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Home/ Questions/Q 6857083
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T01:58:13+00:00 2026-05-27T01:58:13+00:00

I have a method name open_file declare as below. ifstream& open_file(ifstream &in, const string

  • 0

I have a method name open_file declare as below.

ifstream& open_file(ifstream &in, const string &filename)
{
    in.close();
    in.clear();

    in.open(filename.c_str());

    return in;
}

I want to assign its return value to variable in main() method:

int main()
{
    ifstream val1;
    ifstream val2 = open_file(val1, "test.cpp");

    return 0;
}

I can’t compile the code. My questions are:

  1. Can I assign return value from open_file method to variable in main(), and if so how to do that?
  2. If I can’t assign return value from open_file method to variable, what’s the difference if I change its return type to void?
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  1. Editorial Team
    Editorial Team
    2026-05-27T01:58:14+00:00Added an answer on May 27, 2026 at 1:58 am
    ifstream val2 = open_file(val1, "test.cpp");
    

    This won’t compile because it attempts to make a copy of stream object, which is disabled by having made the copy-constructor private (see this).

    Do this:

    ifstream & val2 = open_file(val1, "test.cpp");
    //val1 and val2 is same here, as val2 is just a reference to val1
    

    But then, why would you even do that? You can simply write:

    open_file(val1, "test.cpp");
    //use val1 here - no need to define val2
    

    Since the returned value is ignored, it is better if you make the return type void. That is less confusing.

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