Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
509 views
Welcome To Ask or Share your Answers For Others

1 Answer

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);

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share

548k questions

547k answers

4 comments

86.3k users

...