Preprocessor equality test, is this standard?

Question

I had envisaged one of these in the project preferences

• TESTING = HOST
• TESTING = TARGET

Answer 1

I had the same problem. It used to be the compiler would error if the #if parameter was not defined which is what I wanted. I learned the hard way this is no longer the case. The solution I came up with is as follows:

#define BUILDOPT 3

#if 3/BUILDOPT == 1
    <compile>
#endif

If BUILDOPT is not defined you get a compile error for divide by 0. If BUILDOPT != 3 && > 0 the #if fails.

Answer 2

My final implementation is to set BUILDOPT to 1 to enable the code, > 1 to disable. Then the #if becomes #if 1/BUILDOPT==1.

Answer 3

#if TESTING==HOST

If TESTING is not defined, then

it is equivalent to:

#if 0==HOST

From the C Standard:

(C99, 6.10.1p4) "After all replacements due to macro expansion and the defined unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0""

Answer 4

And note that you can do this:

#ifndef TESTING
    ...
#elif TESTING == HOST
    ...
#elif TESTING == TARGET
    ...
#else
    #error "Unexpected value of TESTING."
#endif

Also:

#if defined(TESTING) && TESTING == HOST
    ...
#endif

If you want to collapse the tests. The parenthesis are optional (#if defined TESTING is valid) but I think it's clearer to include them, especially when you start adding additional logic.

Answer 5

The original C preprocessors required explicit #ifdef validation before using a symbol. It is a relatively recent innovation (perhaps driven by scripting languages like Javascript) to assume that undefined symbols have a default value.

Why don't you always insure the symbol is defined?:

#ifndef TESTING
 #define TESTING  (default value)
#endif

#if TESTING==HOST
  ...
#elif TESTING==TARGET
 ...
#else
 ...
#endif

Alternative, maybe force a selection?:

#ifndef TESTING
 #error You must define a value for TESTING
#endif