How can I define a function with a parameter dependent on a symbol debug?

Question

I defined with -D compiler option a symbol debug: -DDEBUG_VALUE. I would like a function in which the presence of a parameter depends on the definition or less of the symbol debug flag.

Namely if DEBUG_VALUE is defined, I have

my_function(int parameter1, int  my_parameter_dependent)

Otherwise

my_function(int parameter1)

In this way

my_function(int parameter1  #ifdef DEBUG_VALUE , int my_parameter_dependent #endif)

I get

error: stray ‘#’ in program
error: expected ‘,’ or ‘...’ before ‘ifdef’

How can I solve it?

(I'm on a C++ compiler on a Unix system.)

I'm pretty sure preprocessor directives have to be on their own line — Justin, Jun 16 '17 at 17:24

score 5 · Accepted Answer · edited May 08 '23 at 20:14

5

You can declare the function differently...

 #if defined( DEBUG_VALUE )
     void my_function( int parameter1, int my_parameter_dependent );
 #else
     void my_function( int parameter1 );
 #endif

Create an embedded macro

 # if defined( DEBUG_VALUE )
         #define DEPENDENT_PARAM( x )   x
 # else
         #define DEPENDENT_PARAM( x )
 #endif
 void my_function( int parameter1  DEPENDENT_PARAM(, int my_parameter_dependent) );

This means that the text within the macro is munched by the preprocessor, and is hidden

Or you can declare debug data

  #if defined( DEBUG_VALUE )
      #define EXTRA_DEBUG  , int my_parameter_dependent
  #else
      #define EXTRA_DEBUG
  #endif
  void my_function( int parameter1 EXTRA_DEBUG );

They all have their merits, depending on flexibility and how many functions are changed.

edited May 08 '23 at 20:14

Peter Mortensen

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answered Jun 16 '17 at 17:39

mksteve

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`DEPENDENT_PARAM` strikes me as a good solution to this particular problem (though under the circumstances, I think a different name like `DEBUG_PARAM` would probably be better). – Jerry Coffin Jun 16 '17 at 17:41
I'd rather call it `DEBUG_ONLY`, since its use isn't limited to parameters. In fact, having `_PARAM` there in the name would be confusing to me, because I might be tempted to omit the comma delimiter. (Warning: MFC has such a `DEBUG_ONLY` macro, so beware of possible name clashes.) – Cody Gray - on strike Jun 16 '17 at 18:03

score 4 · Answer 2 · answered Jun 16 '17 at 17:26

4

You can't embed a preprocessor macro within a line. They require a dedicated line of their own. So you have to break this declaration up onto separate lines:

#ifdef DEBUG_VALUE
    void my_function(int parameter1, int my_parameter_dependant);
#else
    void my_function(int parameter1);
#endif

Or, if you want to get clever and DRY, take advantage of C++'s great flexibility with regard to statements and whitespace:

void my_function(int parameter1
#ifdef DEBUG_VALUE
                , int my_parameter_dependant
#endif
                );

answered Jun 16 '17 at 17:26

Cody Gray - on strike

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2

A 3rd way is with an appropriately defined helper macro: `void my_function(int parameter1 DEBUG_VALUE_ONLY(int my_parameter_dependent))`. – 1201ProgramAlarm Jun 16 '17 at 17:29
1

Hmm yeah, the only thing better than 1 macro is 2 macros! – Cody Gray - on strike Jun 16 '17 at 17:33
@CodyGray Excellent, your second solution. I knew the first solution but it did not matter because I would have to replicate many lines of code. The second way works. Thanks. – Nick Jun 16 '17 at 17:35
I strongly recommend against the second version; it tends to reduce understanding even more than the first. Regardless, preprocessor trickery like this can easily get hard to understand, especially once you start wanting yet another configuration – Justin Jun 16 '17 at 17:37
@Justin I use it in debug mode alone. When the code will be ready for release this preprocessor directive will disappear. – Nick Jun 16 '17 at 17:40

How can I define a function with a parameter dependent on a symbol debug?

2 Answers2