C ++ Primer第五版书（if语句）是不是正确？ [英] C++ Primer fifth edtion book (if statement) is this not correct?

问题描述

#include <iostream>

int main()
{
    int currVal = 0, val = 0;

    if (std::cin >> currVal) {
        int cnt = 1; 

        while (std::cin >> val) { 
            if (val == currVal)
                ++cnt; 
            else 
            {
                std::cout << currVal << " occurs " << cnt << " times" << std::endl;
                currVal = val; 
                cnt = 1; 
            }
        } 

        std::cout << currVal << " occurs " << cnt << " times" << std::endl;
    } 

    return 0;
}

如果我们给这个程序以下输入：

If we give this program the following input:

42 42 42 42 42 55 55 62 100 100 100

42 42 42 42 42 55 55 62 100 100 100

输出应为

42发生5次

42 occurs 5 times

55发生2次

62发生1次

100发生3次

所有这些代码都应该正常工作。对于由此带来的不便，（有一个大脑放屁）

All this code should work properly. Sorry for the inconvenience.(had a brain fart)

推荐答案

This 看起来按预期工作。

有一个很好的理由做 std :: cin>>在如果语句或 while 循环的条件中的任何提取运算符>> 返回您正在操作的流。因此， std :: cin>>>

There is a very good reason to do std::cin >> whatever in the condition of an if statement or while loop. The extraction operator >> returns the stream that you're operating on. So the value of the std::cin >> whatever expression is std::cin itself.

一个流，例如 c code> std :: cin ，可以转换为 bool ，以检查流是否处于有效状态。请考虑：

A stream, such as std::cin, can be converted to bool for the sake of checking whether the stream is in a valid state. Consider:

if (std::cin >> whatever)

这可以理解为从用户输入变量 whatever ，然后检查输入是否有效。

This can be understood as taking input from the user into the variable whatever and then checking if the input was valid or not. If not, the if block will not execute.

同样地，

while (std::cin >> whatever)

这将循环，从用户每次的输入，直到用户输入无效的东西。

This will loop, taking input from the user each time, until the user enters something invalid.

什么是无效的输入？如果在上面的例子中无论是 int ，并且用户输入文本 blah ，这将失败，因为它不是有效的整数。

What is an invalid input? Well if whatever in the above examples was an int, and the user entered the text blah, this would fail because it's not a valid integer.

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