如何从boost的gzip_decompressor()获取gzip_params [英] How to get gzip_params from boost's gzip_decompressor()

问题描述

我正在通过以下链接使用boost gzip_decompressor(): 我如何阅读使用Boost IOStreams的界面逐行处理Gzip文件?

I am using the boost gzip_decompressor() from the following link: How can I read line-by-line using Boost IOStreams' interface for Gzip files?

读取gzip文件可以正常工作，但是如何读取gzip_params?我想知道存储在gzip_params.file_name中的原始文件名.

Reading the gzip file works fine, but how do I read the gzip_params? I want to know the original file name that's stored in the gzip_params.file_name.

推荐答案

一个很好的问题.

解决方案是使用component<N, T>获取指向实际解压缩器实例的指针:

The solution is to use component<N, T> to get a pointer to the actual decompressor instance:

在Coliru上直播

#include <iostream>
#include <fstream>
#include <boost/iostreams/filtering_stream.hpp>
#include <boost/iostreams/filter/gzip.hpp>
int main()
{
    std::ifstream file("file.gz", std::ios_base::in | std::ios_base::binary);
    try {
        boost::iostreams::filtering_istream in;
        using gz_t = boost::iostreams::gzip_decompressor;
        in.push(gz_t());
        in.push(file);

        for(std::string str; std::getline(in, str); )
        {
            std::cout << "Processed line " << str << '\n';
        }

        if (gz_t* gz = in.component<0, gz_t>()) {
            std::cout << "Original filename: " << gz->file_name() << "\n";
            std::cout << "Original mtime: " << gz->mtime() << "\n";
            std::cout << "Zip comment: " << gz->comment() << "\n";
        }
    }
    catch(const boost::iostreams::gzip_error& e) {
         std::cout << e.what() << '\n';
    }
}

使用准备样品文件

gzip testj.txt
mv testj.txt.gz file.gz

打印

Processed line Hello world
Original filename: testj.txt
Original mtime: 1518987084
Zip comment: 

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