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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x7c(%rbp)
  jmp    L39
L60:
  mov    -0x7c(%rbp),%eax
  cltq
  movb   $0x2a,-0x70(%rbp,%rax,1)
  addl   $0x1,-0x7c(%rbp)
L39:
  cmpl   $0x60,-0x7c(%rbp)
  jle    L60
  mov    -0x7c(%rbp),%eax
  cltq
  movb   $0x0,-0x70(%rbp,%rax,1)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x70(%rbp),%rax
  mov    $0x5,%edx
  lea    D42(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <string_ncopy@plt>
  mov    %rax,-0x78(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x78(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x70(%rbp),%rax
  mov    $0x5a,%edx
  lea    D42(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <string_ncopy@plt>
  mov    %rax,-0x78(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x78(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x7c(%rbp)
  jmp    L43
L18:
  mov    -0x7c(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  je     L34
  mov    $0x20,%edi
  call   <putchar@plt>
L34:
  mov    -0x7c(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %ecx,%edx
  sub    %eax,%edx
  test   %edx,%edx
  jne    L30
  cmpl   $0x0,-0x7c(%rbp)
  je     L30
  mov    $0xa,%edi
  call   <putchar@plt>
L30:
  mov    -0x7c(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x7c(%rbp)
L43:
  cmpl   $0x61,-0x7c(%rbp)
  jle    L18
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D42
data_type: string
name: parse_data
arguments: data_label: D7
data_type: string
name: parse_data
arguments: data_label: D34
data_type: string

tool: D42: First, solve the problem. Then, write the code


tool: D7: %s

tool: D34: 0x%02x

assistant: ```c
int main(void) {
  char s1[98];
  char *p;
  int i;
  for (i = 0; i < 98 - 1; i++) {
    s1[i] = '*';
  }
  s1[i] = '\0';
  printf("%s\n", s1);
  p = string_ncopy(s1, "First, solve the problem. Then, write the code\n", 5);
  printf("%s\n", s1);
  printf("%s\n", p);
  p = string_ncopy(s1, "First, solve the problem. Then, write the code\n", 90);
  printf("%s", s1);
  printf("%s", p);
  for (i = 0; i < 98; i++) {
    if (i % 10) {
      printf(" ");
    }
    if (!(i % 10) && i) {
      printf("\n");
    }
    printf("0x%02x", s1[i]);
  }
  printf("\n");
  return (0);
}
```