user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movl   $0x1,-0x24(%rbp)
  jmp    L33
L88:
  movl   $0x0,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  add    $0xa,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L71
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x2,%edi
  call   <exit@plt>
L71:
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <random_string@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0xa(%rax),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0xa(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L98
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  jmp    L40
L98:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L40
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L40:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x20(%rbp)
  jmp    L5
L44:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L70
  addl   $0x1,-0x1c(%rbp)
L70:
  addl   $0x1,-0x20(%rbp)
L5:
  mov    -0x24(%rbp),%eax
  add    $0x9,%eax
  cmp    %eax,-0x20(%rbp)
  jle    L44
  cmpl   $0xa,-0x1c(%rbp)
  je     L73
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L73:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x7d0,-0x24(%rbp)
L33:
  cmpl   $0xc34f,-0x24(%rbp)
  jle    L88
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D14
datatype: string
name: retrieve_data
arguments: label: D7
datatype: string
name: retrieve_data
arguments: label: D1
datatype: string
name: retrieve_data
arguments: label: D25
datatype: string
name: retrieve_data
arguments: label: D20
datatype: string

tool: D14: strncpy/2-1.c Failed to allocate memory

tool: D7: strncpy/2-1.c Test failed, string copy failed. Expected string: %s, But got: %s


tool: D1: strncpy/2-1.c Test Failed, return is not as expected. Expected return: %p, But obtained: %p


tool: D25: strncpy/2-1.c Test Failed, The difference in the number of bytes for two strings (s1 and s2) have not been appended with NULL bytes

tool: D20: strncpy/2-1.c  Test Passed, The difference in the number of bytes for two strings (s1 and s2) have been appended with NULL bytes in s2

assistant: ```c
int main(void) {
  int i, j, c;
  char *ret_str;
  for (i = 1; i < 50000; i += 2000) {
    c = 0;
    char *sample_str_1;
    char *sample_str_2 = malloc(i + 10);
    if (sample_str_2 == NULL) {
      printf("strncpy/2-1.c"
             " Failed to allocate memory\n");
      exit(2);
    }
    sample_str_1 = random_string(i);
    ret_str = strncpy(sample_str_2, sample_str_1, i + 10);
    sample_str_2[i + 10] = '\0';
    if (strcmp(sample_str_1, sample_str_2) != 0) {
      printf("strncpy/2-1.c"
             " Test failed, string copy failed. "
             "Expected string: %s, But got: %s\n",
             sample_str_1, sample_str_2);
      exit(2);
    } else if (ret_str != sample_str_2) {
      printf("strncpy/2-1.c"
             " Test Failed, return is not as expected. "
             "Expected return: %p, But obtained: %p\n",
             sample_str_2, ret_str);
      exit(1);
    }
    for (j = strlen(sample_str_1); j < i + 10; j++) {
      if (sample_str_2[j] == '\0') {
        c++;
      }
    }
    if (c != 10) {
      printf("strncpy/2-1.c"
             " Test Failed, The difference in the number of bytes"
             " for two strings (s1 and s2) have not been appended"
             " with NULL bytes\n");
      exit(1);
    }
    free(sample_str_1);
    free(sample_str_2);
  }
  printf("strncpy/2-1.c"
         "  Test Passed, The difference in the number of bytes for two "
         "strings (s1 and s2) have been appended with NULL bytes in s2\n");
  exit(0);
}
```