#if 0 as a define

Question

I need a way to define a FLAGS_IF macro (or equivalent) such that

FLAGS_IF(expression)
<block_of_code>
FLAGS_ENDIF

when compiling in debug (e.g. with a specific compiler switch) compiles to

if (MyFunction(expression))
{
    <block_of_code>
}

whereas in release does not result in any instruction, just as it was like this

#if 0
    <block_of_code>
#endif

In my ignorance on the matter of C/C++ preprocessors I can't think of any naive way (since #define FLAGS_IF(x) #if 0 does not even compile) of doing this, can you help?

I need a solution that:

• Does not get messed up if */ is present inside <block_of_code>
• Is sure to generate 0 instructions in release even inside inline functions at any depth (I guess this excludes if (false){<block_of_code>} right?)
• Is standard compliant if possible

Answer 1

Macros are pretty evil, but there's nothing more evil than obfuscating control statements and blocks with macros. There is no good reason to write code like this. Just make it:

#ifdef DEBUG
  if (MyFunction(expression))
  {
    <block_of_code>
  }
#endif

Answer 2

The following should do what you want:

#ifdef DEBUG
# define FLAGS_IF(x) if (MyFunction((x))) {
# define FLAGS_ENDIF }
#else
# define FLAGS_IF(x) if(0) {
# define FLAGS_ENDIF }
#endif

The if(0) should turn into no instructions, or at least it does so on most compilers.

Edit: Hasturkun commented that you don't really need the FLAGS_ENDIF, so you would instead write your code like this:

FLAGS_IF(expression) {
   <block_of_code>
}

with the follow macros:

#ifdef DEBUG
# define FLAGS_IF(x) if (MyFunction((x)))
#else
# define FLAGS_IF(x) if(0)
#endif