如何在win32上刷新stdlib输出文件？ [英] How do I flush a stdlib output file on win32?

问题描述

我有一个C ++程序，正在写入一个文件在Windows 7上。当我调用 f.flush（） NTFS文件不会变大。有没有办法强制文件被刷新？

I have a C++ program that is writing to a file on Windows 7. When I call f.flush() the NTFS file does not get bigger. Is there any way to force the file to get flushed?

推荐答案

如何获取文件HANDLE fopen FILE结构？

代码如下：

You can look here:
How do I get the file HANDLE from the fopen FILE structure?
the code looks like this:

 FlushFileBuffers((HANDLE) _fileno(_file));

不要在调用FlusFileBuffers之前调用fflush（文件）。

do not forget call fflush(file), before call FlusFileBuffers.

对于std :: fstream和gnu stl，我在我的linux盒子上检查过，在家没有窗口，
但应该与mingw一起工作，可能需要一些修改：

For std::fstream and gnu stl, I checked it on my linux box, have no windows at home, but should work with mingw, may be need some modifications:

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <fstream>
#include <ext/stdio_filebuf.h>


int main(int argc, char *argv[])
{
    assert(argc == 2);
    std::ifstream f(argv[1]);
    assert(!!f);
/*
    //cin, cout etc
    __gnu_cxx::stdio_filebuf<char> *stdio_buf = dynamic_cast<__gnu_cxx::stdio_filebuf<char> *>(f.rdbuf());
    if (stdio_buf != 0) {
        printf("your fd is %d\n", stdio_buf->fd());
        return EXIT_SUCCESS;
    }
*/
    std::basic_filebuf<char> *file_buf = dynamic_cast<std::basic_filebuf<char> *>(f.rdbuf());
    if (file_buf != 0) {
        struct to_get_protected_member : public std::basic_filebuf<char> {
            int fd() { return _M_file.fd(); }
        };
        printf("your fd is %d\n", static_cast<to_get_protected_member *>(file_buf)->fd());
    }
    printf("what is going on?\n");
    return EXIT_FAILURE;
}

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