why answer is7?

Question

What is output of this code?
#define sqr(x) x*x
int main()
{ printf(”٪d”, sqr(3+1);}

Answer 1

#define sqr(x) x*x
// sqr(3+1) = 3 + 1*3 + 1 = 7

#define sqr(x) (x)*(x)
// sqr(3+1) = (3+1)*(3+1) = 16

#define sqr(x) (x*x)
// what is the output... ?

Answer 2

As you define sqr(x) x*x mean which you passing argument that will be multiplied two times

Now sqr(3+1) define it will treat like 3+1*3+1
As multiplication has higher precession that why it will be 3+3+1=7
That is my point of view..
Nika Soltani Tehrani

Answer 3

sqr(3+1) will be interpreted as - 3+1*3+1 = 3+3+1 = 7, as * has higher precedence.
To get ur expected result correct the macro -
#define sqr(x) ((x)*(x))
Observe additional '()' around macro expansion.

Hope this helps...!!!

Answer 4

Thanks a lot guys🙏🙏🙏