user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  lea    D35(%rip),%rsi
  mov    $0x17,%edx
  push   %r13
  lea    D90(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x2b00000017,%rax
  mov    %rsi,0x20(%rsp)
  mov    %rsp,%r12
  lea    0x20(%rsp),%rbx
  movl   $0x2,0x10(%rsp)
  mov    %rax,(%rsp)
  movabs $0x590000000c,%rax
  mov    %rax,0x8(%rsp)
  lea    D64(%rip),%rax
  mov    %rax,0x28(%rsp)
  lea    D100(%rip),%rax
  mov    %rax,0x30(%rsp)
  lea    D2(%rip),%rax
  mov    %rax,0x38(%rsp)
  lea    D11(%rip),%rax
  mov    %rax,0x40(%rsp)
L21:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x14,%rbp
  je     L76
  mov    (%r12,%rbp,1),%edx
  mov    (%rbx,%rbp,2),%rsi
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L76:
  lea    D105(%rip),%r14
  xor    %ebp,%ebp
  lea    D51(%rip),%r13
  mov    %r14,%rdi
  call   <puts@plt>
  lea    D35(%rip),%rsi
  mov    $0x17,%edx
L74:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x14,%rbp
  je     L41
  mov    (%r12,%rbp,1),%edx
  mov    (%rbx,%rbp,2),%rsi
  jmp    L74
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %r14,%rdi
  xor    %ebp,%ebp
  call   <puts@plt>
  lea    D35(%rip),%rsi
  mov    $0x17,%edx
L32:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    $0x14,%rbp
  je     L11
  mov    (%r12,%rbp,1),%edx
  mov    (%rbx,%rbp,2),%rsi
  jmp    L32
  nopl   0x0(%rax)
L11:
  mov    %r14,%rdi
  mov    %r12,%rbp
  call   <puts@plt>
  lea    D35(%rip),%rsi
  mov    $0x17,%edx
  jmp    L18
  nopl   0x0(%rax)
L58:
  mov    0x0(%rbp),%edx
  mov    (%rbx),%rsi
L18:
  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    L58
  mov    %r14,%rdi
  call   <puts@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x50,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D35
data_type: string
name: parse_data
arguments: label: D90
data_type: string
name: parse_data
arguments: label: D64
data_type: string
name: parse_data
arguments: label: D100
data_type: string
name: parse_data
arguments: label: D2
data_type: string
name: parse_data
arguments: label: D11
data_type: string
name: parse_data
arguments: label: D105
data_type: string
name: parse_data
arguments: label: D51
data_type: string

tool: D35: Alan

tool: D90: %s is %d years old.


tool: D64: Frank

tool: D100: Many

tool: D2: John

tool: D11: Lisa

tool: D105: ----------------------

tool: D51: %s is %d years old,


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