#include<stdio.h>
#include<conio.h>
#define PROD(x) (x*x)
void main()
{
clrscr();
int p=3,k;
k=PROD(p+1); //here i think value 3+1=4 would be passed to macro
printf("\n%d",k);
getch();
}
In my opinion, the output should be 16, but I get 7.
Can anyone please tell me why?
Macros are expanded, they don't have values passed to them. Have look what your macro expands to in the statement that assigns to k.
k=(p+1*p+1);
Prefer functions to macros, if you have to use a macro the minimum you should do is to fully parenthesise the parameters. Note that even this has potential surprises if users use it with expressions that have side effects.
#define PROD(x) ((x)*(x))
The preprocessor expands PROD(p+1) as follows:
k = (p+1*p+1);
With p=3, this gives: 3+1*3+1 = 7.
You should have written your #define as follows:
#define PROD(x) ((x)*(x))
The problem here is that PROD is a macro and will not behave exactly like you intend it to. Hence, it will look like this:
k = p+1*p+1
Which of course means you have:
k = 3+1*3+1 = 7
#define PROD(x) (x*x)
PROD(3+1) is changed by the preprocessor to 3+1*3+1
macro are not function . These are replaced by name
It will be p+1*p+1
This is what compiler is going to see after preprocessors does its job: k= p+1*p+1. When p = 3, this is evaluated as k = 3+(1*3)+1. Hence 7.
This is exactly why you should use functions instead of macros. A function only evaluates each parameter once. Why not try
int prod(int x)
{ return x * x; }
and see the difference!
void main()...