user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D53(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x8(,%rax,8),%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%esi
  pxor   %xmm0,%xmm0
  movsd  D89(%rip),%xmm5
  mov    %rax,%rdi
  cvtsi2sd %esi,%xmm0
  divsd  %xmm0,%xmm5
  test   %esi,%esi
  js     L5
  lea    0x1(%rsi),%ecx
  cmp    $0x2,%esi
  jle    L22
  mov    %ecx,%edx
  movapd %xmm5,%xmm4
  movdqa D80(%rip),%xmm2
  movdqa D61(%rip),%xmm6
  shr    $0x2,%edx
  movapd D79(%rip),%xmm3
  unpcklpd %xmm4,%xmm4
  shl    $0x5,%rdx
  add    %rax,%rdx
L14:
  movdqa %xmm2,%xmm0
  add    $0x20,%rax
  paddd  %xmm6,%xmm2
  cvtdq2pd %xmm0,%xmm1
  mulpd  %xmm4,%xmm1
  pshufd $0xee,%xmm0,%xmm0
  cvtdq2pd %xmm0,%xmm0
  mulpd  %xmm4,%xmm0
  addpd  %xmm3,%xmm1
  addpd  %xmm3,%xmm0
  movups %xmm1,-0x20(%rax)
  movups %xmm0,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L14
  mov    %ecx,%eax
  and    $0xfffffffc,%eax
  and    $0x3,%ecx
  je     L5
L45:
  pxor   %xmm0,%xmm0
  movsd  D79(%rip),%xmm1
  movslq %eax,%rdx
  cvtsi2sd %eax,%xmm0
  lea    0x0(,%rdx,8),%rcx
  mulsd  %xmm5,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdi,%rdx,8)
  lea    0x1(%rax),%edx
  cmp    %edx,%esi
  jl     L5
  pxor   %xmm0,%xmm0
  add    $0x2,%eax
  cvtsi2sd %edx,%xmm0
  mulsd  %xmm5,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rdi,%rcx,1)
  cmp    %eax,%esi
  jl     L5
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm5,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rdi,%rcx,1)
L5:
  movapd %xmm5,%xmm0
  call   <intCompSimp@plt>
  lea    D58(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L68
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L22:
  xor    %eax,%eax
  jmp    L45
L68:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D53
data_type: string
name: unpack_data
arguments: data_tag: D89
data_type: byte[8]
name: unpack_data
arguments: data_tag: D80
data_type: byte[16]
name: unpack_data
arguments: data_tag: D61
data_type: byte[16]
name: unpack_data
arguments: data_tag: D79
data_type: f64[2]
name: unpack_data
arguments: data_tag: D58
data_type: string
name: unpack_data
arguments: data_tag: D2
data_type: string

tool: D16: dame el numero de partes a dividir el intervalo:

tool: D53: %d

tool: {"D89": [0x00,0x00,0x00,0x00,0x00,0x00,0x14,0x40]}

tool: {"D80": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D61": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D79": [2.0,2.0]}

tool: D58: estimacion de la integral--> %lf


tool: D2: PAUSE

assistant: ```c
int main() {
  int n;
  int i;
  double a = 2;
  double b = 7;
  double *X;
  double h;
  double integral;
  printf("dame el numero de partes a dividir el intervalo:");
  scanf("%d", &n);
  X = (double *)malloc((n + 1) * sizeof(double));
  h = (b - a) / n;
  for (i = 0; i < n + 1; i++)
    X[i] = a + h * i;
  integral = intCompSimp(X, h, n);
  printf("estimacion de la integral--> %lf\n", integral);
  system("PAUSE");
  return 0;
}
```