Step 2: A Short Word About Memory

So, how much memory should our processor actually have? And how should we store it? Well, minimum, MIPS processors have 31 general-purpose registers, and one $zero register that is meant to store the number 0 at all times. A register is a location in memory that a processor can access quickly. We can represent these 32 registers as a list with 32 items in Scratch. As for the rest of the memory, simulating a processor moving chunks of data in and out of its cache in Scratch would be pretty pointless and would actually slow things down, rather than speed them up. So instead, the physical memory will be represented as five lists containing 131,072 elements each, where each element will be a 32-bit integer, giving us about 2.6MB of memory. A contiguous block of memory like these lists is usually called a “page”, and the size of the data that the instruction set works with (in this case 32 bits) is usually called a “word”.

(131 072 words / page) * (32 bits / word) * (1 MB / 8e6 bits) * (5 pages) ≈ 2.62 MB

Step 3: Visits from a Magical ELF

So, how do we get machine code in here? We can’t just import a file into Scratch. But we *can *import text! So, I wrote a program in C to take a binary executable file, and convert every 32 bytes of the file into an integer. C, by default, was reading each byte in little-endian, so I had to introduce a function to flip the endianness. Then, I can save the machine code of a program as a text file (a list of integers), and then import it into my proc:memory:program variable.

#include <stdio.h>

unsigned int flip_endian(unsigned int value) {
    return ((value >> 24) & 0xff) | ((value >> 8) & 0xff00) | ((value << 8) & 0xff0000) | ((value << 24) & 0xff000000);
}

int main(int argc, char* argv[]) {
    if (argc != 3 && argc != 2) {
        printf("Usage: %s <input file> <output file?>\n", argv[0]);
        return 1;
    }

    FILE* in = fopen(argv[1], "r");
    if (!in) {
        perror("fopen");
        return 1;
    }

    unsigned int value;

    FILE* out = argc == 3 ? fopen(argv[2], "w") : stdout;
    if (!out) {
        perror("fopen");
        return 1;
    }

    while (fread(&value, sizeof(value), 1, in) == 1) {
        fprintf(out, "%u\n", flip_endian(value));
    }

    fclose(in);
    if (out != stdout) {
        fclose(out);
    }
    return 0;
}
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The Fibonacci program from above, now loaded into Scratch, as a list of 2,040 32-bit words.

Okay, so now that we can import the data into Scratch, we can just set the program counter (the integer keeping track of the current instruction) to the top of the list, and start executing instructions, right?

Wrong.

I didn’t realize this going into this project, but the first several bytes of an executable file *aren’t *instructions, but a header identifying what type of executable file it is. On Windows, it’ll usually be the PE, or Portable Executable, format, and on UNIX-based systems (the version we’ll be using) it’ll be the ELF format. So, how do we actually know where the code starts? On Linux, we can use the builtin readelf utility to actually see what’s in the ELF header, and the Linux Foundation has a page detailing the ELF header standard. So, we can use the LF page to figure out which bytes mean what, and the readelf command to “check our work”.

 $ readelf -h fibonacci
ELF Header:
  Magic:   7f 45 4c 46 01 02 01 00 00 00 00 00 00 00 00 00
  Class:                             ELF32
  Data:                              2's complement, big endian
  Version:                           1 (current)
  OS/ABI:                            UNIX - System V
  ABI Version:                       0
  Type:                              EXEC (Executable file)
  Machine:                           MIPS R3000
  Version:                           0x1
  Entry point address:               0x4012cc
  Start of program headers:          52 (bytes into file)
  Start of section headers:          7596 (bytes into file)
  Flags:                             0x1001, noreorder, o32, mips1
  Size of this header:               52 (bytes)
  Size of program headers:           32 (bytes)
  Number of program headers:         5
  Size of section headers:           40 (bytes)
  Number of section headers:         14
  Section header string table index: 13
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The output of the command  raw `hexdump -C fibonacci` endraw , which shows the contents of the file  raw `fibonacci` endraw  in hexadecimal, with arrows pointing out what different parts of the ELF header represent.

