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Home/ Questions/Q 8924947
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T07:33:49+00:00 2026-06-15T07:33:49+00:00

I am trying to write to file and stdout at the same time within

  • 0

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

test.h

 #pragma once 

#include <iostream>

using  std::ofstream;

class OutputAndConsole:public ofstream
{
public:
    std::string fileName;        
    OutputAndConsole(const std::string& fileName):ofstream(fileName),fileName(fileName){
    };
    template <typename T>
    OutputAndConsole& operator<<(T var);
};


template <typename T>
OutputAndConsole& OutputAndConsole::operator<<(T var)
{
    std::cout << var;
    ofstream::operator << (var);
    return (*this);
};

test.cpp

  OutputAndConsole file("output.txt");
  file << "test" ;

The output in the file is

01400930

but in the console is

test

I debugged the code it looks like it is entering into

_Myt& __CLR_OR_THIS_CALL operator<<(const void *_Val)

What am I doing wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T07:33:51+00:00Added an answer on June 15, 2026 at 7:33 am

    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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