为什么在输入为int的情况下cin期望为int时将相应的int变量更改为零？ [英] Why does cin, expecting an int, change the corresponding int variable to zero in case of invalid input?

问题描述

我对C ++完全陌生，正在尝试编写一个非常基本的程序，但是在初始化整数时遇到问题。我将其简化为一个仍然有问题的小型程序：

I am completely new to C++ and am trying to write an extremely basic program, but I am having issues with initializing an integer. I have stripped it down to a very small program that still has the issue:

#include <iostream>
using namespace std;

int main()
{
    cout << "Please enter your age\n";
    int age = -1;
    cin >> age;
    cout <<"\n\n Your age is " << age << "\n\n";
}

我读到如果我尝试输入字符串，例如 abc 到 age 变量，则输入应该失败并且该值应保留为空，因此应打印您的年龄是-1 。

I read that if I attempt to input a string, e.g. abc to the age variable, then the input should fail and the value should be left alone and therefore it should print Your age is -1.

但是，当我运行此程序并键入 abc ，然后显示您的年龄是0 。为什么？

However, when I run this program and type abc, then it prints Your age is 0. Why?

推荐答案

您要观察的行为在2011年发生了变化。

The behavior you want to observe changed in 2011. Until then:

如果提取失败（例如，如果在预期数字的位置输入了字母），则值将保持不变，并设置故障位。

If extraction fails (e.g. if a letter was entered where a digit is expected), value is left unmodified and failbit is set.

但是从C ++ 11开始：

But since C++11:

如果提取失败，则将值写入零，并且故障位已设置。 [...]

If extraction fails, zero is written to value and failbit is set. [...]

（来自 cppr 。）

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