user: 这段汇编代码用C语言表示是什么样的?

```asm
<merge>:
  endbr64
  push   %rbp
  movslq %esi,%rax
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  mov    %edx,%ebx
  mov    %ebx,%r10d
  sub    %ebx,%ecx
  sub    %eax,%r10d
  mov    %ecx,%r12d
  sub    $0x28,%rsp
  lea    0x1(%r10),%r9d
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x38(%rbp)
  xor    %edx,%edx
  movslq %r9d,%rdx
  mov    %rsp,%rsi
  lea    0xf(,%rdx,4),%rdx
  mov    %rdx,%rcx
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L130
L27:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L27
L130:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L159
L209:
  movslq %r12d,%r14
  mov    %rsp,%rsi
  mov    %rsp,%r8
  lea    0xf(,%r14,4),%rdx
  mov    %rdx,%rcx
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L226
L78:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L78
L226:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L147
L58:
  mov    %rsp,%rcx
  test   %r9d,%r9d
  jle    L9
  mov    %r10d,%edx
  mov    %r8,%rdi
  lea    0x0(%r13,%rax,4),%rsi
  mov    %r9d,-0x48(%rbp)
  lea    0x4(,%rdx,4),%rdx
  mov    %rcx,-0x50(%rbp)
  mov    %r10d,-0x44(%rbp)
  call   <memcpy@plt>
  test   %r12d,%r12d
  mov    -0x44(%rbp),%r10d
  mov    -0x50(%rbp),%rcx
  mov    -0x48(%rbp),%r9d
  mov    %rax,%r8
  jle    L54
  add    $0x1,%ebx
  mov    %rcx,%rdi
  mov    %r9d,-0x48(%rbp)
  xor    %r15d,%r15d
  movslq %ebx,%rbx
  lea    0x0(,%r14,4),%rdx
  mov    %r10d,-0x44(%rbp)
  lea    0x0(%r13,%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
  mov    $0x1,%eax
  nopl   0x0(%rax)
L202:
  mov    (%r8,%r15,4),%edi
  mov    (%rcx,%r11,4),%esi
  movslq %eax,%r14
  cmp    %esi,%edi
  jg     L148
L19:
  mov    %edi,-0x4(%r13,%rax,4)
  lea    0x1(%rdx),%r15d
  add    $0x1,%rax
  cmp    %edx,%r10d
  jle    L105
  cmp    %ebx,%r12d
  jle    L105
  movslq %r15d,%r15
  mov    (%rcx,%r11,4),%esi
  movslq %eax,%r14
  mov    (%r8,%r15,4),%edi
  mov    %r15,%rdx
  cmp    %esi,%edi
  jle    L19
L148:
  mov    %esi,-0x4(%r13,%rax,4)
  add    $0x1,%ebx
  add    $0x1,%rax
  cmp    %edx,%r9d
  jle    L49
  cmp    %ebx,%r12d
  jle    L49
  movslq %ebx,%r11
  jmp    L202
  xchg   %ax,%ax
L159:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L209
  nopl   0x0(%rax,%rax,1)
L49:
  mov    %edx,%r15d
L105:
  cmp    %r15d,%r9d
  mov    %r9d,-0x44(%rbp)
  jle    L142
  movslq %r14d,%rax
  sub    %r15d,%r10d
  mov    %rcx,-0x50(%rbp)
  lea    0x0(%r13,%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    %r14d,%r9d
  sub    %r15d,%r9d
  movslq %r9d,%r14
L142:
  cmp    %ebx,%r12d
  jle    L64
L86:
  sub    $0x1,%r12d
  lea    0x0(%r13,%r14,4),%rdi
  sub    %ebx,%r12d
  movslq %ebx,%rbx
  lea    0x4(,%r12,4),%rdx
  lea    (%rcx,%rbx,4),%rsi
  call   <memcpy@plt>
L64:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L178
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L9:
  test   %r12d,%r12d
  jle    L54
  add    $0x1,%ebx
  lea    0x0(,%r14,4),%rdx
  mov    %rcx,%rdi
  xor    %r14d,%r14d
  movslq %ebx,%rbx
  lea    0x0(%r13,%rbx,4),%rsi
  xor    %ebx,%ebx
  call   <memcpy@plt>
  mov    %rax,%rcx
  jmp    L86
  nopl   (%rax)
L54:
  xor    %r14d,%r14d
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  jmp    L105
L147:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L58
L178:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void merge(int a[], int l, int m, int r) {
  int l1 = m - l + 1;
  int l2 = r - m;
  int i = 0, j = 0, k = 0;
  int a1[l1], a2[l2];
  for (i = 0; i < l1; i++)
    a1[i] = a[l + i];
  for (i = 0; i < l2; i++)
    a2[i] = a[m + 1 + i];
  i = 0;
  j = 0;
  k = 0;
  while ((i < l1) && (j < l2)) {
    if (a1[i] <= a2[j])
      a[k++] = a1[i++];
    else
      a[k++] = a2[j++];
  }
  while (i < l1)
    a[k++] = a1[i++];
  while (j < l2)
    a[k++] = a2[j++];
}
```