user: 下面的汇编指令对应的C源代码是什么:

```asm
<a_minus>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  mov    -0x40(%rbp),%rsi
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <Equalize@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strdup@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x2c(%rbp)
  subl   $0x1,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L76
L70:
  mov    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L68
  mov    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x37,%eax
  jmp    L2
L68:
  mov    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
L2:
  mov    -0x18(%rbp),%rcx
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  cmp    $0x39,%dl
  jle    L39
  mov    -0x18(%rbp),%rcx
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    $0x37,%edx
  jmp    L58
L39:
  mov    -0x18(%rbp),%rcx
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    $0x30,%edx
L58:
  sub    %edx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x28(%rbp),%eax
  add    %eax,-0x24(%rbp)
  addl   $0x10,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0xf(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x4,%eax
  mov    %eax,-0x28(%rbp)
  subl   $0x1,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  cltd
  shr    $0x1c,%edx
  add    %edx,%eax
  and    $0xf,%eax
  sub    %edx,%eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x9,-0x24(%rbp)
  jle    L7
  mov    -0x24(%rbp),%eax
  add    $0x37,%eax
  mov    %eax,%edx
  jmp    L65
L7:
  mov    -0x24(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edx
L65:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  subl   $0x1,-0x2c(%rbp)
L76:
  cmpl   $0x0,-0x2c(%rbp)
  je     L30
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  jne    L70
L30:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

assistant: ```c
char *a_minus(char *pcL, char *pcR) {
  char *pcResult;
  char *pcNewLeft, *pcNewRight;
  int iDigitDiff;
  int i, iCarry;
  Equalize(pcL, pcR, &pcNewLeft, &pcNewRight);
  pcResult = my_strdup(pcNewLeft);
  i = strlen(pcNewLeft);
  i--;
  iCarry = 0;
  while ((i != 0) && (pcL[i] != 'x')) {
    iDigitDiff = (((pcNewLeft[i]) > '9') ? ((pcNewLeft[i]) - 'A' + 10) : ((pcNewLeft[i]) - '0')) - (((pcNewRight[i]) > '9') ? ((pcNewRight[i]) - 'A' + 10) : ((pcNewRight[i]) - '0'));
    iDigitDiff = iDigitDiff + iCarry;
    iDigitDiff += 0x10;
    iCarry = iDigitDiff / 0x10;
    iCarry--;
    iDigitDiff = iDigitDiff % 0x10;
    pcResult[i] = (((iDigitDiff) > 9) ? ((iDigitDiff)-10 + 'A') : ((iDigitDiff) + '0'));
    i--;
  }
  free(pcNewLeft);
  free(pcNewRight);
  return pcResult;
}
```