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Asked: May 11, 20262026-05-11T01:27:25+00:00 2026-05-11T01:27:25+00:00

In ANSI C++, how can I assign the cout stream to a variable name?

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In ANSI C++, how can I assign the cout stream to a variable name? What I want to do is, if the user has specified an output file name, I send output there, otherwise, send it to the screen. So something like:

ofstream outFile; if (outFileRequested)      outFile.open('foo.txt', ios::out); else     outFile = cout;  // Will not compile because outFile does not have an                       // assignment operator  outFile << 'whatever' << endl; 

I tried doing this as a Macro function as well:

#define OUTPUT outFileRequested?outFile:cout  OUTPUT << 'whatever' << endl; 

But that gave me a compiler error as well.

I supposed I could either use an IF-THEN block for every output, but I’d like to avoid that if I could. Any ideas?

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  1. 2026-05-11T01:27:26+00:00Added an answer on May 11, 2026 at 1:27 am

    Use a reference. Note that the reference must be of type std::ostream, not std::ofstream, since std::cout is an std::ostream, so you must use the least common denominator.

    std::ofstream realOutFile;  if(outFileRequested)     realOutFile.open('foo.txt', std::ios::out);  std::ostream & outFile = (outFileRequested ? realOutFile : std::cout); 
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