Initial commit: implementation of the virtual machine

This commit is contained in:
Dimitri Lozeve 2021-01-13 09:16:21 +01:00
commit 87a815ac27
6 changed files with 262 additions and 0 deletions

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.gitignore vendored Normal file
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*\~
.gdb_history
vm
synacor-challenge.tgz

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Makefile Normal file
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CC = gcc
CFLAGS = -Wall -Werror -pedantic -O3
.PHONY: all
all: vm
vm: src/vm.c
$(CC) $^ -o $@ $(CFLAGS)
clean:
rm -f vm

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README.org Normal file
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* Synacor Challenge
Solutions to the [[https://challenge.synacor.com/][Synacor Challenge]]
The virtual machine is implemented in [[src/vm.c]]. Compile it with =make=
and run it on the challenge program:
#+begin_src sh
./vm challenge.bin
#+end_src

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arch-spec Normal file
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== Synacor Challenge ==
In this challenge, your job is to use this architecture spec to create a
virtual machine capable of running the included binary. Along the way,
you will find codes; submit these to the challenge website to track
your progress. Good luck!
== architecture ==
- three storage regions
- memory with 15-bit address space storing 16-bit values
- eight registers
- an unbounded stack which holds individual 16-bit values
- all numbers are unsigned integers 0..32767 (15-bit)
- all math is modulo 32768; 32758 + 15 => 5
== binary format ==
- each number is stored as a 16-bit little-endian pair (low byte, high byte)
- numbers 0..32767 mean a literal value
- numbers 32768..32775 instead mean registers 0..7
- numbers 32776..65535 are invalid
- programs are loaded into memory starting at address 0
- address 0 is the first 16-bit value, address 1 is the second 16-bit value, etc
== execution ==
- After an operation is executed, the next instruction to read is immediately after the last argument of the current operation. If a jump was performed, the next operation is instead the exact destination of the jump.
- Encountering a register as an operation argument should be taken as reading from the register or setting into the register as appropriate.
== hints ==
- Start with operations 0, 19, and 21.
- Here's a code for the challenge website: SJIDydAvKFEH
- The program "9,32768,32769,4,19,32768" occupies six memory addresses and should:
- Store into register 0 the sum of 4 and the value contained in register 1.
- Output to the terminal the character with the ascii code contained in register 0.
== opcode listing ==
halt: 0
stop execution and terminate the program
set: 1 a b
set register <a> to the value of <b>
push: 2 a
push <a> onto the stack
pop: 3 a
remove the top element from the stack and write it into <a>; empty stack = error
eq: 4 a b c
set <a> to 1 if <b> is equal to <c>; set it to 0 otherwise
gt: 5 a b c
set <a> to 1 if <b> is greater than <c>; set it to 0 otherwise
jmp: 6 a
jump to <a>
jt: 7 a b
if <a> is nonzero, jump to <b>
jf: 8 a b
if <a> is zero, jump to <b>
add: 9 a b c
assign into <a> the sum of <b> and <c> (modulo 32768)
mult: 10 a b c
store into <a> the product of <b> and <c> (modulo 32768)
mod: 11 a b c
store into <a> the remainder of <b> divided by <c>
and: 12 a b c
stores into <a> the bitwise and of <b> and <c>
or: 13 a b c
stores into <a> the bitwise or of <b> and <c>
not: 14 a b
stores 15-bit bitwise inverse of <b> in <a>
rmem: 15 a b
read memory at address <b> and write it to <a>
wmem: 16 a b
write the value from <b> into memory at address <a>
call: 17 a
write the address of the next instruction to the stack and jump to <a>
ret: 18
remove the top element from the stack and jump to it; empty stack = halt
out: 19 a
write the character represented by ascii code <a> to the terminal
in: 20 a
read a character from the terminal and write its ascii code to <a>; it can be assumed that once input starts, it will continue until a newline is encountered; this means that you can safely read whole lines from the keyboard and trust that they will be fully read
noop: 21
no operation

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#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#define MEM_SIZE (2 << 15)
#define STACK_SIZE 10000
#define val(r) ((r >> 15) ? reg[r & 7] : r)
size_t read_program(uint16_t mem[static 1], char *filename) {
FILE *fp = fopen(filename, "rb");
if (fp == NULL) {
fprintf(stderr, "Could not open file %s\n", filename);
exit(EXIT_FAILURE);
}
int c = 0;
size_t n = -1;
int i = 0;
while ((c = fgetc(fp)) != EOF) {
if (i == 0) {
n++;
mem[n] = c;
} else {
mem[n] = (c << 8) | mem[n];
}
i = (i + 1) % 2;
}
fclose(fp);
return n + 1;
}
void run(uint16_t mem[static 1]) {
uint16_t reg[8] = {0}; // registers
uint16_t stack[STACK_SIZE] = {0}; // stack
size_t stack_top = 0;
size_t pc = 0;
for (;;) {
uint16_t opcode = mem[pc];
uint16_t a = mem[pc + 1];
uint16_t b = mem[pc + 2];
uint16_t c = mem[pc + 3];
switch (opcode) {
case 0:
return;
case 1:
reg[a & 7] = val(b);
pc += 3;
break;
case 2:
stack[stack_top] = val(a);
stack_top++;
pc += 2;
break;
case 3:
stack_top--;
reg[a & 7] = stack[stack_top];
pc += 2;
break;
case 4:
reg[a & 7] = val(b) == val(c);
pc += 4;
break;
case 5:
reg[a & 7] = val(b) > val(c);
pc += 4;
break;
case 6:
pc = val(a);
break;
case 7:
if (val(a)) {
pc = val(b);
} else {
pc += 3;
}
break;
case 8:
if (!val(a)) {
pc = val(b);
} else {
pc += 3;
}
break;
case 9:
reg[a & 7] = (val(b) + val(c)) % 32768;
pc += 4;
break;
case 10:
reg[a & 7] = (val(b) * val(c)) % 32768;
pc += 4;
break;
case 11:
reg[a & 7] = val(b) % val(c);
pc += 4;
break;
case 12:
reg[a & 7] = val(b) & val(c);
pc += 4;
break;
case 13:
reg[a & 7] = val(b) | val(c);
pc += 4;
break;
case 14:
reg[a & 7] = ~val(b) & 0x7FFF;
pc += 3;
break;
case 15:
reg[a & 7] = mem[val(b)];
pc += 3;
break;
case 16:
mem[val(a)] = val(b);
pc += 3;
break;
case 17:
stack[stack_top] = pc + 2;
stack_top++;
pc = val(a);
break;
case 18:
if (!stack_top) {
return;
}
stack_top--;
pc = stack[stack_top];
break;
case 19:
putchar(val(a));
pc += 2;
break;
case 20:
reg[a & 7] = getchar();
pc += 2;
break;
case 21:
pc++;
break;
}
}
}
int main(int argc, char *argv[]) {
if (argc < 2) {
printf("Usage: %s <program>\n", argv[0]);
return EXIT_SUCCESS;
}
uint16_t mem[MEM_SIZE] = {0};
read_program(mem, argv[1]);
run(mem);
return EXIT_SUCCESS;
}