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

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x10,%ecx
  mov    $0x10,%edx
  mov    $0x6,%esi
  lea    D113(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2,%edx
  mov    $0x3,%esi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xf,%edx
  mov    $0x5,%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D49(%rip),%rbp
L15:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L15
  mov    $0xa,%esi
  lea    D99(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%edx
  mov    $0x3,%esi
  lea    D91(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10,%edx
  mov    $0x6,%esi
  lea    D59(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%ecx
  mov    $0x6,%edx
  mov    $0x4,%esi
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%ecx
  mov    $0x7,%edx
  mov    $0x4,%esi
  lea    D105(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  lea    D64(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D112(%rip),%r12
  lea    D93(%rip),%rbp
L2:
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x6,%ebx
  jne    L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D113
type_infered: string
name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D30
type_infered: string
name: parse_data
arguments: tag: D49
type_infered: string
name: parse_data
arguments: tag: D99
type_infered: string
name: parse_data
arguments: tag: D91
type_infered: string
name: parse_data
arguments: tag: D59
type_infered: string
name: parse_data
arguments: tag: D47
type_infered: string
name: parse_data
arguments: tag: D105
type_infered: string
name: parse_data
arguments: tag: D64
type_infered: string
name: parse_data
arguments: tag: D112
type_infered: string
name: parse_data
arguments: tag: D93
type_infered: string

tool: D113: i = %d,j = %d,m = %d


tool: D7: i = 0,i = %d (i++,i++,i++) = %d


tool: D30: 
i = 3,i = %d (++i,i++,i+10) = %d


tool: D49: %d 

tool: D99: %d

tool: D91: i = 0,i = %d (i++)+(i++)+(i++) = %d



tool: D59: i = 3,i = %d ++i+++i+++i = %d


tool: D47: i = 3 d= 3,i = %d i+++d = %d d = %d


tool: D105: i = 3 d= 3,i = %d i+++++d = %d d = %d



tool: D64: First=%d

tool: D112: Fourth=%d

tool: D93: Third=%d

assistant: ```c
void main() {
  int i = 3;
  int j;
  int m;
  int n;
  int d = 3;
  j = (++i) + (++i) + (++i);
  m = (i, j);
  printf("i = %d,j = %d,m = %d\n", i, j, m);
  i = 0;
  m = (i++, i++, i++);
  printf("i = 0,i = %d (i++,i++,i++) = %d\n", i, m);
  i = 3;
  n = (++i, i++, i + 10);
  printf("\ni = 3,i = %d (++i,i++,i+10) = %d\n", i, n);
  for (i = 0; i < 10; i++) {
    printf("%d ", i);
  }
  printf("%d", i);
  i = 0;
  j = (i++) + (i++) + (i++);
  printf("i = 0,i = %d (i++)+(i++)+(i++) = %d\n\n", i, j);
  i = 3;
  j = ++i + ++i + ++i;
  printf("i = 3,i = %d ++i+++i+++i = %d\n", i, j);
  i = 3;
  m = i++ + d;
  printf("i = 3 d= 3,i = %d i+++d = %d d = %d\n", i, m, d);
  i = 3;
  d = 3;
  m = i++ + ++d;
  printf("i = 3 d= 3,i = %d i+++++d = %d d = %d\n\n", i, m, d);
  for (i = 0, printf("First=%d", i); i <= 5; i++, printf("Third=%d", i)) {
    printf("Fourth=%d", i);
  }
}
```