user: 这段汇编代码用C语言表示是什么样的？

```asm
<merge>:
  endbr64
  push   %r15
  lea    (%rsi,%rcx,1),%rax
  mov    %r8,%r15
  push   %r14
  imul   %r8,%rax
  mov    %rsi,%r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,0x28(%rsp)
  mov    %rax,%rdi
  mov    %rdx,0x18(%rsp)
  mov    %r9,0x30(%rsp)
  mov    %rax,0x38(%rsp)
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L134
L74:
  mov    %r14,%rax
  mov    %r12,%rcx
  imul   %r15,%rax
  imul   %r15,%rcx
  mov    %rax,0x10(%rsp)
  mov    %rcx,0x8(%rsp)
  test   %rax,%rax
  je     L68
  test   %rcx,%rcx
  je     L120
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  jmp    L156
  xchg   %ax,%ax
L106:
  call   <memcpy@plt>
  add    %r15,%rbx
  add    %r15,%r12
  cmp    0x10(%rsp),%rbp
  jae    L23
L92:
  cmp    0x8(%rsp),%rbx
  jae    L84
L156:
  mov    0x18(%rsp),%rax
  lea    (%rax,%rbx,1),%rsi
  mov    0x28(%rsp),%rax
  mov    %rsi,0x20(%rsp)
  lea    (%rax,%rbp,1),%r14
  mov    0x30(%rsp),%rax
  mov    %r14,%rdi
  call   *%rax
  mov    0x20(%rsp),%rsi
  lea    0x0(%r13,%r12,1),%rdi
  mov    %r15,%rdx
  test   %eax,%eax
  jns    L106
  mov    %r14,%rsi
  add    %r15,%rbp
  add    %r15,%r12
  call   <memcpy@plt>
  cmp    0x10(%rsp),%rbp
  jb     L92
  nopl   0x0(%rax)
L23:
  lea    0x0(%r13,%r12,1),%rbp
  sub    %rbx,%rbp
  cmp    0x8(%rsp),%rbx
  jae    L19
  nopw   0x0(%rax,%rax,1)
L169:
  mov    0x18(%rsp),%rax
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r15,%rdx
  lea    (%rax,%rbx,1),%rsi
  add    %r15,%rbx
  call   <memcpy@plt>
  cmp    0x8(%rsp),%rbx
  jb     L169
L19:
  mov    0x38(%rsp),%rdx
  mov    0x28(%rsp),%rdi
  mov    %r13,%rsi
  call   <memcpy@plt>
  test   %r13,%r13
  jne    L43
L85:
  add    $0x48,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L84:
  mov    %r12,0x20(%rsp)
  mov    %r12,%rax
L123:
  mov    0x28(%rsp),%rcx
  mov    %rax,%r12
  lea    (%rcx,%rbp,1),%r14
  sub    %rax,%r14
  nopw   0x0(%rax,%rax,1)
L58:
  lea    (%r14,%r12,1),%rsi
  lea    0x0(%r13,%r12,1),%rdi
  mov    %r15,%rdx
  add    %r15,%r12
  call   <memcpy@plt>
  mov    %rbp,%rax
  sub    0x20(%rsp),%rax
  add    %r12,%rax
  cmp    %rax,0x10(%rsp)
  ja     L58
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L134:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L74
  nopl   0x0(%rax)
L43:
  mov    %r13,%rdi
  call   <free@plt>
  jmp    L85
L68:
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  jmp    L23
L120:
  movq   $0x0,0x20(%rsp)
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  mov    0x20(%rsp),%rax
  jmp    L123
```

assistant: ```c
int merge(void *const part1, const size_t p1Len, const void *const part2, const size_t p2Len, const size_t size, int (*const cmp)(const void *const, const void *const)) {
  char *tmp = NULL;
  char *p1 = NULL, *p2 = NULL;
  size_t index = 0, index2 = 0, index3 = 0;
  p1 = (char *)part1;
  p2 = (char *)part2;
  if (!(tmp = (char *)malloc((p1Len + p2Len) * size))) {
    abort();
  }
  index = index2 = index3 = 0;
  while ((index < p1Len * size) && (index2 < p2Len * size)) {
    if (cmp(&p1[index], &p2[index2]) < 0) {
      memcpy(&tmp[index3], &p1[index], size);
      index += size;
    } else {
      memcpy(&tmp[index3], &p2[index2], size);
      index2 += size;
    }
    index3 += size;
  }
  while (index < p1Len * size) {
    memcpy(&tmp[index3], &p1[index], size);
    index += size;
    index3 += size;
  }
  while (index2 < p2Len * size) {
    memcpy(&tmp[index3], &p2[index2], size);
    index2 += size;
    index3 += size;
  }
  memcpy(p1, tmp, (p1Len + p2Len) * size);
  if (tmp) {
    free(tmp);
    tmp = NULL;
  }
  return 1;
}
```