user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x29,-0x30(%rbp)
  movl   $0x11,-0x2c(%rbp)
  movl   $0x17,-0x28(%rbp)
  movl   $0x53,-0x24(%rbp)
  movl   $0x49,-0x20(%rbp)
  movl   $0x5f,-0x1c(%rbp)
  movl   $0x6,-0x18(%rbp)
  movl   $0x2,-0x14(%rbp)
  movl   $0x59,-0x10(%rbp)
  movl   $0x35,-0xc(%rbp)
  movl   $0x0,-0x44(%rbp)
  jmp    L88
L92:
  cmpl   $0x0,-0x44(%rbp)
  jne    L72
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x40(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x34(%rbp)
  jmp    L100
L72:
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x40(%rbp)
  jge    L22
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x40(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x38(%rbp)
L22:
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x3c(%rbp)
  jle    L100
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x34(%rbp)
L100:
  addl   $0x1,-0x44(%rbp)
L88:
  cmpl   $0x9,-0x44(%rbp)
  jle    L92
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x44(%rbp)
  jmp    L12
L21:
  cmpl   $0x9,-0x44(%rbp)
  je     L54
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L59
L54:
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L59:
  addl   $0x1,-0x44(%rbp)
L12:
  cmpl   $0x9,-0x44(%rbp)
  jle    L21
  mov    -0x38(%rbp),%esi
  mov    -0x40(%rbp),%ecx
  mov    -0x34(%rbp),%edx
  mov    -0x3c(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L76
  call   <__stack_chk_fail@plt>
L76:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D24
type_infered: string
name: parse_data
arguments: data_label: D3
type_infered: string
name: parse_data
arguments: data_label: D6
type_infered: string
name: parse_data
arguments: data_label: D22
type_infered: string

tool: D24: Vetor: 

tool: D3: %d 

tool: D6: %d


tool: D22: Menor: %d. Posicao: %d.
Maior: %d. Posicao: %d.


assistant: ```c
void main() {
  int V[10] = {41, 17, 23, 83, 73, 95, 6, 2, 89, 53}, i, maior, menor, maipos, menpos;
  for (i = 0; i < 10; i++) {
    if (i == 0) {
      maior = V[i];
      maipos = i;
      menor = V[i];
      menpos = i;
    } else {
      if (V[i] > maior) {
        maior = V[i];
        maipos = i;
      }
      if (V[i] < menor) {
        menor = V[i];
        menpos = i;
      }
    }
  }
  printf("Vetor: ");
  for (i = 0; i < 10; i++)
    if (i != 9)
      printf("%d ", V[i]);
    else
      printf("%d\n", V[i]);
  printf("Menor: %d. Posicao: %d.\n"
         "Maior: %d. Posicao: %d.\n",
         menor, menpos, maior, maipos);
}
```