user: 这段汇编代码用C语言表示是什么样的?

```asm
<cvt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  movsd  %xmm0,-0x38(%rbp)
  mov    %edi,-0x3c(%rbp)
  mov    %esi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %r8d,-0x54(%rbp)
  mov    %r9,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x66,-0x54(%rbp)
  jne    L165
  movl   $0x3,-0x24(%rbp)
  jmp    L5
L165:
  cmpl   $0x65,-0x54(%rbp)
  je     L62
  cmpl   $0x45,-0x54(%rbp)
  jne    L90
L62:
  addl   $0x1,-0x3c(%rbp)
L90:
  movl   $0x2,-0x24(%rbp)
L5:
  pxor   %xmm0,%xmm0
  comisd -0x38(%rbp),%xmm0
  jbe    L33
  movsd  -0x38(%rbp),%xmm0
  movq   D7(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,-0x38(%rbp)
  mov    -0x48(%rbp),%rax
  movb   $0x2d,(%rax)
  jmp    L49
L33:
  mov    -0x48(%rbp),%rax
  movb   $0x0,(%rax)
L49:
  lea    -0x20(%rbp),%r8
  lea    -0x28(%rbp),%rcx
  mov    -0x50(%rbp),%rdx
  mov    -0x3c(%rbp),%esi
  mov    -0x24(%rbp),%edi
  mov    -0x38(%rbp),%rax
  movq   %rax,%xmm0
  mov    $0x1,%eax
  call   <__dtoa@plt>
  cltq
  mov    %rax,-0x10(%rbp)
  cmpl   $0x67,-0x54(%rbp)
  je     L123
  cmpl   $0x47,-0x54(%rbp)
  jne    L29
L123:
  mov    -0x40(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L117
L29:
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  cmpl   $0x66,-0x54(%rbp)
  jne    L150
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jne    L76
  pxor   %xmm0,%xmm0
  ucomisd -0x38(%rbp),%xmm0
  jp     L37
  pxor   %xmm0,%xmm0
  ucomisd -0x38(%rbp),%xmm0
  je     L76
L37:
  mov    $0x1,%eax
  sub    -0x3c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x50(%rbp),%rax
  mov    %edx,(%rax)
L76:
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  cltq
  add    %rax,-0x18(%rbp)
L150:
  pxor   %xmm0,%xmm0
  ucomisd -0x38(%rbp),%xmm0
  jp     L10
  pxor   %xmm0,%xmm0
  ucomisd -0x38(%rbp),%xmm0
  jne    L125
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L125:
  jmp    L10
L69:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movb   $0x30,(%rax)
L10:
  mov    -0x20(%rbp),%rax
  cmp    %rax,-0x18(%rbp)
  ja     L69
L117:
  mov    -0x20(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x60(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_infered: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
char *cvt(value, ndigits, flags, sign, decpt, ch, length)
double value;
int ndigits, flags, *decpt, ch, *length;
char *sign;
{
  int mode, dsgn;
  char *digits, *bp, *rve;
  if (ch == 'f') {
    mode = 3;
  } else {
    if (ch == 'e' || ch == 'E') {
      ndigits++;
    }
    mode = 2;
  }
  if (value < 0) {
    value = -value;
    *sign = '-';
  } else
    *sign = '\000';
  digits = __dtoa(value, mode, ndigits, decpt, &dsgn, &rve);
  if ((ch != 'g' && ch != 'G') || flags & 0x001) {
    bp = digits + ndigits;
    if (ch == 'f') {
      if (*digits == '0' && value)
        *decpt = -ndigits + 1;
      bp += *decpt;
    }
    if (value == 0)
      rve = bp;
    while (rve < bp)
      *rve++ = '0';
  }
  *length = rve - digits;
  return (digits);
}
```