user: 请从给定的汇编指令推导出C源代码:

```asm
<id>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    $0x0,%eax
  call   <getuid@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <getpwuid@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L34
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L34:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <geteuid@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <getpwuid@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L29
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L29:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <getgid@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <getgrgid@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L11
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L11:
  mov    $0x0,%eax
  call   <getegid@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <getgrgid@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L26
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L26:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D51
data_type: string
name: retrieve_data
arguments: data_tag: D24
data_type: string
name: retrieve_data
arguments: data_tag: D57
data_type: string
name: retrieve_data
arguments: data_tag: D42
data_type: string
name: retrieve_data
arguments: data_tag: D21
data_type: string

tool: D51: uid:%i

tool: D24: (%s)

tool: D57: euid:%i

tool: D42: gid:%i

tool: D21: egid:%i

assistant: ```c
int id(char **command, struct tm start) {
  struct passwd *pwd = NULL;
  struct group *gr = NULL;
  uid_t uid = getuid();
  printf("uid:%i", uid);
  pwd = getpwuid(uid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  uid_t euid = geteuid();
  printf("euid:%i", euid);
  pwd = getpwuid(euid);
  if (pwd)
    printf("(%s)", pwd->pw_name);
  printf(" ");
  gid_t gid = getgid();
  printf("gid:%i", gid);
  gr = getgrgid(gid);
  if (gr)
    printf("(%s)", gr->gr_name);
  gid_t egid = getegid();
  printf("egid:%i", egid);
  gr = getgrgid(egid);
  if (gr)
    printf("(%s)", gr->gr_name);
  printf("\n");
  return 1;
}
```