I don’t understand, y output is 1

Question

# include <stdio.h>

int main(){
int x, s;
x = scanf(“%d”, &x);
s = x * x;
printf(“%d”, s);
return 0;
}

The output is 1, but I write 2 into scanf
WHY?!

Answer 1

Probably because scanf returns 1. But either way, just scan it, you don't set x equal to the scanf function.

https://code.sololearn.com/cjRc22Ri14AE/?ref=app

Answer 2

Look at the fix. I took away "x =" on line 5

Answer 3

Матвей Барканов
"Ok, but how the meaning of scanf put into x?"

scanf("%d", &x 👈HERE)

Answer 4

Ok, but how the meaning of scanf put into x?

Answer 5

Oh, great! It’s working! Thank you! But y scanf returned 1?

Answer 6

I don't know, please use google. Plus, thats not the correct way to use scanf. The way i posted is. So use it right and you'll have no issues

Answer 7

Because he used assignment operator on scanf , so c compiler gets it as there is only one value entered so x=1 therefore after multiplication it stays as 1 so prints as 1 !!

Answer 8

scanf returns how many number of values it read.
int n = scanf("%d %d", &x, &y) ;

printf returns how many character it displayes on the output..

int y = printf("n=%d", n); //n=2

printf("%d", y) ; // 3

Hope it helps..

https://code.sololearn.com/cjh3dwj9kA34/?ref=app

Answer 9

Hii