C-preprocessor #define SQR(x) ( x * x ) [closed] - c

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I think I'm going crazy. I'm learning C-preprocessor and I can't figure this problem out, I defined SQR(x) (X*X) purposely without brackets
#include <stdio.h>
#define SQR(x) (x * x)
int main()
{
int counter; /* counter for loop */
for(counter = 0; counter < 5; ++counter)
{
printf("x %d, x squared %d\n",counter+1, SQR(counter+1))
}
return (0);
}
In one moment in loop counter = 3. Then it should be 3 + 1 * 3 + 1 and that is equal to 7, but the output says 5.
Here is output of the code above:
x 1, x squared 1
x 2, x squared 3
x 3, x squared 5
x 4, x squared 7
x 5, x squared 9
Process returned 0 (0x0) execution time : 0.020 s
Press any key to continue.
I'm clearly missing something but I can't figure it out.

Answer 1

When counter=3, it prints as 4 because you print counter+1. Use:
printf("x %d, x squared %d\n",counter, SQR(counter+1));
to see what the actual value of counter is.

Answer 2

Aside from the most obvious error pointed out in Ignaus' answer, you have a second, more insidious and more dangerous error here. Do the substitution that the preprocessor will do on SQR:
SQR(counter+1)
becomes:
counter+1 * counter+1
which will be executed as if it were written like this:
counter + (1 * counter) + 1
That is obviously incorrect and not what you intended. With macros, you should always enclose the arguments in parentheses. So your macro should look like this:
#define SQR(x) ((x) * (x))
Now do the substitution again and you get:
((counter+1) * (counter+1))
which is correct. Make it a habit to always include macro arguments in parentheses to avoid these sorts of errors.