出于无法理解的原因，我的代码编写了两次 [英] My code is written twice for uncomprehensible reasons

问题描述

这段代码是用C ++编写的，由于我不太理解的原因，它被编写了两次.我希望输入一个随机字符后，它将显示一次字符，并且字符串也将显示在其下.但是我没有得到这个作为输出.我想念什么?

This code is written in C++ and for reasons that I don't quite understand it is written twice. I would expect that after inputting a random char it would display the char once and the String lower to it once as well. But I don't get this as output. What am I missing?

解决方案:添加cin.ignore()语句也会忽略读入的返回值.使我的代码一次遍历循环.

Solution: Adding a cin.ignore() statement disregards the return that is read in as well. Making my code go through the loop once.

#include <iostream>

using std::cin;
using std::cout;
using std::endl;

int main()
{
    char letter;

    letter = cin.get();
    while (letter!= 'X')
    {
        cout << letter << endl;
        cout << "this will be written twice for ununderstandable reasons";                
        letter = cin.get();
    }
}

示例:如果我要使用cmd scrn c 编写代码，则会得到一个 c 返回+两次短语，出于不可理解的原因，该短语将被编写两次.所以我认为是输出

Example: If I were to write in cmd scrn c, I'd get a c back + twice the phrase this will be written twice for ununderstandable reasons. So what I thought to be the output

c
this will be written twice for ununderstandable reasons

实际上是

c
this will be written twice for ununderstandable reasons
this will be written twice for ununderstandable reasons

正如大家已经提到的，

推荐答案

，每次您按Enter键， cin 都会添加换行标记 \ n .另一种解决方案是将 cin.ignore(); 放在每个 cin.get(); 之后.

as everyone already mentioned, cin will append the newline marker \n every time you hit enter. another solution is to place cin.ignore(); after every cin.get();.

#include <iostream>

using std::cin;
using std::cout;
using std::endl;

int main()
{
    char letter;

    letter = cin.get();
    cin.ignore();
    while (letter!= 'X')
    {
          cout<<letter<<endl;
          cout<<"this will be written twice for ununderstandable reasons";
          letter= cin.get();
          cin.ignore();
          }
}

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