user: 这段汇编代码用C语言表示是什么样的？

```asm
<calc_string>:
  endbr64
  push   %r15
  pxor   %xmm0,%xmm0
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xf8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x70(%rsp)
  movl   $0x0,0xe0(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x80(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movaps %xmm0,0xc0(%rsp)
  movaps %xmm0,0xd0(%rsp)
  call   <strlen@plt>
  test   %eax,%eax
  jle    L95
  mov    %eax,%r13d
  mov    %r15,%r12
  xor    %ebp,%ebp
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L98:
  and    $0xfffffffd,%eax
  cmp    $0x2d,%al
  je     L179
  add    $0x1,%r12
  cmp    %ebp,%r13d
  je     L95
L39:
  movzbl (%r12),%eax
  movslq %ebp,%rbx
  add    $0x1,%ebp
  lea    -0x2a(%rax),%edx
  cmp    $0x1,%dl
  ja     L98
L179:
  lea    0x10(%rsp),%r14
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <strncpy@plt>
  movzbl (%r12),%r12d
  cmp    %ebx,%r13d
  jle    L151
  movslq %ebp,%rax
  mov    %rax,0x8(%rsp)
  jmp    L170
  nopl   (%rax)
L70:
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jle    L151
L170:
  cmpb   $0x3d,(%r15,%rbx,1)
  jne    L70
  mov    0x8(%rsp),%rax
  mov    %ebx,%edx
  lea    0x80(%rsp),%rdi
  add    $0x1,%rbx
  sub    %ebp,%edx
  lea    (%r15,%rax,1),%rsi
  call   <strncpy@plt>
  cmp    %ebx,%r13d
  jg     L170
L151:
  cmp    $0x2d,%r12b
  je     L197
  jg     L56
  cmp    $0x2a,%r12b
  je     L124
  cmp    $0x2b,%r12b
  jne    L95
  mov    %r14,%rdi
  call   <atoi@plt>
  lea    0x80(%rsp),%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    %ebx,%eax
  nopl   0x0(%rax,%rax,1)
L95:
  mov    0xe8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L85
  add    $0xf8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L56:
  cmp    $0x2f,%r12b
  jne    L95
  lea    0x80(%rsp),%rbp
  mov    %rbp,%rdi
  call   <atoi@plt>
  test   %eax,%eax
  je     L95
  mov    %r14,%rdi
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  jmp    L95
  nopw   0x0(%rax,%rax,1)
L124:
  mov    %r14,%rdi
  call   <atoi@plt>
  lea    0x80(%rsp),%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  imul   %ebx,%eax
  jmp    L95
  nop
L197:
  mov    %r14,%rdi
  call   <atoi@plt>
  lea    0x80(%rsp),%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  sub    %r8d,%eax
  jmp    L95
L85:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int calc_string(const char *s) {
  char buf1[100] = {0};
  char oper1 = 0;
  char buf2[100] = {0};
  int len = strlen(s);
  int i;
  for (i = 0; i < len; i++) {
    if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/') {
      strncpy(buf1, s, i);
      oper1 = s[i];
      break;
    }
  }
  int start = i + 1;
  for (; i < len; i++) {
    if (s[i] == '=') {
      strncpy(buf2, &s[start], i - start);
    }
  }
  switch (oper1) {
  case '+':
    return atoi(buf1) + atoi(buf2);
  case '-':
    return atoi(buf1) - atoi(buf2);
  case '*':
    return atoi(buf1) * atoi(buf2);
  case '/': {
    int a = atoi(buf2);
    if (a)
      return atoi(buf1) / atoi(buf2);
    else
      return 0;
  }
  }
}
```