What exactly does an #if 0 ..... #endif block do?

Question

Not only does it not get executed, it doesn't even get compiled.

#if is a preprocessor command, which gets evaluated before the actual compilation step. The code inside that block doesn't appear in the compiled binary.

It's often used for temporarily removing segments of code with the intention of turning them back on later.

It's identical to commenting out the block, except with one important difference: Nesting is not a problem. Consider this code:

foo();
bar(x, y); /* x must not be NULL */
baz();

If I want to comment it out, I might try:

/*
foo();
bar(x, y); /* x must not be NULL */
baz();
*/

Bzzt. Syntax error! Why? Because block comments do not nest, and so (as you can see from SO's syntax highlighting) the */ after the word "NULL" terminates the comment, making the baz call not commented out, and the */ after baz a syntax error. On the other hand:

#if 0
foo();
bar(x, y); /* x must not be NULL */
baz();
#endif

Works to comment out the entire thing. And the #if 0s will nest with each other, like so:

#if 0
pre_foo();
#if 0
foo();
bar(x, y); /* x must not be NULL */
baz();
#endif
quux();
#endif

Although of course this can get a bit confusing and become a maintenance headache if not commented properly.

It permanently comments out that code so the compiler will never compile it.

The coder can later change the #ifdef to have that code compile in the program if he wants to.

It's exactly like the code doesn't exist.

I'd like to add on for the #else case:

#if 0
   /* Code here will NOT be complied. */
#else
   /* Code will be compiled. */
#endif


#if 1
   /* Code will be complied. */
#else
   /* Code will NOT be compiled. */
#endif