user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D25(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D13(%rip),%r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%r14
  call   <puts@plt>
  lea    -0x3c(%rbp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rcx
  mov    %rax,%rsi
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L33
L88:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L88
L33:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L71
L68:
  lea    0x3(%rsp),%r13
  mov    %r13,%rax
  and    $0xfffffffffffffffc,%r13
  shr    $0x2,%rax
  mov    %rax,-0x48(%rbp)
  test   %esi,%esi
  je     L25
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  cmpl   $0x0,-0x3c(%rbp)
  jle    L4
  mov    %r13,%r15
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L56:
  mov    %r15,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %ebx,%eax
  jg     L56
  sub    $0x1,%eax
  jne    L4
  mov    -0x48(%rbp),%rax
  lea    D79(%rip),%rdi
  mov    0x0(,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
L13:
  mov    %r14,%rsp
L34:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L122
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L25:
  lea    D43(%rip),%rdi
  call   <puts@plt>
  jmp    L13
L4:
  lea    D15(%rip),%r12
  mov    %r12,%rdi
  call   <puts@plt>
  mov    -0x3c(%rbp),%esi
  mov    %r13,%rdi
  call   <ReverseSort@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  cmpl   $0x0,-0x3c(%rbp)
  jle    L73
  xor    %ebx,%ebx
  lea    D42(%rip),%r12
  nopl   (%rax)
L99:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x3c(%rbp)
  jg     L99
L73:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r14,%rsp
  jmp    L34
L71:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L68
L122:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
datatype: string
name: retrieve_data
arguments: data_tag: D13
datatype: string
name: retrieve_data
arguments: data_tag: D0
datatype: string
name: retrieve_data
arguments: data_tag: D6
datatype: string
name: retrieve_data
arguments: data_tag: D79
datatype: string
name: retrieve_data
arguments: data_tag: D43
datatype: string
name: retrieve_data
arguments: data_tag: D15
datatype: string
name: retrieve_data
arguments: data_tag: D42
datatype: string

tool: D25: Please write here the array size 

tool: D13: %d

tool: D0: The dimension of the array is %d


tool: D6: Please type the %d elements one by one and push ENTER


tool: D79: The reversed array is %d


tool: D43: An empty array has not elements. We can not reverse an empty array

tool: D15: Loading... 

tool: D42: %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;
    }
  }
}
```