trying to write std:out and file at the same time

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清歌不尽
清歌不尽 2021-01-06 06:52

I am trying to write to file and stdout at the same time within c++ by overloading ofstream

test.h

 #pragma once 

#include 

using           


        
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  • I'm not going to comment on why your approach doesn't work, mainly because it cannot be patched up to work properly. The main problem is that you can't use your stream an pass it to something which expected an std::ostream& and still write to both streams. However, there is a relatively simple although not necessarily obvious approach to implement what you actually want: You'd derive a new stream buffer, i.e., a class derived from std::streambuf, and override its overflow() and sync() functions. Here is the complete code for a simple demo:

    #include <streambuf>
    
    struct teebuf
        : std::streambuf
    {
        std::streambuf* sb1_;
        std::streambuf* sb2_;
    
        teebuf(std::streambuf* sb1, std::streambuf* sb2)
            : sb1_(sb1), sb2_(sb2) {
        }
        int overflow(int c) {
            typedef std::streambuf::traits_type traits;
            bool rc(true);
            if (!traits::eq_int_type(traits::eof(), c)) {
                traits::eq_int_type(this->sb1_->sputc(c), traits::eof())
                    && (rc = false);
                traits::eq_int_type(this->sb2_->sputc(c), traits::eof())
                    && (rc = false);
            }
            return rc? traits::not_eof(c): traits::eof();
        }
        int sync() {
            bool rc(true);
            this->sb1_->pubsync() != -1 || (rc = false);
            this->sb2_->pubsync() != -1 || (rc = false);
            return rc? 0: -1;
        }
    };
    
    #include <fstream>
    #include <iostream>
    
    int main()
    {
        std::ofstream fout("tee.txt");
        teebuf        sbuf(fout.rdbuf(), std::cout.rdbuf());
        std::ostream  out(&sbuf);
        out << "hello, world!\n";
    }
    

    Obviously, the creation of tee-stream could be packaged up nicely but how this looks exactly doesn't really matter. The important thing is that it is possible to create a custom destination (or source) for IOStreams and that it does not involve any attempt to inherit from std::ostream. The only reason to inherit from std::ostream (or std::istream) is to make the initialization of a stream with a custom stream buffer easier.

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  • 2021-01-06 07:54

    The problem

    ofstream::operator << (var);
    

    It's your use of ofstream::operator<< as a qualified function call. You're mandating that the function lookup locate a member function of ofstream; the best match that's a member is the one for void*, whereas the specialisation for char* that prints the actual string contents is a free function (i.e. not a member function).

    You'll find the same problem if you do this with cout, too:

    std::cout.operator<<(var);
    

    The solution

    This might do it:

    static_cast<ofstream&>(*this) << var;
    

    because you're still using normal operator syntax (with all the overload resolution that this entails), but doing so with an ofstream as the LHS operand.

    I haven't actually tested it, though.

    Conclusion

    As an aside, your operator<< ought to be a free function too, in order to fit in with this convention.

    So:

    struct OutputAndConsole : std::ofstream
    {
        OutputAndConsole(const std::string& fileName)
           : std::ofstream(fileName)
           , fileName(fileName)
        {};
    
        const std::string fileName;
    };
    
    template <typename T>
    OutputAndConsole& operator<<(OutputAndConsole& strm, const T& var)
    {
        std::cout << var;
        static_cast<std::ofstream&>(strm) << var;
        return strm;
    };
    

    I also took the liberty of making some minor syntax adjustments.

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