user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D109(%rip),%rcx
  push   %r13
  lea    D22(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x50,%rsp
  movdqa D20(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    D91(%rip),%rax
  mov    %rsp,%r12
  lea    0x20(%rsp),%rbx
  movq   %rax,%xmm1
  movaps %xmm0,(%rsp)
  lea    D92(%rip),%rax
  movq   %rcx,%xmm0
  punpcklqdq %xmm1,%xmm0
  movq   %rax,%xmm2
  lea    D18(%rip),%rcx
  movl   $0x2,0x10(%rsp)
  lea    D4(%rip),%rax
  movaps %xmm0,0x20(%rsp)
  movq   %rcx,%xmm0
  mov    %rax,0x40(%rsp)
  punpcklqdq %xmm2,%xmm0
  movaps %xmm0,0x30(%rsp)
L41:
  mov    (%r12,%rbp,4),%edx
  mov    (%rbx,%rbp,8),%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x5,%rbp
  jne    L41
  lea    D84(%rip),%r13
  xor    %ebp,%ebp
  lea    D49(%rip),%r14
  mov    %r13,%rdi
  call   <puts@plt>
L2:
  mov    (%r12,%rbp,4),%edx
  mov    (%rbx,%rbp,8),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x5,%rbp
  jne    L2
  mov    %r13,%rdi
  xor    %ebp,%ebp
  lea    D32(%rip),%r14
  call   <puts@plt>
L27:
  mov    (%r12,%rbp,4),%edx
  mov    (%rbx,%rbp,8),%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x5,%rbp
  jne    L27
  mov    %r13,%rdi
  mov    %r12,%rbp
  lea    D98(%rip),%r13
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L16:
  mov    0x0(%rbp),%ecx
  mov    (%rbx),%rsi
  mov    %rbp,%r8
  mov    %rbx,%rdx
  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    L16
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x50,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D109
type: string
name: parse_data
arguments: label: D22
type: string
name: parse_data
arguments: label: D20
type: byte[16]
name: parse_data
arguments: label: D91
type: string
name: parse_data
arguments: label: D92
type: string
name: parse_data
arguments: label: D18
type: string
name: parse_data
arguments: label: D4
type: string
name: parse_data
arguments: label: D84
type: string
name: parse_data
arguments: label: D49
type: string
name: parse_data
arguments: label: D32
type: string
name: parse_data
arguments: label: D98
type: string

tool: D109: Alan

tool: D22: %s has %i years alive.


tool: {"D20": [0x17,0x00,0x00,0x00,0x2b,0x00,0x00,0x00,0x0c,0x00,0x00,0x00,0x59,0x00,0x00,0x00]}

tool: D91: Frank

tool: D92: Jhon

tool: D18: Mary

tool: D4: Lisa

tool: D84: ---

tool: D49: %s has %i years old again.


tool: D32: %s has %i years old.


tool: D98: %s (%p) has %i (%p) years so far.


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