user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D6(%rip),%r14
  push   %r13
  lea    D37(%rip),%r13
  push   %r12
  mov    $0x4,%r12d
  push   %rbp
  lea    D30(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  movl   $0x3,0xc(%rsp)
L171:
  mov    %r14,%rdi
  mov    %r12d,%ebx
  call   <printf@plt>
  cmp    $0x1,%r12d
  je     L185
  mov    %r12d,%r15d
L109:
  mov    %r13,%rdi
  sub    $0x1,%r15d
  call   <printf@plt>
  cmp    $0x1,%r15d
  jne    L109
  lea    D94(%rip),%rdi
  call   <printf@plt>
  cmp    $0x4,%r12d
  je     L121
  nopl   (%rax)
L167:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x4,%ebx
  jne    L167
  mov    0xc(%rsp),%ecx
  test   %ecx,%ecx
  je     L175
L121:
  subl   $0x1,0xc(%rsp)
  sub    $0x1,%r12d
  jmp    L171
L185:
  lea    D94(%rip),%rdi
  call   <printf@plt>
  jmp    L167
L175:
  mov    %r14,%rdi
  mov    $0x7,%r13d
  lea    D37(%rip),%r12
  call   <printf@plt>
  movl   $0x6,0xc(%rsp)
  xchg   %ax,%ax
L24:
  mov    %r14,%rdi
  mov    %r13d,%ebx
  call   <printf@plt>
  cmp    $0x1,%r13d
  je     L174
  mov    %r13d,%r15d
  nopl   0x0(%rax)
L177:
  mov    %r12,%rdi
  sub    $0x1,%r15d
  call   <printf@plt>
  cmp    $0x1,%r15d
  jne    L177
  lea    D94(%rip),%rdi
  call   <printf@plt>
  cmp    $0x7,%r13d
  je     L113
  nopl   0x0(%rax)
L57:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7,%ebx
  jne    L57
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  je     L93
L113:
  subl   $0x1,0xc(%rsp)
  sub    $0x1,%r13d
  jmp    L24
L174:
  lea    D94(%rip),%rdi
  call   <printf@plt>
  jmp    L57
L93:
  mov    %r14,%rdi
  mov    $0xb,%r13d
  lea    D37(%rip),%r12
  call   <printf@plt>
  movl   $0xa,0xc(%rsp)
  xchg   %ax,%ax
L14:
  mov    %r14,%rdi
  mov    %r13d,%ebx
  call   <printf@plt>
  cmp    $0x1,%r13d
  je     L152
  mov    %r13d,%r15d
  nopl   0x0(%rax)
L124:
  mov    %r12,%rdi
  sub    $0x1,%r15d
  call   <printf@plt>
  cmp    $0x1,%r15d
  jne    L124
  lea    D94(%rip),%rdi
  call   <printf@plt>
  cmp    $0xb,%r13d
  je     L156
  nopl   0x0(%rax)
L63:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xb,%ebx
  jne    L63
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L98
L156:
  subl   $0x1,0xc(%rsp)
  sub    $0x1,%r13d
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L152:
  lea    D94(%rip),%rdi
  call   <printf@plt>
  jmp    L63
L98:
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type: string
name: retrieve_data
arguments: label: D37
type: string
name: retrieve_data
arguments: label: D30
type: string
name: retrieve_data
arguments: label: D94
type: string

tool: D6: 


tool: D37:  

tool: D30: **

tool: D94: *

assistant: ```c
int main() {
  int i, p;
  char c;
  p = 0;
  for (i = 4; i >= 1; i--) {
    p = i;
    printf("\n");
    while (p >= 2) {
      printf(" ");
      p--;
    }
    printf("*");
    p = i;
    while (p <= 3) {
      printf("**");
      p++;
    }
  }
  printf("\n");
  for (i = 7; i >= 1; i--) {
    p = i;
    printf("\n");
    while (p >= 2) {
      printf(" ");
      p--;
    }
    printf("*");
    p = i;
    while (p <= 6) {
      printf("**");
      p++;
    }
  }
  printf("\n");
  for (i = 11; i >= 1; i--) {
    p = i;
    printf("\n");
    while (p >= 2) {
      printf(" ");
      p--;
    }
    printf("*");
    p = i;
    while (p <= 10) {
      printf("**");
      p++;
    }
  }
  printf("\n");
  return 0;
}
```