user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xb4(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L228
L229:
  addl   $0x1,-0xb4(%rbp)
L228:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    -0xb4(%rbp),%edx
  movslq %edx,%rdx
  mov    %eax,-0xb0(%rbp,%rdx,4)
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0xa,%eax
  jne    L229
  subl   $0x1,-0xb4(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L26
L152:
  subl   $0x1,-0xb4(%rbp)
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0x20,%eax
  jne    L116
  addl   $0x1,-0xb4(%rbp)
  jmp    L191
L116:
  cmpl   $0x0,-0xb4(%rbp)
  je     L184
L113:
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0x20,%eax
  je     L118
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0xa,%eax
  jne    L152
  jmp    L118
L184:
  nop
L191:
  jmp    L118
L64:
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0xb4(%rbp)
L118:
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0x20,%eax
  je     L57
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0xa,%eax
  jne    L64
L57:
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0xa,%eax
  je     L79
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0x20,%eax
  jne    L51
L79:
  subl   $0x1,-0xb4(%rbp)
  jmp    L81
L14:
  subl   $0x1,-0xb4(%rbp)
  cmpl   $0x0,-0xb4(%rbp)
  je     L25
L51:
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0x20,%eax
  je     L153
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0xa,%eax
  jne    L14
  jmp    L153
L25:
  nop
  jmp    L153
L108:
  mov    $0x20,%edi
  call   <putchar@plt>
  subl   $0x1,-0xb4(%rbp)
L153:
  mov    -0xb4(%rbp),%eax
  cltq
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    $0x20,%eax
  je     L108
L26:
  cmpl   $0x0,-0xb4(%rbp)
  jg     L113
L81:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L100
  call   <__stack_chk_fail@plt>
L100:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D11
type_infered: string
name: extract_symbolic_data
arguments: data_label: D26
type_infered: string
name: extract_symbolic_data
arguments: data_label: D23
type_infered: string

tool: D11: Enter a sentence: 

tool: D26: Words reversed: 

tool: D23: i = %d


assistant: ```c
int main(void) {
  int words[40];
  int i = 0;
  printf("Enter a sentence: ");
  while ((words[i] = getchar()) != '\n') {
    i += 1;
  }
  i -= 1;
  printf("Words reversed: ");
  while (i > 0) {
    while (words[i] != ' ' && words[i] != '\n') {
      i -= 1;
      if (words[i] == ' ') {
        i += 1;
        break;
      }
      if (i == 0) {
        break;
      }
    }
    while (words[i] != ' ' && words[i] != '\n') {
      printf("%c", words[i]);
      i += 1;
    }
    if (words[i] == '\n' || words[i] == ' ') {
      i -= 1;
      break;
    }
    while (words[i] != ' ' && words[i] != '\n') {
      i -= 1;
      if (i == 0) {
        break;
      }
    }
    while (words[i] == ' ') {
      printf(" ");
      i -= 1;
    }
  }
  printf("\n");
  printf("i = %d\n", i);
  return 0;
}
```