user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%esi
  lea    D44(%rip),%rdi
  lea    D88(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0x8(%rsp)
  movabs $0x53202e412064655a,%rax
  mov    %rax,0xc(%rsp)
  xor    %eax,%eax
  movl   $0x776168,0x14(%rsp)
  call   <printf@plt>
  mov    $0x14,%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%edx
  mov    $0xa,%esi
  xor    %eax,%eax
  lea    D60(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D80(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xc,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D36(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D44
type_infered: string
name: extract_symbolic_data
arguments: tag: D88
type_infered: string
name: extract_symbolic_data
arguments: tag: D23
type_infered: string
name: extract_symbolic_data
arguments: tag: D58
type_infered: string
name: extract_symbolic_data
arguments: tag: D60
type_infered: string
name: extract_symbolic_data
arguments: tag: D80
type_infered: string
name: extract_symbolic_data
arguments: tag: D31
type_infered: string
name: extract_symbolic_data
arguments: tag: D22
type_infered: string
name: extract_symbolic_data
arguments: tag: D36
type_infered: string

tool: D44: The size of an int: %ld


tool: D88: The number of chars: %ld


tool: D23: The size of areas (int[]): %ld


tool: D58: The number of ints in areas: %ld


tool: D60: The first area is %d, the 2nd %d.


tool: D80: The size of a char: %ld


tool: D31: The size of name (char[]): %ld


tool: D22: The size of full name (char[]): %ld


tool: D36: name="%s" and full_name="%s"


assistant: ```c
int main(int argc, char *argv[]) {
  int areas[] = {10, 12, 13, 14, 20};
  char name[] = "Zed";
  char full_name[] = {'Z', 'e', 'd', ' ', 'A', '.', ' ', 'S', 'h', 'a', 'w', '\0'};
  printf("The size of an int: %ld\n", sizeof(int));
  printf("The size of areas (int[]): %ld\n", sizeof(areas));
  printf("The number of ints in areas: %ld\n", sizeof(areas) / sizeof(int));
  printf("The first area is %d, the 2nd %d.\n", areas[0], areas[1]);
  printf("The size of a char: %ld\n", sizeof(char));
  printf("The size of name (char[]): %ld\n", sizeof(name));
  printf("The number of chars: %ld\n", sizeof(name) / sizeof(char));
  printf("The size of full name (char[]): %ld\n", sizeof(full_name));
  printf("The number of chars: %ld\n", sizeof(full_name) / sizeof(char));
  printf("name=\"%s\" and full_name=\"%s\"\n", name, full_name);
  return 0;
}
```