Now, there’s a lot of really interesting stuff here, but to save some time, the *really *important data here (besides the entry point, of course) are the section headers. Oversimplifying greatly, in order for our program to run correctly, we need to take certain chunks of the file and place them in certain parts of memory so our code can access them.

Standard program memory layout

Using the readelf utility, we can actually see all of the sections in the file:

 $ readelf -S fibonacci
There are 14 section headers, starting at offset 0x1dac:

Section Headers:
  [Nr] Name              Type            Addr     Off    Size   ES Flg Lk Inf Al     
  [ 0]                   NULL            00000000 000000 000000 00      0   0  0
  [ 1] .MIPS.abiflags    MIPS_ABIFLAGS   004000d8 0000d8 000018 18   A  0   0  8
  [ 2] .reginfo          MIPS_REGINFO    004000f0 0000f0 000018 18   A  0   0  4
  [ 3] .note.gnu.build-i NOTE            00400108 000108 000024 00   A  0   0  4
  [ 4] .text             PROGBITS        00400130 000130 001200 00  AX  0   0 16
  [ 5] .rodata           PROGBITS        00401330 001330 000020 00   A  0   0 16
  [ 6] .bss              NOBITS          00411350 001350 000010 00  WA  0   0 16
  [ 7] .comment          PROGBITS        00000000 001350 000029 01  MS  0   0  1
  [ 8] .pdr              PROGBITS        00000000 00137c 000440 00      0   0  4
  [ 9] .gnu.attributes   GNU_ATTRIBUTES  00000000 0017bc 000010 00      0   0  1
  [10] .mdebug.abi32     PROGBITS        00000000 0017cc 000000 00      0   0  1
  [11] .symtab           SYMTAB          00000000 0017cc 000380 10     12  14  4
  [12] .strtab           STRTAB          00000000 001b4c 0001db 00      0   0  1
  [13] .shstrtab         STRTAB          00000000 001d27 000085 00      0   0  1
Key to Flags:
  W (write), A (alloc), X (execute), M (merge), S (strings), I (info),
  L (link order), O (extra OS processing required), G (group), T (TLS),
  C (compressed), x (unknown), o (OS specific), E (exclude),
  p (processor specific)
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Going through all the details of the ELF format could be its own multi-part write-up, but using the Linux Foundation page on section headers, I was able to decipher the section header bytes of the program, and copy all the important bytes from the proc:memory:program variable to the correct places in memory, by checking whether or not the section header had the ALLOCATE flag set.

The Scratch routine that loads the program into memory.

Step 4: Cycling Through Instructions

Fast-forwarding about a week to the point where all of the important instructions have been implemented, let’s take a look at the steps the processor (or really, any processor) needs to take in order to understand just one instruction, using 0x8D02002A (2365718570) as an example.

The first step is called **INSTRUCTION FETCH. **The current instruction is retrieved from the address stored in the proc:program_counter variable.

The next step is INSTRUCTION DECODE, where the instruction is decoded into its separate parts (see Step 1).

Finally, we reach EXECUTE, which, in my Scratch processor, is pretty much just a big if statement.

In this case, the INSTRUCTION DECODE step revealed that the opcode is 35, which means 0x8D02002A is a lw (load word) instruction. Therefore, based off the values in proc:instr:rs, proc:instr:rt, and proc:instr:immediate, the instruction 0x8D02002A actually means lw $2, 0x2a($8) , or in other words, lw $v0, 42($t0).

The Scratch routines for Instruction Fetch, Instruction Decode, and Execute.

And here is the code that handles the lw instruction:

Step 5: Hello, World?

Okay, home stretch. Now, we just need to be able to do the bare minimum and create a “Hello, World” program in C, and run it in Scratch, and the last two weeks of my life will have been validated.

So, will this work?

#include <stdio.h>

int main() {
  printf("Hello, world!");
  return 0;
}
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Three changes. First of all, the MIPS linker uses start to find the entry point of the program, much the same way you use main in C, or "main__" in Python. So, that’s an easy fix.

