缓冲区如何工作？ [英] How buffers are working ?

问题描述

我打开一个包含两行的文件（sample.txt），现在如果我在while循环中声明字符串，我在屏幕上得到两行但是如果我在循环中声明字符串，我在控制台上获得第二行两次。任何人都可以解释发生了什么。缓冲区在其中扮演什么角色？

I am opening a file(sample.txt) containing two lines, now if I declare string in while loop, i get two lines on screen but if I declare string outside while loop i am getting second line twice on console. Can any one explain whats happening . Is buffers playing any role in it?

int main()
{
    ifstream itf("Sample.txt");
    if(!itf)
    {
        cout << "Not able to open file :("<<endl;
        exit(1);
    }

    string str;
    while(itf)
    {

        getline(itf, str);
        cout <<str;
    }

    return 0;
}

推荐答案

我认为缓冲区运行正常。

Buffers are working fine, I suppose.

引用：

但如果我声明字符串外部while循环我在控制台上获得第二行两次

but if I declare string outside while loop i am getting second line twice on console

这不应该发生，事实上它不会发生在我运行您发布的代码的系统上。

This shouldn't happen, and in fact it doesn't happen on my system running the code you posted.

更改 cout 运算符以在最后添加换行符，因此：

Change your cout operator to put a newline at the end, thus:

cout <<str << endl;

这应该是你的输出清除。

which should make your output clearer.

您应该在 getline 之后和<之前检查 itf code> COUT 。当它读取第二行时，应该没有错误，因此在循环顶部检查的条件（itf）仍应返回 true 。循环然后第三次运行，但由于没有其他内容可读， getline 什么都不做，显然也没有重置 str 。以下cout然后再次打印未更改的 str ，虽然有错误！



作为一般规则，在使用结果之前，请务必检查读取操作是否成功！

You should check itf after getline and before the cout. When it reads the second line, there should be no error, and therefore the condition (itf) checked at the top of your loop should still return true. The loop then runs a third time, but since there is nothing else to read, getline does nothing and it apparently also does not reset str. The following cout then simply prints the unchanged str again, although there was an error!

As a general rule, always check the success of read operations before using the result!

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