user: 下面的汇编指令对应的C源代码是什么:

```asm
<subtraction>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x12(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  mov    -0x40(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  movl   $0x0,-0x1c(%rbp)
  jmp    L82
L76:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  mov    %dl,-0x12(%rbp,%rax,1)
  addl   $0x1,-0x1c(%rbp)
L82:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L76
  movl   $0x0,-0x1c(%rbp)
  jmp    L105
L149:
  mov    -0x1c(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jb     L129
  mov    -0x1c(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%ecx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x40(%rbp),%rax
  add    %rsi,%rax
  sub    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L79
L129:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L14
L96:
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  test   %al,%al
  je     L24
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %dl,-0x12(%rbp,%rax,1)
  mov    -0x1c(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%ecx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x40(%rbp),%rax
  add    %rsi,%rax
  sub    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L79
L24:
  mov    -0x18(%rbp),%eax
  cltq
  movb   $0xff,-0x12(%rbp,%rax,1)
  addl   $0x1,-0x18(%rbp)
L14:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L96
L79:
  addl   $0x1,-0x1c(%rbp)
L105:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L149
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L67
L0:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L45
  subl   $0x1,-0x2c(%rbp)
  subl   $0x1,-0x18(%rbp)
L67:
  cmpl   $0x0,-0x18(%rbp)
  jns    L0
  jmp    L28
L45:
  nop
L28:
  mov    -0x2c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L77
  call   <__stack_chk_fail@plt>
L77:
  leave
  ret
```

assistant: ```c
int subtraction(unsigned char *a, int a_len, unsigned char *b, unsigned char *subres) {
  int i, j, k;
  unsigned char diversion[10];
  bzero(diversion, sizeof(diversion));
  bzero(subres, sizeof(subres));
  for (i = 0; i <= a_len; i++)
    diversion[i] = a[i];
  for (i = 0; i <= a_len; i++) {
    if (diversion[i] >= b[i])
      subres[i] = diversion[i] - b[i];
    else {
      j = i + 1;
      while (j <= a_len) {
        if (diversion[j] > 0) {
          diversion[j]--;
          subres[i] = 256 - b[i] + diversion[i];
          break;
        } else
          diversion[j] = 255;
        j++;
      }
    }
  }
  for (j = a_len; j >= 0; j--) {
    if (subres[j] > 0)
      break;
    a_len--;
  }
  return a_len;
}
```