#include <stdio.h>

int __start() {
  printf("Hello, world!");
  return 0;
}
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Next, we need some way to actually see this output in Scratch. We *could *make some intricate array of text sprites, but the simpler solution is just to use a list.

stdout as a list.

Finally, we can’t use stdio.h.

Yeah, basically, implementing floating point registers and multiprocessor instructions would have been more trouble than it was worth, so I skipped it, but the standard library kind of expects all that to be there. So, we need to make printf ourselves.

Putting the complications of variadic arguments and text formatting aside, how can you actually print a string using MIPS? The TL;DR is you put the address of the string in a certain register, and then a special “print string” value in another register, and then execute the syscall (“system call”) instruction, and let the OS/CPU handle the rest.

The exact special values and registers to use are implementation-dependent, and can be implemented pretty much any way you see fit, but I chose to replicate MARS’ (a very popular MIPS simulator) implementation. With MARS, the address of the string goes in $a0, and the value 4 goes in $v0 to say “hey, I want to print a string!”

The function that handles syscalls on the left, and the function that handles printing an individual character on the right.

And with C, we can use a feature called “inline assembly” to inject assembly code directly into our compiled output. Putting it all together we get this:

#define puts _puts

void _puts(const char *s) {
 __asm__(
  "li $v0, 4\n"
  "syscall\n"
  :
  : "a"(s)
 );
}

int __start() {
 puts("Hello, World!\n");
 return 0;
}
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And when we run it, we get this:

Hello, World!

Conclusion

You can view the final product here: https://scratch.mit.edu/projects/1000840481/.

I wanted to keep this read under 15 minutes, so I had to skip over **a lot **of details. Some parts of the Scratch code had to be cut out of the screenshots for simplicity’s sake and I ran into a lot of silly and not-so-silly mistakes. If you’re curious how I was able to get Connect Four working (with minimax and alpha-beta pruning), the source code is on my Github. Here’s a quick list of some of the other problems I ran into in development:

 * The fact that my computer is little-endian, but MIPS is big-endian caused more issues than I'd like to admit
 * The `mult` instruction in MIPS is 32-bit multiplication, and multiplying two 32-bit integers can result in a 64-bit integer.
   Javascript (and as a result, Scratch) is incapable of storing a 64-bit integer without losing precision.
 * The `u` in the `addu` instruction and the `u` in the `sltu` instruction both stand for "unsigned", but mean completely different things.
 * As you may have noticed, functions in Scratch don't have return values. This was quite annoying.
 * Any branch instruction (like "jump", "jump register", "branch on equals") in MIPS will also execute the instruction immediately after it, **regardless** of if the branch was taken or not.
   So, instead of updating the program count directly, the next address needs to be put in the "branch delay slot" and the program counter should only be updated after the *next* instruction.
 * Lists in Scratch are one-indexed.
 * All of a sudden, Scratch stopped letting me save the project to the cloud. It took awhile before I realized that lists filled with over 100,000 items wasn't something Scratch's servers were particularly excited to store.
 * I had to design my own `malloc` in C, which was fun, but also very difficult to debug in Scratch.
 * When I tried making syscalls that asked the user for input, all of the letters ended up capitalized. It turns out that in Scratch a lowercase `"a"` and a capital `"A"` are considered equal.
   I thought this was an unsolvable problem for awhile, before I realized that the names of sprites' costumes in Scratch are actually case-sensitive.
   So every time I try to convert a character to its ASCII value, I tell the processor sprite to switch to, for example, the `"a"` costume or the `"A"` costume, and then retrieve the costume number.
 * I made another syscall to print emojis to the `stdout`, but some emojis are considered two characters long and other emojis are considered one character long.
 * Compiling any code that calls `malloc` with -O1 crashes the CPU. I still have no idea why this is the case.
 * Endianness is really hard to get right. I know I said this in the beginning of the list, but it's worth repeating.
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With all that said, I’m really happy with the way this project turned out. If you found this interesting, please check out sharc, my graphics engine built completely in Typescript: https://www.sharcjs.org. Because clearly, if there’s one thing I know how to make, it’s questionable decisions.

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