How do I declare a debug only statement

Question

In C# I can use the following code to have code which only executes during debug build, how can I do the same in Xcode?

#if DEBUG
{
    // etc etc
}
#endif


        

Answer 1

There is a very useful debugging technote: Technical Note TN2124 Mac OS X Debugging Magic http://developer.apple.com/technotes/tn2004/tn2124.html#SECENV which contains lots of useful stuff for debugging your apps.

Tony

Answer 2

The NDEBUG symbol should be defined for you already in release mode builds

#ifndef NDEBUG
/* Debug only code */    
#endif 

By using NDEBUG you just avoid having to specify a -D DEBUG argument to the compiler yourself for the debug builds

Answer 3

DEBUG is now defined in "debug mode" by default under Project/Preprocessor Macros. So testing it always works unless you have a very old project.

However I hate the fact that it messes up the code indentation and not particularly compact. That is why I use another macro which makes life easier.

#ifdef DEBUG
#define DEBUGMODE YES
#else
#define DEBUGMODE NO
#endif

So testing the DEBUGMODE value is much more compact:

if (DEBUGMODE) {
//do this
} else {
//do that
}

My favourite:

NSTimeInterval updateInterval = DEBUGMODE?60:3600; 

Answer 4

You can use

#ifdef DEBUG
    ....
#endif

You'll need to add DEBUG=1 to the project's preprocessor symbol definitions in the Debug configuration's settings as that's not done for you automatically by Xcode.

I personally prefer doing DEBUG=1 over checking for NDEBUG=0, since the latter implies that the default build configuration is with debug information which you then have to explicitly turn off, whereas 'DEBUG=1' implies turning on debug only code.