Please explain this answer 😓

Question

https://code.sololearn.com/cPiIT6Fz0H6l/?ref=app

Answer 1

A macro is just text replacement, so sqr( 3 + 1 ) is expanded to 3 + 1 * 3 + 1 = 7. Use parentheses if you have to worry about something like that happening, e.g. #define sqr( x ) ( x ) * ( x ).

Answer 2

#include <stdio.h>
#define sqr(x) x*x
int main() {
printf("%d",sqr(3+1));
return 0;
}

when the preprocessor goes through before the program is compiled, it will replace the call to the macro sqr(x) with its definition

#include <stdio.h>
#define sqr(x) x*x
int main() {
printf("%d",3+1*3+1); // order of operations
return 0;
}

...
3+1*3+1
3+3+1
7

to fix this, you must put variables in parenthesis when defining a macro

#define sqr(x) ((x)*(x))

Answer 3

Hey Martin Taylor ,
But your explanation also too good 😊

Answer 4

Thanks Slick 😊

Answer 5

Thanks Shadow 😊