接收到错误的数据类型时如何获取正确的数据类型

Question

#include<iostream>
#include<string>
using namespace std;

int main() {
    while (1) {
        int num;
        cin >> num;
        if (typeid(num).name() != "int") {
            cout << "please input integer" << endl;
            continue;
        }
    }
}

When I wrote the code like this, I thought that if the wrong data type was input, the while statement would be executed again by continue , and the variable num would be input again.

However, an infinite loop occurs instead. I would be grateful if you could help me with why.

Answer 1

You are not checking whether cin >> num; succeeds or fails. And if it does fail, you are not clear() 'ing the error before continuing.

You can't use typeid() to check for success/failure of >> . That is not what it is meant for. typeid(num) is evaluated at compile-time, producing a std::type_info that describes int itself. An int will always be an int , regardless of what is assigned to num at runtime. Besides, in your example, typeid(num).name() != "int" will always evaluate as true, since you are comparing a const char* pointer from name() to a string literal that is clearly stored elsewhere in memory, so the two memory addresses don't evaluate as equal. And since you are doing that comparison inside a while(1) loop, that is why "please input integer" is printed out endlessly. You don't have any viable error handling in your loop.

You have to look at cin 's actual state to know whether >> fails, eg:

#include <iostream>
#include <string>
#include <limits>
using namespace std;

int main() {
    while (1) {
        int num;
        // cin >> num;
        // if (cin.fail()) {
        if (!(cin >> num)) {
            cout << "please input integer" << endl;
            cin.clear();
            cin.ignore(numeric_limit<streamsize>::max(), '\n');
            continue;
        }
        // use num as needed...
    }
}