user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jle    L4
  mov    %rsi,%r12
  mov    %edi,%ebp
  mov    $0x0,%ebx
  lea    D119(%rip),%r13
L12:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L12
L4:
  movl   $0x17,(%rsp)
  movl   $0x2b,0x4(%rsp)
  movl   $0xc,0x8(%rsp)
  movl   $0x59,0xc(%rsp)
  movl   $0x2,0x10(%rsp)
  lea    D3(%rip),%rax
  mov    %rax,0x20(%rsp)
  lea    D116(%rip),%rax
  mov    %rax,0x28(%rsp)
  lea    D81(%rip),%rax
  mov    %rax,0x30(%rsp)
  lea    D60(%rip),%rax
  mov    %rax,0x38(%rsp)
  lea    D69(%rip),%rax
  mov    %rax,0x40(%rsp)
  mov    $0x4,%ebx
  lea    D103(%rip),%rbp
L13:
  mov    (%rsp,%rbx,4),%edx
  mov    0x20(%rsp,%rbx,8),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%rbx
  cmp    $0xffffffffffffffff,%rbx
  jne    L13
  lea    D121(%rip),%rdi
  call   <puts@plt>
  mov    $0x4,%ebx
  lea    D106(%rip),%rbp
L36:
  mov    (%rsp,%rbx,4),%edx
  mov    0x20(%rsp,%rbx,8),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%rbx
  cmp    $0xffffffffffffffff,%rbx
  jne    L36
  lea    D121(%rip),%rdi
  call   <puts@plt>
  mov    $0x4,%ebx
  lea    D23(%rip),%rbp
L60:
  mov    (%rsp,%rbx,4),%edx
  mov    0x20(%rsp,%rbx,8),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%rbx
  cmp    $0xffffffffffffffff,%rbx
  jne    L60
  lea    D121(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%rbp
  mov    %rsp,%rbx
  lea    D1(%rip),%r12
L41:
  mov    (%rbx),%edx
  mov    0x0(%rbp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rbp
  add    $0x4,%rbx
  mov    %rsp,%rdx
  mov    %rbx,%rax
  sub    %rdx,%rax
  cmp    $0x10,%rax
  jle    L41
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D119
type_infered: string
name: extract_symbolic_data
arguments: data_label: D3
type_infered: string
name: extract_symbolic_data
arguments: data_label: D116
type_infered: string
name: extract_symbolic_data
arguments: data_label: D81
type_infered: string
name: extract_symbolic_data
arguments: data_label: D60
type_infered: string
name: extract_symbolic_data
arguments: data_label: D69
type_infered: string
name: extract_symbolic_data
arguments: data_label: D103
type_infered: string
name: extract_symbolic_data
arguments: data_label: D121
type_infered: string
name: extract_symbolic_data
arguments: data_label: D106
type_infered: string
name: extract_symbolic_data
arguments: data_label: D23
type_infered: string
name: extract_symbolic_data
arguments: data_label: D1
type_infered: string

tool: D119: the argument %d is %s.


tool: D3: Alan

tool: D116: Frank

tool: D81: Mary

tool: D60: John

tool: D69: Lisa

tool: D103: %s has %d years alive.


tool: D121: ---

tool: D106: %s is %d years old.


tool: D23: %s is %d years old again.


tool: D1: %s lived %d years so far.


assistant: ```c
int main(int argc, char *argv[]) {
  char **commandArgs = argv;
  for (int k = 0; k < argc; k++) {
    printf("the argument %d is %s.\n", k, *(commandArgs + k));
  }
  int ages[] = {23, 43, 12, 89, 2};
  char *names[] = {"Alan", "Frank", "Mary", "John", "Lisa"};
  int count = sizeof(ages) / sizeof(int);
  int i = 0;
  for (i = count - 1; i >= 0; i--) {
    printf("%s has %d years alive.\n", names[i], ages[i]);
  }
  printf("---\n");
  int *cur_age = ages;
  char **cur_name = names;
  for (i = count - 1; i >= 0; i--) {
    printf("%s is %d years old.\n", *(cur_name + i), *(cur_age + i));
  }
  printf("---\n");
  for (i = count - 1; i >= 0; i--) {
    printf("%s is %d years old again.\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 lived %d years so far.\n", *cur_name, *cur_age);
  }
  return 0;
}
```