user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D64(%rip),%rcx
  push   %r14
  lea    D49(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x16,%ebp
  push   %rbx
  sub    $0x58,%rsp
  movdqa D15(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  movl   $0x2,0x10(%rsp)
  lea    0x20(%rsp),%rbx
  movq   %rax,%xmm1
  movaps %xmm0,(%rsp)
  mov    %rbx,%r13
  mov    %rbx,%r12
  movq   %rcx,%xmm0
  lea    D26(%rip),%rax
  lea    D35(%rip),%rcx
  punpcklqdq %xmm1,%xmm0
  movq   %rax,%xmm2
  lea    D27(%rip),%rax
  movaps %xmm0,0x20(%rsp)
  movq   %rcx,%xmm0
  mov    %rax,0x40(%rsp)
  punpcklqdq %xmm2,%xmm0
  movaps %xmm0,0x30(%rsp)
L29:
  mov    (%r12),%rsi
  mov    %ebp,%edx
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x8,%r12
  call   <printf@plt>
  cmp    $0x1b,%ebp
  jne    L29
  lea    D79(%rip),%r14
  xor    %ebp,%ebp
  mov    %rsp,%r12
  mov    %r14,%rdi
  lea    D77(%rip),%r15
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  mov    (%r12,%rbp,4),%edx
  mov    (%rbx,%rbp,8),%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x6,%rbp
  jne    L2
  mov    %r14,%rdi
  mov    $0x16,%ebp
  lea    D89(%rip),%r15
  call   <puts@plt>
  nopl   0x0(%rax,%rax,1)
L13:
  mov    0x0(%r13),%rsi
  mov    %ebp,%edx
  mov    %r15,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x8,%r13
  call   <printf@plt>
  cmp    $0x1b,%ebp
  jne    L13
  mov    %r14,%rdi
  mov    %r12,%rbp
  lea    D72(%rip),%r13
  call   <puts@plt>
  xchg   %ax,%ax
L8:
  mov    0x0(%rbp),%edx
  mov    (%rbx),%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  add    $0x8,%rbx
  call   <printf@plt>
  mov    %rbp,%rax
  sub    %r12,%rax
  cmp    $0x10,%rax
  jle    L8
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D64
type_infered: string
name: unpack_data
arguments: tag: D49
type_infered: string
name: unpack_data
arguments: tag: D15
type_infered: byte[16]
name: unpack_data
arguments: tag: D1
type_infered: string
name: unpack_data
arguments: tag: D26
type_infered: string
name: unpack_data
arguments: tag: D35
type_infered: string
name: unpack_data
arguments: tag: D27
type_infered: string
name: unpack_data
arguments: tag: D79
type_infered: string
name: unpack_data
arguments: tag: D77
type_infered: string
name: unpack_data
arguments: tag: D89
type_infered: string
name: unpack_data
arguments: tag: D72
type_infered: string

tool: D64: Alan

tool: D49: %s is %d years old


tool: {"D15": [0x16,0x00,0x00,0x00,0x24,0x00,0x00,0x00,0x1b,0x00,0x00,0x00,0x17,0x00,0x00,0x00]}

tool: D1: Frank

tool: D26: John

tool: D35: Mary

tool: D27: Lisa

tool: D79: ---

tool: D77: %s is %d years alive


tool: D89: %s has lived %d years


tool: D72: %s's age is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int ages[] = {22, 36, 27, 23, 2};
  char *names[] = {"Alan", "Frank", "Mary", "John", "Lisa"};
  int count = sizeof(ages) / sizeof(int);
  int i = 0;
  char **cur_name = names;
  int *cur_age = ages;
  for (; i < count; i++) {
    printf("%s is %d years old\n", *(cur_name + i), *cur_age + i);
  }
  printf("---\n");
  for (i = 0; i < count + 1; i++) {
    printf("%s is %d years alive\n", *(cur_name + i), *(cur_age + i));
  }
  printf("---\n");
  for (i = -1; i++, i < count;) {
    printf("%s has lived %d years\n", *(cur_name + i), *cur_age + i);
  }
  printf("---\n");
  for (cur_age = ages, cur_name = names; (cur_age - ages) < count; cur_age++, cur_name++) {
    printf("%s's age is %d\n", *cur_name, *cur_age);
  }
  return 0;
}
```