“printf” only printing variable addresses

Question

so here is my code:

#include 
main(){
int hi;
hi = 3;
printf(\"%d\",&hi);
}

and the output is: \"2686748\"

im using

Answer 1

If you intend to print the value of hi, just pass it to printf, not its address:

printf("%d", hi);

You may be confusing printf with scanf, the latter requiring all of its arguments to be pointers.

Answer 2

The "%d" tells the printf you are putting in an integer. The integer you are giving it is &hi which is the address of hi. If you want the value of hi just use that