Using boost::iostreams::tee_device?(使用 boost::iostreams::tee_device?)
问题描述
有人可以帮我吗?
我正在尝试执行以下操作:
I am trying to do something like the following:
#include <boost/iostreams/tee.hpp>
#include <boost/iostreams/stream.hpp>
#include <sstream>
#include <cassert>
namespace io = boost::iostreams;
typedef io::stream<io::tee_device<std::stringstream, std::stringstream> > Tee;
std::stringstream ss1, ss2;
Tee my_split(ss1, ss2); // redirects to both streams
my_split << "Testing";
assert(ss1.str() == "Testing" && ss1.str() == ss2.str());
但它不会在 VC9 中编译:
But it won't compile in VC9:
c:liboost_current_versionoostiostreamsstream.hpp(131) : error C2665: 'boost::iostreams::tee_device<Sink1,Sink2>::tee_device' : none of the 2 overloads could convert all the argument types
有人用过这个吗?我知道我可以让自己的班级来做这件事,但我想知道我做错了什么.
Has anyone gotten this to work? I know I could make my own class to do it, but I want to know what I am doing wrong.
谢谢
推荐答案
您使用 constructor-forwarding version 的 io::stream
,它自己构造一个 tee-stream 并将所有参数转发给它.C++03 在将参数转发给函数时只有有限的能力(需要的重载量很容易呈指数增长).它 (io::stream
) 做了以下限制:
You use the constructor-forwarding version of io::stream
, which construct a tee-stream itself and forward all arguments to that. C++03 has only limited capabilities when it comes to forwarding arguments to functions (amount of overloads needed easily grow exponentially). It (io::stream
) makes the following restrictions:
这些成员中的每一个都构造了一个流的实例,并将其与从给定的参数列表构造的 Device T 的一个实例相关联.所涉及的 T 构造函数必须按值或常量引用获取所有参数.
Each of these members constructs an instance of stream and associates it with an instance of the Device T constructed from the given lists of arguments. The T constructors involved must take all arguments by value or const reference.
好吧,但是 tee_device
构造函数说
Well, but the tee_device
constructor says
基于给定的一对接收器构造 tee_device 的实例.如果相应的模板参数是流或流缓冲区类型,则每个函数参数都是非常量引用,否则为常量引用.
Constructs an instance of tee_device based on the given pair of Sinks. Each function parameter is a non-const reference if the corresponding template argument is a stream or stream buffer type, and a const reference otherwise.
这当然行不通.io::stream
提供了另一个以 T
作为第一个参数的构造函数.这在这里有效(至少编译.不过,断言失败.我没有使用过 boost::iostreams
,所以我无能为力)
That won't work, of course. io::stream
provides another constructor that takes a T
as first argument. This works here (Compiles, at least. The assertion fails, though. I've not worked with boost::iostreams
so i can't help with that)
namespace io = boost::iostreams;
typedef io::tee_device<std::stringstream, std::stringstream> TeeDevice;
typedef io::stream< TeeDevice > TeeStream;
std::stringstream ss1, ss2;
TeeDevice my_tee(ss1, ss2);
TeeStream my_split(my_tee);
my_split << "Testing";
assert(ss1.str() == "Testing" && ss1.str() == ss2.str());
在调用 flush()
或流式传输 <<<;std::flush
,断言通过.
After calling flush()
or streaming << std::flush
, the assertion passes.
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本文标题为:使用 boost::iostreams::tee_device?
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