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我正在构建一个客户端/服务器应用程序,用于通过局域网发送文件。这是服务器应用程序,当我要获取文件名时,我的代码出现以下错误。

error C2679: binary '=' : no operator found which takes a right-hand operand of type 'boost::filesystem::path' (or there is no acceptable conversion)

#include "stdafx.h"
#ifdef _WIN32 
# define _WIN32_WINNT 0x0501 
#endif 
#include <boost/asio.hpp> 
#include <boost/array.hpp> 
#include <boost/filesystem.hpp> 
#include <boost/filesystem/path.hpp>
#include <string> 
#include <fstream> 
#include <sstream> 
#include <iostream> 

std::string filename; 
std::string file; 
boost::asio::io_service io_service; 
boost::asio::ip::tcp::endpoint endpoint(boost::asio::ip::tcp::v4(), 9999); 
boost::asio::ip::tcp::acceptor acceptor(io_service, endpoint); 
boost::asio::ip::tcp::socket sock(io_service); 
boost::array<char, 4096> buffer; 

void read_handler(const boost::system::error_code &ec, std::size_t bytes_transferred) { 
    if (!ec || ec == boost::asio::error::eof){ 
        std::string data(buffer.data(), bytes_transferred); 
        if (filename.empty()) { 
            std::istringstream iss(data); 
            std::getline(iss, filename); 
            file = data; 
            file.erase(0, filename.size() + 1); 
            filename = boost::filesystem::path(filename).filename(); 
        } 
        else  
            file += data; 
        if (!ec) 
            boost::asio::async_read(sock, boost::asio::buffer(buffer), read_handler); 
        else { 
         //code
    } 
} 

//代码

4

1 回答 1

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只需更改此行:

filename = boost::filesystem::path(filename).filename(); 

对此:

filename = boost::filesystem::path(filename).filename().string();

基本上,编译器告诉您,std::string它没有定义任何将 aboost::filesystem::path作为参数的赋值运算符(或者它无法进行任何转换来提供可用作赋值运算符参数的类型)。幸运的是,boost::filesystem::path提供了一个返回字符串的函数!

于 2013-07-14T06:16:33.390 回答