user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L28
  mov    (%rsi),%rsi
  lea    D58(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
L28:
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  test   %eax,%eax
  jle    L60
L67:
  mov    0x8(%rbx),%rdi
  lea    D80(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D20(%rip),%rbx
  call   <atoi@plt>
  lea    D3(%rip),%rdi
  movl   $0xffffffff,0x4(%rsp)
  mov    %eax,%r13d
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L51:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    $0x1,%eax
  ja     L51
  mov    %r13d,%esi
  lea    D105(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x7fffffff,%esi
  lea    D103(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  je     L24
  cmp    $0x2,%eax
  jne    L20
  lea    0x7fffffff(%r13),%r14d
  mov    %r13d,%edx
  mov    $0x7fffffff,%esi
  xor    %eax,%eax
  mov    %r14d,%ecx
  lea    D101(%rip),%rdi
  call   <printf@plt>
L20:
  sub    $0x8,%rsp
  mov    %r13d,%ecx
  mov    %r14d,%r9d
  mov    %r13d,%r8d
  push   %r14
  mov    $0x7fffffff,%edx
  mov    $0x7fffffff,%esi
  xor    %eax,%eax
  lea    D68(%rip),%rdi
  call   <printf@plt>
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    $0x7fffffff,%edi
  call   <checkForOverflow@plt>
  pop    %rdx
  pop    %rcx
  test   %rax,%rax
  je     L62
  lea    D13(%rip),%rdi
  call   <puts@plt>
L57:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L62:
  lea    D31(%rip),%rdi
  call   <puts@plt>
  jmp    L57
L24:
  imul   $0x7fffffff,%r13d,%r14d
  mov    %r13d,%edx
  mov    $0x7fffffff,%esi
  xor    %eax,%eax
  lea    D82(%rip),%rdi
  mov    %r14d,%ecx
  call   <printf@plt>
  jmp    L20
L60:
  mov    (%rbx),%rsi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L67
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D58
datatype: string
name: extract_symbolic_data
arguments: tag: D80
datatype: string
name: extract_symbolic_data
arguments: tag: D20
datatype: string
name: extract_symbolic_data
arguments: tag: D3
datatype: string
name: extract_symbolic_data
arguments: tag: D105
datatype: string
name: extract_symbolic_data
arguments: tag: D103
datatype: string
name: extract_symbolic_data
arguments: tag: D101
datatype: string
name: extract_symbolic_data
arguments: tag: D68
datatype: string
name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D31
datatype: string
name: extract_symbolic_data
arguments: tag: D82
datatype: string

tool: D58: usage: %s [POSITIVE NUMBER]


tool: D80: >> 

tool: D20: %d

tool: D3: 
Enter
1: Multiplication
2: Addition

tool: D105: You entered %d


tool: D103: The largest signed 32-bit integer is 0x7FFFFFFF = %d


tool: D101: %d + %d = %d


tool: D68: In hex:
%d = 0x%x
%d = 0x%x
%d = 0x%x


tool: D13: 
Overflow exists.

tool: D31: 
Overflow does not exist.

tool: D82: %d x %d = %d


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    printf("usage: %s [POSITIVE NUMBER]\n", argv[0]);
    exit(0);
  }
  if (atoi(argv[1]) < 1) {
    printf("usage: %s [POSITIVE NUMBER]\n", argv[0]);
    exit(0);
  }
  int baseNum = 0x7FFFFFFF;
  int x = atoi(argv[1]);
  int ans;
  int temp = -1;
  printf("\nEnter\n1: Multiplication\n2: Addition\n");
  do {
    printf(">> ");
    scanf("%d", &temp);
  } while ((temp != 1) && (temp != 2));
  printf("You entered %d\n", x);
  printf("The largest signed 32-bit integer is 0x7FFFFFFF = %d\n", baseNum);
  switch (temp) {
  case 1:
    ans = x * baseNum;
    printf("%d x %d = %d\n", baseNum, x, ans);
    break;
  case 2:
    ans = x + baseNum;
    printf("%d + %d = %d\n", baseNum, x, ans);
  }
  printf("In hex:\n%d = 0x%x\n%d = 0x%x\n%d = 0x%x\n", baseNum, baseNum, x, x, ans, ans);
  if (checkForOverflow(baseNum, x, ans)) {
    printf("\nOverflow exists.\n");
  } else {
    printf("\nOverflow does not exist.\n");
  }
}
```