…或者换句话说，这是进入嵌入式系统的最愚蠢的方法。
在这里观看实际操作！
目标
目标很简单。用C或C++编写一些代码，并能够在Scratch中执行。老实说，我觉得这个想法很有趣：最快的编程语言之一是最慢的编程语言之一。我有一种感觉这是可能的，但我不太确定如何实现。在这个过程中，我学到的关于汇编语言、进程内存和可执行文件的知识比我预想的要多得多，我希望你能学到新的东西，同时我也回顾一下我的旅程。
第0步：制定计划
我的第一个想法是把我用 C 编写的代码分成几部分，然后使用 Scratch 将这些部分重新组合在一起。例如，C 中的 while 循环可能会变成 Scratch 中的重复直到块：
为了让C编译器理解代码，它首先需要生成一个AST（抽象语法树），它是源代码中每个重要符号的树表示。例如，左括号、变量名称或 return 关键字都可能被转换为不同的节点。然而，在查看了一个简单的斐波那契数列程序的 AST 之后……
步骤 0b：制定（更好的）计划
好吧，所以这是不可能的。但是，如果我们不是尝试重新编译源代码，而是沿着阶梯往下走一步：汇编呢？为了让程序运行，首先需要将其编译为汇编语言。在我的计算机上，它是 x86-64asm。由于汇编没有任何复杂的嵌套结构、类甚至变量，因此尝试解析汇编指令列表（理论上）应该比尝试解析 AST 的意大利面怪物（例如上面的指令）更容易。这是相同的斐波那契数列，但采用 x86 汇编语言。
哦，兄弟。好吧，也许情况并没有那么糟糕。总共有多少条指令？
步骤 0c：完整计划
值得庆幸的是，x86 并不是唯一的汇编语言。作为大学课程的一部分，我了解了 MIPS，这是一种汇编语言（过度简化），在 20 世纪 90 年代到 2000 年代初的一些视频游戏机和超级计算机中使用，至今仍有一些用途。从 x86 切换到 MIPS 使指令计数从*未知*减少到 50 左右。
使用 32 位版本的 MIPS，该汇编代码可以转换为机器代码，其中每条指令都根据处理器架构设置的指导原则转换为处理器可以理解的 32 位整数。网上有一本关于 MIPS 指令集架构的书，所以如果我获取机器代码，然后准确地模拟 MIPS 处理器的功能，那么我应该能够在 Scratch 中运行我的 C 代码！
现在一切都已经解决了，我们可以开始了。
第一步：我们还不能开始
好吧，已经有问题了。通常，如果你有一个整数，并且想要从中提取一系列位，则可以计算 num & mask，其中 mask 是一个整数，其中每个重要位都是 1，每个不重要位都是 0。
001000 01001 01000 1111111111111100
& 000000 00000 00000 1111111111111111
--------------------------------------
000000 0000 000000 1111111111111100
登录后复制
问题？ Scratch 中没有 & 运算符。
现在，我*可以*简单地逐位检查这两个数字，并检查两位的四种可能组合中的每一种，但这会非常慢；毕竟，这需要对*每个*指令执行多次。相反，我想出了一个更好的计划。
首先，我编写了一个快速的 Python 脚本来计算 0 到 255 之间的每个 x 和每个 y 的 x 和 y。
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)
登录后复制
现在，例如，为了计算两个 32 位整数的 x 和 y，我们可以执行以下操作：
Split x and y into four 8-bit integers (or bytes).
Check what first_byte_in_x & first_byte_in_y is by looking in the table generated from the Python script.
Similarly, look up what second_byte_in_x & second_byte_in_y is, and the third bytes, and the fourth bytes.
Take the results of each of these calculations, and put them together to get the result of x & y .
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.
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”.
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;
}
登录后复制
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
登录后复制
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.
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)
登录后复制
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.
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).
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;
}
登录后复制
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;
}
登录后复制
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.
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!”
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;
}
登录后复制
And when we run it, we get this:
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.
登录后复制
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.
以上是用儿童编程语言构建计算机的详细内容。更多信息请关注PHP中文网其他相关文章！
