[priyabc]

#define square(x) x*x

main()
{
 int i;
 i=64/square(4);
 printf("%d",i);
}

the result obtained for this
64

how it is so??

i thought 64/square(4) will be 64/4*4 = 64/16 which will be 4... but the result is not so why.. what happens

[DaWei]

64/4*4

64/4 = 16

16*4 = 64

If you're going to use macros, you better learn that you need to parenthesize the hell out of them. You would be much wiser to write the function.

[kaustubh]

hi priyabc,
you should try something like this

#define square(x) x*x

main()
{
 int i;
 i=(64)/(square(4));
 printf("%d",i);
}

this mean (64) / (4 * 4) = (64)/(16) = 4

[DaWei]

You guys don't listen well. If you must use a macro, do it right. Also, kaustub, the C standard defines main as returning an int. Please don't teach bad habits.

#include <stdio.h>

#define square(x) (x*x)

int main ()
{
    int i = 64 / square(4);
    printf ("%d", i);
    return 0;
}

[kaustubh]

hi dawei,
she must be beginner. Most books of beginners written about 5 to 10 years ago didn't contain standard c or c++ . I will try to use known c /c++ standards in posts in future.

[DaWei]

The C standards were issued in 1989/90 and 1999. There are lots of compilers that don't completely conform and lots of books and tutorials (including by MS) that teach faulty C.