user: 这段汇编代码用C语言表示是什么样的:

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %edx,%r10d
  sub    %edx,%ecx
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  movslq %esi,%r12
  push   %rbx
  sub    %r12d,%r10d
  mov    %edx,%ebx
  lea    0x1(%r10),%r9d
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %r9d,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L94
L12:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L12
L94:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L24
L198:
  movslq %r13d,%r15
  mov    %rsp,%rcx
  mov    %rsp,%r8
  lea    0xf(,%r15,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L168
L75:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L75
L168:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L43
L192:
  mov    %rsp,%rcx
  test   %r9d,%r9d
  jle    L7
  mov    %r10d,%eax
  mov    %r8,%rdi
  mov    %r9d,-0x48(%rbp)
  lea    0x4(,%rax,4),%rdx
  movslq %r12d,%rax
  mov    %rcx,-0x50(%rbp)
  lea    (%r14,%rax,4),%rsi
  mov    %r10d,-0x44(%rbp)
  call   <memcpy@plt>
  test   %r13d,%r13d
  mov    -0x44(%rbp),%r10d
  mov    -0x50(%rbp),%rcx
  mov    -0x48(%rbp),%r9d
  mov    %rax,%r8
  jle    L112
  add    $0x1,%ebx
  mov    %rcx,%rdi
  mov    %r9d,-0x48(%rbp)
  add    $0x1,%r12d
  movslq %ebx,%rbx
  lea    0x0(,%r15,4),%rdx
  mov    %r10d,-0x44(%rbp)
  xor    %r15d,%r15d
  lea    (%r14,%rbx,4),%rsi
  mov    %rax,-0x50(%rbp)
  xor    %ebx,%ebx
  call   <memcpy@plt>
  mov    -0x44(%rbp),%r10d
  mov    -0x50(%rbp),%r8
  xor    %edx,%edx
  mov    -0x48(%rbp),%r9d
  mov    %rax,%rcx
  xor    %r11d,%r11d
  movslq %r12d,%rax
  nopw   0x0(%rax,%rax,1)
L219:
  mov    (%r8,%r15,4),%edi
  mov    (%rcx,%r11,4),%esi
  movslq %eax,%r12
  cmp    %esi,%edi
  jg     L205
L180:
  mov    %edi,-0x4(%r14,%rax,4)
  lea    0x1(%rdx),%r15d
  add    $0x1,%rax
  cmp    %edx,%r10d
  jle    L89
  cmp    %ebx,%r13d
  jle    L89
  movslq %r15d,%r15
  mov    (%rcx,%r11,4),%esi
  movslq %eax,%r12
  mov    (%r8,%r15,4),%edi
  mov    %r15,%rdx
  cmp    %esi,%edi
  jle    L180
L205:
  mov    %esi,-0x4(%r14,%rax,4)
  add    $0x1,%ebx
  add    $0x1,%rax
  cmp    %edx,%r9d
  jle    L155
  cmp    %ebx,%r13d
  jle    L155
  movslq %ebx,%r11
  jmp    L219
  xchg   %ax,%ax
L24:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L198
  nopl   0x0(%rax,%rax,1)
L155:
  mov    %edx,%r15d
L89:
  cmp    %r15d,%r9d
  mov    %r9d,-0x44(%rbp)
  jle    L132
  movslq %r12d,%rax
  sub    %r15d,%r10d
  mov    %rcx,-0x50(%rbp)
  lea    (%r14,%rax,4),%rdi
  movslq %r15d,%rax
  lea    0x4(,%r10,4),%rdx
  lea    (%r8,%rax,4),%rsi
  call   <memcpy@plt>
  mov    -0x44(%rbp),%r9d
  mov    -0x50(%rbp),%rcx
  add    %r12d,%r9d
  sub    %r15d,%r9d
  movslq %r9d,%r12
L132:
  cmp    %ebx,%r13d
  jle    L70
L58:
  sub    $0x1,%r13d
  lea    (%r14,%r12,4),%rdi
  sub    %ebx,%r13d
  movslq %ebx,%rbx
  lea    0x4(,%r13,4),%rdx
  lea    (%rcx,%rbx,4),%rsi
  call   <memcpy@plt>
L70:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L80
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L7:
  test   %r13d,%r13d
  jle    L112
  add    $0x1,%ebx
  mov    %rcx,%rdi
  lea    0x0(,%r15,4),%rdx
  movslq %ebx,%rbx
  lea    (%r14,%rbx,4),%rsi
  xor    %ebx,%ebx
  call   <memcpy@plt>
  mov    %rax,%rcx
  jmp    L58
  nopw   0x0(%rax,%rax,1)
L112:
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  jmp    L89
L43:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L192
L80:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void merge(int A[], int p, int q, int r) {
  int n1 = q - p + 1, n2 = r - q;
  int B[n1], C[n2];
  for (int i = 0; i < n1; i++) {
    B[i] = A[i + p];
  }
  for (int j = 0; j < n2; j++) {
    C[j] = A[q + 1 + j];
  }
  int i, j, k;
  i = j = 0;
  k = p;
  while (i < n1 && j < n2) {
    if (B[i] <= C[j])
      A[k++] = B[i++];
    else
      A[k++] = C[j++];
  }
  while (i < n1) {
    A[k++] = B[i++];
  }
  while (j < n2) {
    A[k++] = C[j++];
  }
}
```