Pls give explanation for the following

Question

#define square(x) (x*x)
int main()
{
int x,y=1;
x=square(y+1);
printf("%d",x);
return 0;
}

The output of this code is 3

Answer 1

Basically macro code is replaced in preprocessing phase.
So the square(y+1) becomes (y+1 * y+1)
After that by operator precedence 1+1*1+1 gives result 3

Answer 2

1+1*1+1
1*1--> 1
1+1+1 - - > 3