print only if _DEBUG defined: c

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");

Try this:

#ifdef DEBUG
#define Print(s) printf("%s", s)
#else
#define Print(s)
#endif

Then:

#define DEBUG
Print("Inside main in debug mode");

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.

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.

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 */

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

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.