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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    D19(%rip),%rdi
  call   <puts@plt>
  lea    -0x3c(%rbp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@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
L61:
  cmp    %rdx,%rsp
  je     L105
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L61
L105:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L60
  orq    $0x0,-0x8(%rsp,%rax,1)
L60:
  mov    %rsp,%r12
  test   %esi,%esi
  jle    L106
  mov    %r12,%r13
  mov    $0x0,%ebx
  lea    D50(%rip),%r15
  lea    D31(%rip),%r14
L84:
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    -0x3c(%rbp),%esi
  add    $0x4,%r13
  cmp    %ebx,%esi
  jg     L84
L106:
  mov    %r12,%r13
  lea    -0x1(%rsi),%eax
  lea    0x4(%r12,%rax,4),%rdi
  mov    $0x0,%r10d
  mov    $0x1,%r9d
L67:
  test   %esi,%esi
  jle    L91
  mov    %r13,%rax
  mov    %r10d,%r8d
  jmp    L94
L134:
  add    $0x4,%rax
  cmp    %rdi,%rax
  je     L169
L94:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L134
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  mov    %r9d,%r8d
  jmp    L134
L169:
  test   %r8d,%r8d
  jne    L67
L91:
  mov    -0x3c(%rbp),%eax
  test   %eax,%eax
  jle    L0
  mov    $0x0,%ebx
  lea    D18(%rip),%r14
L147:
  mov    0x0(%r13),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  mov    -0x3c(%rbp),%eax
  add    $0x4,%r13
  cmp    %ebx,%eax
  jg     L147
L31:
  test   $0x1,%al
  je     L150
  mov    $0x2,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  cltq
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%r12,%rax,4),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D40(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L108:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L0:
  mov    $0x0,%ebx
  jmp    L31
L150:
  lea    -0x1(%rbx),%eax
  mov    $0x2,%ecx
  cltd
  idiv   %ecx
  cltq
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%r12,%rax,4),%xmm0
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  cltq
  pxor   %xmm1,%xmm1
  cvtsi2ssl (%r12,%rax,4),%xmm1
  addss  %xmm1,%xmm0
  mulss  D49(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D35(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L108
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D19
type_guessed: string
name: unpack_data
arguments: data_tag: D31
type_guessed: string
name: unpack_data
arguments: data_tag: D50
type_guessed: string
name: unpack_data
arguments: data_tag: D18
type_guessed: string
name: unpack_data
arguments: data_tag: D40
type_guessed: string
name: unpack_data
arguments: data_tag: D49
type_guessed: byte[4]
name: unpack_data
arguments: data_tag: D35
type_guessed: string

tool: D19: hvor mange tal skal du bruge?

tool: D31: %d

tool: D50: Tal nr. %d:
 

tool: D18: %d


tool: D40: 
Medianen af dine tal = %.1f


tool: {"D49": [0x00,0x00,0x00,0x3f]}

tool: D35: 
Medianen for dine tal = %.2f


assistant: ```c
int main() {
  int swapped, r, i;
  int tal, antal;
  printf("hvor mange tal skal du bruge?\n");
  scanf("%d", &tal);
  int u[tal];
  for (i = 0; i < tal; i++) {
    printf("Tal nr. %d:\n ", i);
    scanf("%d", &u[i]);
  }
  while (1) {
    swapped = 0;
    for (r = 0; r < tal; r++) {
      if (u[r] > u[r + 1]) {
        int temp = u[r];
        u[r] = u[r + 1];
        u[r + 1] = temp;
        swapped = 1;
      }
    }
    if (swapped == 0) {
      break;
    }
  }
  for (r = 0; r < tal; r++) {
    printf("%d\n", u[r]);
  }
  if ((tal % 2) != 0) {
    float median = u[r / 2];
    printf("\nMedianen af dine tal = %.1f\n", median);
  } else if (tal % 2 == 0) {
    float d = (u[(r - 1) / 2]);
    float c = (u[r / 2]);
    float a = (d + c) / 2;
    printf("\nMedianen for dine tal = %.2f\n", a);
  }
  return 0;
}
```