C++ 函数中的无限循环

Question

我写了一个小程序，把两个整数相加：

#include <iostream>
using namespace std;

int inpInt() {            //ask for an input in integer format
    int number = 0;
    cin >> number;       
    while (!cin) {        //keep asking if the input was not an integer
        cin >> number;
    }
    return number;
}

int main() {
    int summand1, summand2, sum;
    cout << "Summand 1: "; summand1 = inpInt();
    cout << "Summand 2: "; summand2 = inpInt();
    sum = summand1 + summand2;
    cout << "Sum: " << sum << "\n";
}

问题是，当我不输入整数时，我陷入了一个无限循环，我既不能输入整数也不能输入其他任何东西。奇怪的是，当我在 main() 中包含整数测试循环时，代码可以工作。

非常感谢您的帮助！

Answer 1

您必须明确清除 cin 并忽略换行符（按回车时会出现换行符）

int inpInt() {            //ask for an input in integer format
    int number = 0;
    while (!(cin >> number)) {
        cin.clear(); // clear cin
        cin.ignore(numeric_limits<streamsize>::max(), '\n'); // ignore new lines. see: http://www.cplusplus.com/reference/istream/istream/ignore/
    }
    return number;
}

int main() {
    int summand1, summand2, sum;
    cout << "Summand 1: "; summand1 = inpInt();
    cout << "Summand 2: "; summand2 = inpInt();
    sum = summand1 + summand2;
    cout << "Sum: " << sum << "\n";
}

Answer 2

好的。非常感谢您的建议。我的问题是我对 istreams 的了解还不够。清除 cin 是一种解决方案。

另一个解决方案，正如 Myname 所建议的，是使用 isdigit。程序现在从 cin 中读取一个字符串，用 isdigit 对其进行测试，然后用 atoi 将其转换为 int。万一另一个新手阅读了这个讨论，这是有效的代码：

#include <iostream>
using namespace std;

bool checkInt(string str) {
   for (int i = 0; i < str.length(); i++) {
       if (isdigit(str[i]) == false) {
       return false;
    }
   }
   return true;
}

int getInteger () {
    string inpStr;
    bool checkIntFailed = true;
    while (checkIntFailed) {
        cin >> inpStr;
        if (checkInt(inpStr))
            checkIntFailed = false;
        else cout << "Error: no integer. Try again: ";

    }
    return atoi(inpStr.c_str());
}

int main () {
    int iSummand1, iSummand2, iSum;
    cout << "1st summand: "; iSummand1 = getInteger();
    cout << "2nd summand: "; iSummand2 = getInteger();
    iSum = iSummand1 + iSummand2,
    cout << "Sum: " << iSum << "\n";
}

周末愉快！

Answer 3

可以试试这个将数字转换为 char 并检查它的 ascii 代码，因为数字从 48 开始在 ascii 中。

#include <iostream>
using namespace std;

int inpInt() {            //ask for an input in integer format
    char number = 48;
    cin >> number;
    
    int c = char(number);
    while (c < 48 || c > 58) {       
        cin >> number;
        c = char(number);
    }
    return number;
}

int main() {
    int summand1, summand2, sum;
    cout << "Summand 1: "; summand1 = inpInt();
    cout << "Summand 2: "; summand2 = inpInt();
    sum = summand1 + summand2;
    cout << "Sum: " << sum << "\n";
}

Answer 4

尝试使用函数 isdigit()。

#include <iostream>
using namespace std;

int inpInt() {            //ask for an input in integer format
    int number = 0;
    cin >> number;       
    while (isdigit(number))
     {        //keep asking if the input was not an integer
        cin >> number;
    }
    return number;
}

int main() {
    int summand1, summand2, sum;
    cout << "Summand 1: "; summand1 = inpInt();
    cout << "Summand 2: "; summand2 = inpInt();
    sum = summand1 + summand2;
    cout << "Sum: " << sum << "\n";
}