Question

Suppose I have two C++ functions for debug output:

void Trace( const wchar_t* format, ... )
{
    va_list args;
    va_start( args, format );
    VarArgTrace( format, args );
    va_end( args );
}

void VarArgTrace( const wchar_t* format, va_list args )
{
    WCHAR buffer[1024];
    //use ::_vsnwprintf_s to format the string
    ::OutputDebugStringW( buffer );
}

the above uses Win32 OutputDebugStringW(), but it doesn’t really matter. Now I want to optimize the formatting so that when there’s no debugger attached formatting is not done (I measured – speedup is significant):

void Trace( const wchar_t* format, ... )
{
    if( !IsDebuggerPresent() ) {
        return;
    }
    //proceed as previously
    va_list args;
    .....
 }

will the fact that I return early once IsDebuggerPresent() returns null affect anything except that formatting will be skipped?

I mean I no longer call va_start and va_end – will this matter? Will skipping va_start and va_end cause any unexpected behavior changes?

Answer 1

The only requirement on an early return is that if you have used (executed) va_start(), you must use va_end() before you return.

If you flout this rule, you’ll get away with it on most systems, but some system somewhere needs the va_end(), so don’t risk omitting it. It is undefined behaviour to omit it.

Other than that rule, it is up to you how you handle your return. Your proposed early return is not a problem.