print only if _DEBUG defined: c - c

Question

I want to print info only if _DEBUG is defined
#define DEBUG(y) y == true ? #define _DEBUG true : #define _DEBUG false
#ifdef _DEBUG
#define Print(s) printf(s);
#endif
Getting Error:
error: '#' is not followed by a macro parameter
Any suggestion how to achieve this with pre-processor directives?
I intend to use it from my main as:
DEBUG(true);
Print("Inside main in debug mode");

Answer 1

You cannot redefine a MACRO at run-time.
Neither you can have a #define inside of another #define, like you try in the first line of your code.
You can do something like this:
#ifdef _DEBUG
#define Print(s) printf("%s", s)
#else
#define Print(s)
#endif
And use it from your main as:
#define _DEBUG
Print("Inside main in debug mode");
#undef _DEBUG
Print("Inside main debug mode off");
If you really need to switch debug on and off at run-time, your can do something like this:
void PrintIf(BOOL dbg, char * msg)
{
if (dbg)
{
printf("%s", msg)
}
}
And use it like this
y = TRUE;
PrintIf(y,"Inside main in debug mode");
y = FALSE;
PrintIf(y,"Inside main debug mode off");

Answer 2

Try this:
#ifdef DEBUG
#define Print(s) printf("%s", s)
#else
#define Print(s)
#endif
Then:
#define DEBUG
Print("Inside main in debug mode");

Answer 3

I intend to use it from my main as:
DEBUG(y);
Print("Inside main in debug mode");
Sorry, but ifdef are compile time (not run-time). You could use a global bool and runtime checking to enable and disable debug.

Answer 4

You can't create preprocessor statements with macros as you are trying to do; it doesn't work and isn't allowed. For conditional printing, see C #define macro for debug printing.

Answer 5

The problem occurs here:
#define DEBUG(y) y == true ? #define _DEBUG true : #define _DEBUG false
When introducing #defines, the definition # occurs at the beginning of the line, not later (although) preprocessors generally allow one or two line indents.) You need to rewrite your #define eliminating the ternary operator simply as:
#ifdef _DEBUG
#define Print(s) printf(s);
#endif
While you may extend you defines with macros, you often introduce additional errors. It is generally better to stick to wrapping your _DEBUG code simply in #ifdef statements:
#ifdef _DEBUG
fprintf (stderr, "your error messages\n"); // using standard printf/fprintf instead of macros
...
#endif /* _DEBUG */

Answer 6

Macros are substituted at preprocessing stage and
#define DEBUG(y) y == true ? #define _DEBUG true : #define _DEBUG false
this statement will be evaluated at compile time.
Conditional operator (ternary operator) are evaluated at compile time. So you are getting this error and # operator must always be used at the beginning of the statement that is the second mistake you are doing.
You can better use it this way
#define DEBUG
printf ("true");
#else
printf ("false");
You can also define this macro dynamically by using the gcc option -D
gcc -D DEBUG filename.c -o outputFile

Answer 7

The first line is incorrect:
#define DEBUG(y) y == true ? #define _DEBUG true : #define _DEBUG false
You cannot use #define inside preprocessor directive (like another #define)
And it does not make sense, since preprocessing happens before the real compilation (so before run time, when your y has some value). Read the cpp preprocessor documentation. Recall that sometimes the preprocessor is even a different program (/lib/cpp) but is today the first phase of most C compilers.
You could ask for the preprocessed form of your source code (e.g. with gcc -C -E source.c > source.i if using GCC) and look at that form with a pager (less source.i) or your editor.