cin>> val有时会读取0，具体取决于Ctrl-Z [英] cin >> val sometimes reads 0 depending on Ctrl-Z

问题描述

我试图使用MinGW编译器在Windows中用C ++编写代码，我的代码计算并打印给定输入集中出现次数的连续次数.代码如下:

I was trying to write a code in C++ in Windows using MinGW compiler, my code counts and prints the number of consecutive times a number occurs in a given set of input. The code is as follows:

#include <iostream>
int main()
{
    int c_val = 0,val = 0,cnt = 1;
    std::cin>>c_val;
    while(std::cin>>val){
        if(val==c_val)
            cnt++;
        else{
            std::cout<<c_val<<" occurs "<<cnt<< " times"<<std::endl;
            c_val = val;
            cnt = 1;
        }
    }
    std::cout<<val<<" occurs "<<cnt<<" times";
}

输入:42 42 42 12 13 13 ^Z(按Enter)

INPUT: 42 42 42 12 13 13 ^Z (press Enter)

输出:

42 occurs 3 times
12 occurs 1 times
0 occurs 2 times

但是如果我在^Z之前按Enter，则它看起来像:

But if I press Enter before ^Z then it looks like:

输入:42 42 42 12 13 13(按Enter)^Z(按Enter)

INPUT: 42 42 42 12 13 13 (press Enter) ^Z (press Enter)

输出:

42 occurs 3 times
12 occurs 1 times
13 occurs 2 times

我想知道为什么我的代码中的变量val在按下回车键后使用^Z键时会存储13的原因，以及如果我同时输入^Z键为什么会存储0的原因用我的输入.

I want to know why the variable val in my code stores 13, when I use ^Z key after pressing my return key and why does it store 0 if I give the ^Z key along with my input.

推荐答案

这是发生的情况.我使用MinGW-w64 4.9.2进行了观察.无论是在Windows控制台中还是在Cygwin(而不是使用cygwin-mingw)下运行可执行文件，其行为都是相同的.

Here is what happens. I observed this using MinGW-w64 4.9.2. The behaviour was the same whether running the executable in a Windows console, or under Cygwin (but not using cygwin-mingw).

• 在行首按^Z设置文件结束条件
• 在其他任何地方按^Z键实际上将ASCII 26字符发送到流中

• Pressing ^Z at the beginning of a line sets the end-of-file condition
• Pressing ^Z anywhere else actually sends ASCII 26 character to the stream

我还观察到:

如果
• cin >> val由于输入不包含数字而失败，则将val设置为0.
如果由于文件结尾而导致输入失败，则
• cin >> val保留val不变.

• cin >> val sets val to 0 if it fails due to the input not containing a number.
• cin >> val leaves val unchanged if input fails due to end-of-file.

根据此线程，它是由以下行为指定的正确行为C ++ 11.

According to this thread that is the correct behaviour specified by C++11.

因此您的结果可以得到解释.输入42 42 42 12 13 13^Z时，与编写42 42 42 12 13 13x相同.读取前六个数字，然后对x进行加密时，cin >> val失败，并将val设置为0.

So your results can be explained. When you input 42 42 42 12 13 13^Z, it is the same as if you had written 42 42 42 12 13 13x. The first six numbers are read, and then when x is encoutered, cin >> val fails and sets val to 0.

但是，当您按Enter键然后按^Z键时，就好像您正在从文件中读取文件一样，并且已到达文件末尾. cin >> val保留val不变，并且仍保留上次成功执行cin >> val之后的值.

But when you press Enter and then ^Z, it is as if you were reading from a file and you reached the end of the file. cin >> val leaves val unchanged and it is still holding the value that it had after the last successful cin >> val.

如果您按照Gautam Jha的建议进行更改，则在两种情况下都将得到13.这是因为他有效地读取了临时int，然后仅在读取成功的情况下才将临时int存储到实数val中，从而避免了失败的读取将val设置为0的情况.

If you make the change suggested by Gautam Jha suggested then you will get 13 in both cases. This is because he effectively reads into a temporary int, and then only stores the temporary int into the real val if the read succeeded, thereby avoiding the behaviour where a failed read sets val to 0.

这可能是所需的行为，尽管您可能还想检查cnt > 0以避免在输入完全为空的情况下出现奇怪的输出.

This is probably the desired behaviour although you might also want to check cnt > 0 to avoid a weird output in the case of totally empty input.

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