Thread: preprocessor problemn

When the preprocessor processes your program, it becomes:

Code:

/* a bunch of included header file code from stdio.h here */
main()
{
 int x=5,y;
 y=4*x-3*x-3;
printf("%d",y);
}

The A*A is replaced with whatever you have in the parentheses after SQRT. In this case x-3 replaces A in both places, so x-3*x-3. 4*5-3*5-3 == 20-15-3 == 2.

If you want to use a preprocessor macro for SQRT, you can not put anything except a single value as the parameter or "strange" results like this will occur. If you want to have a sqrt() function that you can call with x-3 (or any other expression), you need to make it a function.

Originally Posted by jEssYcAt

If you want to use a preprocessor macro for SQRT, you can not put anything except a single value as the parameter or "strange" results like this will occur. If you want to have a sqrt() function that you can call with x-3 (or any other expression), you need to make it a function.

It is possible to reduce the strange effects of the macro, with extra brackets.

Code:

#define SQR(A) ((A)*(A))

As long as the argument of the macro doesn't have side effects, that will work. 4*SQR(x-3) will expand to 4*((x-3)*(x-3)) which is what you want.

The catch is it won't work with expressions that have side effects. For example SQR(++x) will expand to 4*((++x)*(++x)) which modifies x twice and causes undefined behaviour. In such cases, a function is needed (eg so x is only incremented once and its value retrieved, before calling the function).

The thing to remember is that macros are not functions, even if they sort of look like them. They perform text substitution on the code that is actually compiled.