C++判断一个文件是否可以正确打开的代码

/* fopen example */
#include <iostream>
#include <conio.h>
using namespace std;
int main()
{
    FILE* fp;
    fp = fopen("C:/Users/Qin/Desktop/123.txt", "rb");
    if ((fp == NULL))
    {
        printf("
error on open C:/Users/Qin/Desktop/123.txt!");
        _getch();
        exit(1);
    }
    else
    {
        cout<<"the file C:/Users/Qin/Desktop/123.txt can be open correctly!" << endl;
    }
    system("pause");
    return 0;
}