Difference between all the codes and their outputs?

Question

#define SQUARE(n) n * n
main( )
{
int j ;
j = 64 / SQUARE ( 4 ) ;
printf ( "j = %d", j ) ;
}

#define SQUARE(n) (n * n)
main( )
{
int j ;
j = 64 / SQUARE ( 4 ) ;
printf ( "j = %d", j ) ;
}

#define PRODUCT(x) ( x * x )
main( )
{
int i = 3, j ;
j = PRODUCT( i + 1 ) ;
printf ( "\n%d", j ) ;
}

Answer 1

Just do as directed :-

1) square(n) n*n
64/4*4 = 64

2)square(n) (n*n)
64/(4*4) = 4

3) product(x) (x*x)
j = (3+1*3+1)
= 3 + 3 + 1 = 7

4) product(x) (x)*(x)
j = (3+1)*(3+1) = 16

Answer 2

macro is just a textual replacement .
just replace the macro with what you have defined it to.

Answer 3

Sri pranavi Donapati see I have edited the response and thanks for pointing it out.

Answer 4

Arsenic
#define PRODUCT(x) ( x * x )
main( )
{
int i = 3, j ;
j = PRODUCT( i + 1 ) ;
printf ( "\n%d", j ) ;
}

Output is 7

#define PRODUCT(x) ( x) *( x )
main( )
{
int i = 3, j ;
j = PRODUCT( i + 1 ) ;
printf ( "\n%d", j ) ;
}

Output is 16

How?

Answer 5

#define SQUARE(n) n * n
main( )
{
int j ;
j = 64 / SQUARE ( 4 ) ;
printf ( "j = %d", j ) ;
}