如何避免char输入int变量？ [英] How Can I avoid char input for an int variable?

问题描述

下面的程序显示了一个int值被输入并被同时输出。但是，当我输入一个字符时，它进入一个无限循环，显示输入的前一个'int'值。如何避免输入字符？

The program below shows a 'int' value being entered and being output at the same time. However, when I entered a character, it goes into an infinite loop displaying the previous 'int' value entered. How can I avoid a character being entered?

#include<iostream>
using namespace std;

int main(){
int n;

while(n!=0){
            cin>>n;
            cout<<n<<endl;
           }
return 0;
}

推荐答案

循环：

cin进入失败状态，使其忽略对其的进一步调用，直到错误标志和缓冲区重置。

cin goes into a failed state and that makes it ignore further calls to it, till the error flag and buffer are reset.

cin.clear();
cin.ignore(100, '\n'); //100 --> asks cin to discard 100 characters from the input stream.

检查输入是否为数字：

在你的代码中，即使一个非int类型也会被转换为int。没有办法检查输入是否为数字，不输入到一个字符数组，并调用 isdigit（）函数每个数字。

In your code, even a non-int type gets cast to int anyway. There is no way to check if input is numeric, without taking input into a char array, and calling the isdigit() function on each digit.

函数 isdigit（）可用于告诉数字和字母分开。此函数位于< cctype> 头中。

The function isdigit() can be used to tell digits and alphabets apart. This function is present in the <cctype> header.

一个is_int（）函数看起来像这样。

An is_int() function would look like this.

for(int i=0; char[i]!='\0';i++){
    if(!isdigit(str[i]))
    return false;
}
return true;

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