user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %edi,-0x84(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x84(%rbp)
  jg     L5
  mov    $0x1,%eax
  jmp    L6
L5:
  movl   $0x0,-0x78(%rbp)
  movl   $0x0,-0x74(%rbp)
  mov    -0x90(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  mov    %rax,-0x70(%rbp)
  mov    -0x90(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  mov    %rax,-0x68(%rbp)
  jmp    L38
L65:
  mov    -0x70(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,%rsi
  mov    %rsi,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  add    %rax,%rax
  mov    %rcx,%rsi
  sub    %rax,%rsi
  mov    -0x68(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rdi
  sar    $0x3f,%rdi
  sub    %rdi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  add    %rsi,%rdx
  mov    -0x74(%rbp),%eax
  cltq
  add    %rdx,%rax
  cqto
  shr    $0x3f,%rdx
  add    %rdx,%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  mov    %rax,%rcx
  mov    -0x78(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x78(%rbp)
  mov    %ecx,%edx
  cltq
  mov    %edx,-0x60(%rbp,%rax,4)
  mov    -0x70(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,%rsi
  mov    %rsi,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  add    %rax,%rax
  mov    %rcx,%rsi
  sub    %rax,%rsi
  mov    -0x68(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rdi
  sar    $0x3f,%rdi
  sub    %rdi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  add    %rsi,%rdx
  mov    -0x74(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %eax,-0x74(%rbp)
  mov    -0x70(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x70(%rbp)
  mov    -0x68(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x68(%rbp)
L38:
  cmpq   $0x0,-0x70(%rbp)
  jne    L65
  cmpq   $0x0,-0x68(%rbp)
  jne    L65
  cmpl   $0x0,-0x74(%rbp)
  je     L27
  mov    -0x78(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x78(%rbp)
  cltq
  mov    -0x74(%rbp),%edx
  mov    %edx,-0x60(%rbp,%rax,4)
L27:
  subl   $0x1,-0x78(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L60
L37:
  mov    -0x78(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x78(%rbp)
  cltq
  mov    -0x60(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L60:
  cmpl   $0x0,-0x78(%rbp)
  jns    L37
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D18
type_guessed: string
name: disassemble_data
arguments: data_label: D16
type_guessed: string

tool: D18: Sum of two binary numbers: 

tool: D16: %d

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 3)
    return 1;
  long int binary1, binary2;
  int i = 0, remainder = 0, sum[20];
  binary1 = (long int)argv[1][0];
  binary2 = (long int)argv[2][0];
  while (binary1 != 0 || binary2 != 0) {
    sum[i++] = (binary1 % 10 + binary2 % 10 + remainder) % 2;
    remainder = (binary1 % 10 + binary2 % 10 + remainder) / 2;
    binary1 = binary1 / 10;
    binary2 = binary2 / 10;
  }
  if (remainder != 0)
    sum[i++] = remainder;
  --i;
  printf("Sum of two binary numbers: ");
  while (i >= 0)
    printf("%d", sum[i--]);
  printf("\n");
  return 0;
}
```