提高property_tree :: json_parser :: read_json＆安培;输入输出流:: filtering_streambuf [英] boost property_tree::json_parser::read_json & iostreams::filtering_streambuf

问题描述

我试着去阅读瘪JSON和体验型转换的问题，这里是code

Im trying to read deflated json and experiencing type conversion problems, here is the code

boost::iostreams::filtering_streambuf<boost::iostreams::input> in;
std::istringstream iss(std::ios::binary);
iss.rdbuf()->pubsetbuf(buf, len);
iss.imbue( std::locale("ru_RU.CP1251") );
in.push( boost::iostreams::zlib_decompressor() );
in.push( iss );

boost::property_tree::ptree pt;
boost::property_tree::json_parser::read_json(in, pt); // <-- Compile error

编译器说：

的src / ABPacking.cpp：48：错误：调用没有匹配的功能
  read_json（升压::输入输出流:: filtering_streambuf，性病::分配器，
  提高::输入输出流::公共_>＆放;,提振:: property_tree :: ptree中与放大器;）

src/ABPacking.cpp:48: error: no matching function for call to ‘read_json(boost::iostreams::filtering_streambuf, std::allocator, boost::iostreams::public_>&, boost::property_tree::ptree&)’

现在的问题是如何通过的 filtering_streambuf 为 read_json 避免不必要的数据拷贝？

The question is how to pass filtering_streambuf to read_json without unnecessary data copying?

推荐答案

read_json 期望无论是文件名或流的使用JSON内容。你试图通过流的缓存的，它会不知道该怎么办。

read_json expects either a file name or a stream with the JSON content. You're trying to pass a stream buffer, and it won't know what to do with it.

作为一个解决方案，只是通过流缓冲区的的IStream 消耗，并传递给 read_json ：

As a solution, just pass the stream buffer to an istream that consumes it and pass that to read_json:

std::istream input(&in_buf);
read_json(input, pt);

这篇关于提高property_tree :: json_parser :: read_json＆安培;输入输出流:: filtering_streambuf的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！