cin>>不使用getline（） [英] cin>> not work with getline()

问题描述

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

int main () {
  string str;
  int age;
  cout << "Please enter age: ";
  cin>>age;
  cout << "Please enter full name: ";
  getline (cin,str);
  cout << "Thank you, " << str << ".\n";
}

为什么当我使用uperator >>输入整数时，getline ？对于int输入，更好的用法是什么？

Why function getline() not work when I using uperator >> to input integer ? What is better use for int input ?

推荐答案

cin& > age; ，它给你一个空字符串的名称。

You still have a newline in the stream after cin>>age;, which is giving you an empty string for the name.

你可以通过添加另一个 getline（）在获取年龄并丢弃结果后调用。另一个选项是调用 cin.ignore（BIG_NUMBER，'\\\
'）; ，其中 BIG_NUMBER 为MAX_INT

You could solve it by just adding another getline() call after getting the age and throwing away the result. Another options is to call cin.ignore(BIG_NUMBER, '\n');, where BIG_NUMBER is MAX_INT or something.

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