user: 下面的汇编指令对应的C源代码是什么：

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    -0x40(%rbp),%eax
  sub    -0x3c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x44(%rbp),%eax
  sub    -0x40(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L52
L53:
  mov    -0x3c(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x24(%rbp)
L52:
  mov    -0x24(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L53
  movl   $0x0,-0x20(%rbp)
  jmp    L40
L94:
  mov    -0x40(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x20(%rbp)
L40:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L94
  movl   $0x0,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L84
L78:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jg     L68
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  jmp    L84
L68:
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L84:
  mov    -0x24(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  setl   %dl
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  setl   %al
  and    %edx,%eax
  test   %al,%al
  jne    L78
  jmp    L107
L96:
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L107:
  mov    -0x24(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L96
  jmp    L93
L99:
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L93:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L99
  nop
  nop
  leave
  ret
```

assistant: ```c
void merge(int *a, int p, int q, int r) {
  int n1 = q - p + 1;
  int n2 = r - q;
  int *left = malloc(n1 * sizeof(int));
  int *right = malloc(n2 * sizeof(int));
  int i, j, k;
  for (i = 0; i < n1; i++)
    left[i] = a[p + i];
  for (j = 0; j < n2; j++)
    right[j] = a[q + j + 1];
  i = j = 0;
  k = p;
  while (i < n1 & j < n2) {
    if (left[i] <= right[j]) {
      a[k++] = left[i++];
    } else {
      a[k++] = right[j++];
    }
  }
  while (i < n1) {
    a[k++] = left[i++];
  }
  while (j < n2) {
    a[k++] = right[j++];
  }
}
```