user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movq   $0x1,0x18(%rsp)
  movsd  D51(%rip),%xmm1
L7:
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0x80(%rsp,%rax,4)
  movl   $0x0,0x20(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x15,%rax
  jne    L7
  mov    $0x2710,%ebx
  lea    0x18(%rsp),%rbp
  jmp    L52
L41:
  add    $0x4,%rax
  add    $0x4,%rdx
  cmp    %rcx,%rax
  je     L46
L6:
  movss  (%rdx),%xmm1
  comiss %xmm0,%xmm1
  jbe    L41
  comiss -0x4(%rdx),%xmm0
  jbe    L41
  addl   $0x1,(%rax)
  jmp    L41
L46:
  sub    $0x1,%ebx
  je     L27
L52:
  mov    %rbp,%rdi
  call   <expdev@plt>
  lea    0x24(%rsp),%rsi
  lea    0x84(%rsp),%rdx
  lea    0x74(%rsp),%rcx
  mov    %rsi,%rax
  jmp    L6
L27:
  mov    $0x0,%ebp
L72:
  add    (%rsi),%ebp
  add    $0x4,%rsi
  cmp    %rcx,%rsi
  jne    L72
  mov    $0x2710,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <puts@plt>
  mov    $0x4,%ebx
  lea    D62(%rip),%r12
L93:
  pxor   %xmm1,%xmm1
  cvtsi2ssl 0x20(%rsp,%rbx,1),%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2ss %ebp,%xmm0
  divss  %xmm0,%xmm1
  movss  %xmm1,0xc(%rsp)
  movss  0x7c(%rsp,%rbx,1),%xmm0
  movss  0x80(%rsp,%rbx,1),%xmm4
  movss  %xmm0,0x4(%rsp)
  movss  %xmm4,0x8(%rsp)
  addss  %xmm4,%xmm0
  xorps  D56(%rip),%xmm0
  mulss  D46(%rip),%xmm0
  call   <exp@plt>
  pxor   %xmm3,%xmm3
  cvtss2sd %xmm0,%xmm3
  mulsd  D55(%rip),%xmm3
  cvtsd2ss %xmm3,%xmm3
  pxor   %xmm0,%xmm0
  cvtss2sd 0x4(%rsp),%xmm0
  cvtss2sd %xmm3,%xmm3
  pxor   %xmm2,%xmm2
  cvtss2sd 0xc(%rsp),%xmm2
  pxor   %xmm1,%xmm1
  cvtss2sd 0x8(%rsp),%xmm1
  mov    %r12,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    $0x54,%rbx
  jne    L93
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0xe0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D51
type_infered: f64
name: unpack_data
arguments: data_label: D1
type_infered: string
name: unpack_data
arguments: data_label: D21
type_infered: string
name: unpack_data
arguments: data_label: D62
type_infered: string
name: unpack_data
arguments: data_label: D56
type_infered: byte[16]
name: unpack_data
arguments: data_label: D46
type_infered: byte[4]
name: unpack_data
arguments: data_label: D55
type_infered: byte[8]

tool: D51: 20

tool: D1: 
exponential distribution with %7d points


tool: D21:     interval      observed     expected


tool: D62: %6.2f %6.2f %12.6f %12.6f 


tool: {"D56": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D46": [0x00,0x00,0x00,0x3f]}

tool: {"D55": [0x58,0x3a,0xfc,0x3e,0xd2,0x3f,0xb4,0x3f]}

assistant: ```c
int main(void) {
  long idum = (1);
  int i, j, total = 0, x[21];
  float expect, xx, y, trig[21];
  for (i = 0; i <= 20; i++) {
    trig[i] = i / 20.0;
    x[i] = 0;
  }
  for (i = 1; i <= 10000; i++) {
    y = expdev(&idum);
    for (j = 1; j <= 20; j++)
      if ((y < trig[j]) && (y > trig[j - 1]))
        ++x[j];
  }
  for (i = 1; i <= 20; i++)
    total += x[i];
  printf("\nexponential distribution with %7d points\n", 10000);
  printf("    interval      observed     expected\n\n");
  for (i = 1; i <= 20; i++) {
    xx = (float)x[i] / total;
    expect = exp(-(trig[i - 1] + trig[i]) / 2.0);
    expect *= (0.05 * 2.718281828 / (2.718281828 - 1));
    printf("%6.2f %6.2f %12.6f %12.6f \n", trig[i - 1], trig[i], xx, expect);
  }
  return 0;
}
```