user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r13
  lea    D7(%rip),%rsi
  mov    $0x17,%edx
  lea    D54(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x2b00000017,%rax
  mov    %rsp,%r12
  mov    %rsi,0x20(%rsp)
  lea    0x20(%rsp),%rbx
  mov    %rax,(%rsp)
  movabs $0x590000000c,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x1600000002,%rax
  mov    %rax,0x10(%rsp)
  lea    D5(%rip),%rax
  mov    %rax,0x28(%rsp)
  lea    D64(%rip),%rax
  mov    %rax,0x30(%rsp)
  lea    D20(%rip),%rax
  mov    %rax,0x38(%rsp)
  lea    D19(%rip),%rax
  mov    %rax,0x40(%rsp)
  lea    D126(%rip),%rax
  mov    %rax,0x48(%rsp)
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L85:
  mov    (%r12,%rbp,1),%edx
  mov    (%rbx,%rbp,2),%rsi
L46:
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x18,%rbp
  jne    L85
  xor    %eax,%eax
  xor    %ebp,%ebp
  lea    D113(%rip),%r13
  call   <print_break@plt>
  lea    D7(%rip),%rsi
  mov    $0x17,%edx
  jmp    L89
  nopl   0x0(%rax,%rax,1)
L47:
  mov    (%r12,%rbp,1),%edx
  mov    (%rbx,%rbp,2),%rsi
L89:
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x18,%rbp
  jne    L47
  xor    %eax,%eax
  xor    %ebp,%ebp
  lea    D29(%rip),%r13
  call   <print_break@plt>
  lea    D7(%rip),%rsi
  mov    $0x17,%edx
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L78:
  mov    (%r12,%rbp,1),%edx
  mov    (%rbx,%rbp,2),%rsi
L30:
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x18,%rbp
  jne    L78
  xor    %eax,%eax
  mov    %r12,%rbp
  lea    D60(%rip),%r13
  call   <print_break@plt>
  lea    D7(%rip),%rsi
  mov    $0x17,%edx
  jmp    L22
  nopl   0x0(%rax)
L68:
  mov    0x0(%rbp),%edx
  mov    (%rbx),%rsi
L22:
  mov    %r13,%rdi
  add    $0x4,%rbp
  add    $0x8,%rbx
  call   <printf@plt>
  mov    %rbp,%rax
  sub    %r12,%rax
  cmp    $0x14,%rax
  jle    L68
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D7
datatype: string
name: disassemble_data
arguments: tag: D54
datatype: string
name: disassemble_data
arguments: tag: D5
datatype: string
name: disassemble_data
arguments: tag: D64
datatype: string
name: disassemble_data
arguments: tag: D20
datatype: string
name: disassemble_data
arguments: tag: D19
datatype: string
name: disassemble_data
arguments: tag: D126
datatype: string
name: disassemble_data
arguments: tag: D113
datatype: string
name: disassemble_data
arguments: tag: D29
datatype: string
name: disassemble_data
arguments: tag: D60
datatype: string

tool: D7: Alan

tool: D54: %s has %d years alive


tool: D5: Frank

tool: D64: Mary

tool: D20: John

tool: D19: Lisa

tool: D126: Colan

tool: D113: %s is %d years old


tool: D29: %s is %d years old again


tool: D60: %s lived %d years so far.


assistant: ```c
int main(int argc, char *argv[]) {
  int ages[] = {23, 43, 12, 89, 2, 22};
  char *names[] = {"Alan", "Frank", "Mary", "John", "Lisa", "Colan"};
  int count = sizeof(ages) / sizeof(int);
  int i;
  for (i = 0; i < count; ++i) {
    printf("%s has %d years alive\n", names[i], ages[i]);
  }
  print_break();
  int *cur_age = ages;
  char **cur_name = names;
  for (i = 0; i < count; ++i) {
    printf("%s is %d years old\n", *(cur_name + i), *(cur_age + i));
  }
  print_break();
  for (i = 0; i < count; ++i) {
    printf("%s is %d years old again\n", cur_name[i], cur_age[i]);
  }
  print_break();
  for (cur_name = names, cur_age = ages; (cur_age - ages) < count; ++cur_name, ++cur_age) {
    printf("%s lived %d years so far.\n", *cur_name, *cur_age);
  }
}
```