user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%r15
  lea    D50(%rip),%rdi
  call   <puts@plt>
  lea    -0x3c(%rbp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L79:
  cmp    %rdx,%rsp
  je     L20
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L79
L20:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L110
  orq    $0x0,-0x8(%rsp,%rax,1)
L110:
  lea    0x3(%rsp),%r14
  mov    %r14,%rax
  shr    $0x2,%rax
  mov    %rax,-0x48(%rbp)
  and    $0xfffffffffffffffc,%r14
  test   %esi,%esi
  jne    L42
  lea    D25(%rip),%rdi
  call   <puts@plt>
L87:
  mov    %r15,%rsp
L80:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L42:
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%eax
  test   %eax,%eax
  jle    L82
  mov    %r14,%r12
  mov    $0x0,%ebx
  lea    D21(%rip),%r13
L23:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    -0x3c(%rbp),%eax
  add    $0x4,%r12
  cmp    %ebx,%eax
  jg     L23
L82:
  cmp    $0x1,%eax
  jne    L8
  mov    -0x48(%rbp),%rax
  mov    0x0(,%rax,4),%esi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L87
L8:
  lea    D87(%rip),%rbx
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    -0x3c(%rbp),%esi
  mov    %r14,%rdi
  call   <ReverseSort@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  cmpl   $0x0,-0x3c(%rbp)
  jle    L92
  mov    $0x0,%ebx
  lea    D34(%rip),%r12
L54:
  mov    (%r14,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,-0x3c(%rbp)
  jg     L54
L92:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r15,%rsp
  jmp    L80
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D50
datatype: string
name: disassemble_data
arguments: data_tag: D21
datatype: string
name: disassemble_data
arguments: data_tag: D40
datatype: string
name: disassemble_data
arguments: data_tag: D25
datatype: string
name: disassemble_data
arguments: data_tag: D48
datatype: string
name: disassemble_data
arguments: data_tag: D7
datatype: string
name: disassemble_data
arguments: data_tag: D87
datatype: string
name: disassemble_data
arguments: data_tag: D34
datatype: string

tool: D50: Please write here the array size 

tool: D21: %d

tool: D40: The dimension of the array is %d


tool: D25: An empty array has not elements. We can not reverse an empty array

tool: D48: Please type the %d elements one by one and push ENTER


tool: D7: The reversed array is %d


tool: D87: Loading... 

tool: D34: %d 

assistant: ```c
int main() {
  int n;
  int i;
  printf("Please write here the array size \n");
  scanf("%d", &n);
  printf("The dimension of the array is %d\n", n);
  int a[n];
  if (n == 0)
    printf("An empty array has not elements. We can not reverse an empty array\n");
  else {
    printf("Please type the %d elements one by one and push ENTER\n", n);
    for (i = 0; i < n; i++)
      scanf("%d", &a[i]);
    if (n == 1)
      printf("The reversed array is %d\n", a[0]);
    else {
      printf("Loading... \n");
      ReverseSort(a, n);
      printf("Loading... \n");
      for (i = 0; i < n; i++)
        printf("%d ", a[i]);
      printf("\n");
      return 0;
    }
  }
}
```