user: What would the C source look like for the assembly code provided here?

```asm
<dump_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x10(%rbp)
  movl   $0x0,-0x28(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  jmp    L71
L60:
  cmpl   $0xf,-0x24(%rbp)
  jle    L128
  mov    -0x28(%rbp),%edx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x10,-0x2c(%rbp)
  jmp    L96
L128:
  mov    -0x28(%rbp),%edx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x24(%rbp),%edx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
L96:
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x10,-0x28(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L117
L76:
  mov    -0x30(%rbp),%eax
  movzbl -0x20(%rbp,%rax,1),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L41
  mov    $0x20,%edi
  call   <putchar@plt>
L41:
  mov    -0x30(%rbp),%eax
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    $0x7f,%al
  je     L91
  mov    -0x30(%rbp),%eax
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    $0x80,%al
  je     L91
  mov    -0x30(%rbp),%eax
  movzbl -0x20(%rbp,%rax,1),%eax
  test   %al,%al
  js     L130
  mov    -0x30(%rbp),%eax
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    $0x1f,%al
  jg     L130
L91:
  mov    -0x30(%rbp),%eax
  movb   $0x2e,-0x20(%rbp,%rax,1)
L130:
  addl   $0x1,-0x30(%rbp)
L117:
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jb     L76
  cmpl   $0xf,-0x2c(%rbp)
  ja     L78
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  jmp    L146
L103:
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L62
  mov    $0x20,%edi
  call   <putchar@plt>
L62:
  addl   $0x1,-0x30(%rbp)
L146:
  cmpl   $0xf,-0x30(%rbp)
  jbe    L103
  mov    -0x2c(%rbp),%eax
  movb   $0x0,-0x20(%rbp,%rax,1)
L78:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  subl   $0x10,-0x24(%rbp)
L71:
  cmpl   $0x0,-0x24(%rbp)
  jg     L60
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D28
type: string
name: parse_data
arguments: tag: D9
type: string
name: parse_data
arguments: tag: D29
type: string

tool: D28: %08X	

tool: D9: %02X

tool: D29:    

assistant: ```c
void dump_hex(char *str, int len) {
  char d[17];
  d[16] = 0;
  unsigned int i, bytes, index = 0;
  int left_len = len;
  while (left_len > 0) {
    if (left_len / 16 >= 1) {
      memcpy(d, str + index, 16);
      bytes = 16;
    } else {
      memcpy(d, str + index, left_len);
      bytes = left_len;
    }
    printf("%08X\t", index);
    index += 16;
    for (i = 0; i < bytes; i++) {
      printf("%02X", (unsigned char)(d[i]));
      if ((i + 1) % 4 == 0)
        printf(" ");
      if (d[i] == 127 || d[i] < -127 || (d[i] >= 0 && d[i] <= 31))
        d[i] = '.';
    }
    if (bytes < 16) {
      for (i = bytes; i < 16; i++) {
        printf("   ");
        if ((i + 1) % 4 == 0)
          printf(" ");
      }
      d[bytes] = '\0';
    }
    printf("%s\n", d);
    left_len -= 16;
  }
}
```