Hi, I need some help!
I'd like to do a function for debug my code (at compile time) that follows this scheme:
#ifdef DEBUG
my_printf( <list arguments> ) printf( <list arguments> )
#else
my_printf( <list arguments> ) NOTHING
#end
In other words if DEBUG is defined the code is replace with the printf() function else do nothing.
How can I write this macro for my purpose??
Thanks a lots.
#ifdef DEBUG
#define my_printf printf
#else
#define my_printf
#end
Originally Posted by Adak
Thanks but with this define I get a lot of:
where the function is present and DEBUG not defined.Code:cripto.c: In function ‘print_error’: cripto.c:90: warning: left-hand operand of comma expression has no effect cripto.c:90: warning: left-hand operand of comma expression has no effect cripto.c:90: warning: statement with no effect
If you have C99, you can use variadic macro.
#define debug_print(...) fprintf(stderr,__VA_ARGS)
You might also want to read:
Question 10.26
Thanks, I used this:
and it works fine.Code:#ifdef DEBUG_CRIPTO #define lib_debug(...) fprintf(stderr, __VA_ARGS__) #else #define lib_debug(...) #endif
Hrm...
How about:
#ifdef DEBUG
#define my_printf printf
#else
#define my_printf //
#end
(Assuming you have C99.)
For your method, I think you'd have to do something like this:
Then I think you need double parentheses when calling it: my_printf(( "%d %s", 5, "foo" ));
Something like that anyways.
"I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008
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It is #endif not #end
A small addition to Elysia's idea:
Since you might haveCode:#ifdef DEBUG #define my_printf printf #else #define my_printf ;// #end
Of course what Mordor actually did would be the best for C99.Code:if (something) my_printf(...);
testing cpjust idea
this works. This doesn't compileCode:#include <stdlib.h> #include <stdio.h> #define DEBUG #ifdef DEBUG #define my_printf(x) printf(x) #else #define my_printf #endif int main() { printf(("%s", "10")); }
Both give a warning (in GCC). That the left hand operant on the comma has no effect. The error the later gets is thatCode:#include <stdlib.h> #include <stdio.h> #define DEBUG #ifdef DEBUG #define my_printf(x) printf(x) #else #define my_printf #endif int main() { printf(("%d", 10)); }
"passing arg 1 of 'printf' makes pointer from integer without a cast"
What I think happens is that this
so the first one will run. The second I have no idea. Or something equally weird happens.Code:("%s", "hey") --> ("%s" "hey") --> ("%shey")
Conclusion: be careful when being tricky with macros
What maybe you could do is change it to
and always use printf with two inputs and combine multiple of them for more inputsCode:#define my_printf(x, y) printf(x, y)
