cin.get（）循环执行 [英] cin.get() in a loop

问题描述

我试图从标准输入中读取内容。第一行是我将要阅读的行数。我接下来阅读的行将再次打印。以下是代码：

I was trying to read from standard input. The first line is the number of lines that I will read. The lines that I read next will be printed again. Here is the code:

#include <iostream>

using namespace std;

int main()
{
    int n;
    cin >> n;
    for (unsigned int i = 0; i < n; ++i)
    {
        char a[10];
        cin.get (a, 10);
        cout << "String: " << a << endl;
    }
    return 0;
}

当我运行它并给出行数时，程序退出。我还没弄清楚发生了什么事，所以我决定在这里提问。

When I run it and give number of lines, the program exits. I haven't figured out what's going on, so I've decided to ask it here.

预先感谢。

推荐答案

混合格式化和未格式化的输入充满了问题。在您的特定情况下，此行：

Mixing formatted and unformatted input is fraught with problems. In your particular case, this line:

std::cin >> n;

使用您输入的数字，但保留'\n'在输入流中。

consumes the number you typed, but leaves the '\n' in the input stream.

随后，此行：

cin.get (a, 10);

不使用任何数据（因为输入流仍指向'\ n'）。出于相同的原因，下一次调用也不会消耗任何数据，依此类推。

consumes no data (because the input stream is still pointing at '\n'). The next invocation also consumes no data for the same reasons, and so on.

然后问题变成：我如何消耗'＼ ＼n'？有两种方法：

The question then becomes, "How do I consume the '\n'?" There are a couple of ways:

您可以读取一个字符并将其丢弃：

You can read one character and throw it away:

cin.get();

您可以读整行，而不论长度：

You could read one whole line, regardless of length:

std::getline(std::cin, some_string_variable);

您可以忽略当前行的其余部分：

You could ignore the rest of the current line:

std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');

作为一些相关建议，我永远不会使用 std :: istream :: get（char *，streamsize）。我总是喜欢： std :: getline（std :: istream& ;, std :: string&）。

As a bit of related advice, I'd never use std::istream::get(char*, streamsize). I would always prefer: std::getline(std::istream&, std::string&).

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