Im trying to define printf("%d",var) to something more easy,but i cant figure out how.
I want it something like printvar(var),but i cant figure out a code to define it like that =\
Thanks
Im trying to define printf("%d",var) to something more easy,but i cant figure out how.
I want it something like printvar(var),but i cant figure out a code to define it like that =\
Thanks
I'm pretty sure that's how macro functions work.Code:#define printvar(var) printf("%d",var)
Originally Posted by The Jargon File
But when var is no longer an int, what are you going to do then?
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
yes,but that means i would have to name the variable "var".
i want something where i can go,
The code you gave me only defines "var".Code:printvar(var); printvar(var2); printvar(anothervar);
I think you need to think more deeply about how functions (and function-like macros) work. Does this mean I can never call printf with different things inside it?
No, you can use any variable name you want. You can use #define printvar(derp) printf("%d",derp) if you want to, it really doesn't matter.
Is there any way to do that?
I thought preprocessor macros had absolutely no idea about types.
Unless you put something type specific in them that is.... like printf("%d", var) ... the macro doesn't care but printf() does.
So, if I understand it correctly..It would, by definition, be impossible to have the same macro (as here) to printf() different types of variables. Correct?
Correct... You would need to create a macro for each type...
PRINTINT() PRINTULONG() PRINTFLOAT() etc.
Frankly the gain from this over just using printf() directly is almost zero. All you're really doing is obfuscating the behavior of your code.
Incidentally, it would be good to get into the habit of enclosing macro arguments in parentheses to force the correct order evaluation. It makes no difference in this case, but consider the following:
If you then use it in your code with, say, square(a+5) you will get an unexpected result because it will expand in-line to a+5*a+5 rather than (a+5)*(a+5). This is correct:Code:#define square(x) x*x
Code:#define square(x) (x)*(x)
Code:while(!asleep) { sheep++; }
No it isn't. Try "~square(x)". Not only do you need to put each argument in parentheses, the entire expression needs to be in parentheses too.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
Yes, but who would want to do something like that?
Code:while(!asleep) { sheep++; }
We're about 10 posts past where that question was needed.
square is a toy example that nobody would actually use, but generally applying a high precedence operator to the result of a function-like macro is not unheard of.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
