如何使cout表现为二进制模式？ [英] How to make cout behave as in binary mode?

问题描述

每次我做'cout<< endl'或甚至'cout' \ n'然后启动我的程序在Windows下输出到一个文件（a.exe< test.in> result.out）我得到\r\\\
行结尾在结果。出。

Every time I do 'cout << endl' or even 'cout << "\n"' and then launch my program under Windows to output to a file ("a.exe < test.in > result.out") I get "\r\n" line endings in "result.out".

在地球上有一种方法来阻止它这样做，只是输出\\\
在每个cout< \ n'？

Is there on earth a way to stop it doing so and just output "\n" on every 'cout << "\n"'?

提前感谢。

推荐答案

我不相信你可以强制 std :: cout 输出二进制数据。用户通过C ++可以使用的最低级别接口是 streambuf 在 std :: cout.rdbuf（）。你可以很好地假设对象是 basic_filebuf 类型，并且C ++ 11标准的§27.9.1.4部分要求 basic_filebuf 实现单独的文本和二进制模式。

I don't believe you can force std::cout to output binary data. The lowest-level interface available to the user through C++ is the streambuf at std::cout.rdbuf(). You can pretty well assume that object is of type basic_filebuf, and section §27.9.1.4 of the C++11 standard requires that basic_filebuf implement separate text and binary modes.

检查您的平台特定扩展，以打开另一个流 stdout 使用其文件描述符。然后你可以在模式标志中指定 std :: binary 。

Check your platform-specific extensions for a way to open another stream to stdout using its file descriptor. Then you can specify std::binary in the mode flags.

但是，通过第二个转换行尾的程序。

However, it's probably easier to pipe the output through a second program that converts the line endings.

编辑：现在我看到你可能会愿意进一步真正做程序直接输出你想要的。

Now I see that you might be willing to go a little further to truly make the program directly output what you want.

只要C ++接口设置为非缓冲模式，就可以使用C接口输出换行符。

You can use the C interface to output newlines as long as the C++ interface is set to unbuffered mode.

在您的程序开始时， cout 接收任何输出之前，请执行以下操作：

At the very beginning of your program, before cout receives any output, do this:

std::cout.rdbuf()->pubsetbuf( 0, 0 ); // go to unbuffered mode

当要输出换行符时， p>

When you want to output a newline, do it like so:

_write( STDOUT_FILENO, 1, "\n" ); // do not convert line ending

你可以包装这个小... nugget ...自定义操纵： / p>

You can wrap this little… nugget… inside a custom manipulator:

#include <unistd.h>

std::ostream &endln( std::ostream &s ) {
    if ( s->rdbuf() == std::cout.rdbuf() ) { // this is cout or cerr into cout
        _write( STDOUT_FILENO, 1, "\n" ); // do not convert line ending
        return s; // stream is unbuffered, no need to flush
    } else {
        return std::endl( s ); // other streams do convert to \r\n
    }
}

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