Standard no-op output stream

Question

Is there a way to create an ostream instance which basically doesn\'t do anything ?

For example :

std::ostream dummyStream(...);
dummyStream <<         


        

Answer 1

If you are concerned about the overhead of your debugger then you can write a very simple code to void out your debug messages on compilation. This is what I use for my c++ programs.

#include 
#define DEBUGGING // Define this in your config.h or not.
#ifdef DEBUGGING
/*
 * replace std::cout with your stream , you don't need to
 * worry about the context since macros are simply search
 * and replace on compilation.
 */
#define LOG_START std::cout <<
#define LOG_REDIR <<
#define LOG_END   << std::endl;
#else
#define LOG_START (void)
#define LOG_REDIR ;(void)
#define LOG_END   ;
#endif // DEBUGGING

int main(){
LOG_START "This is a log message " LOG_REDIR "Still a log message." LOG_END;
return 0;
}

Now when making your project , check if the user wants to disable the logging , if so , just undefine the DEBUGGING macro or whatever macro you choose to check for.

Now your code will be optimized by the compiler , Because when anything is voided , it will not be included in the resulting binary(most of the time) , making the binary production ready.