linux - Why won't this code permute the value Z? -

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so trying build simple program in c permutes value of z (z equal x + y) every single thing try differently doesn't work. frustrated here. please me understand.

source: 

#include <stdio.h> #include <stdlib.h> int main() {     int x, y, z;     scanf("%d", &x);     scanf("%d", &y);     z = x + y;     printf("%d", &z);     return 0; }

you're printing address of z, not stored value, because you're passing printf pointer z rather value. change printf line to:

printf("%d", z);

scanf returns success value has use way give input. argument you're passing (&x) pointer variable want use storage. that's ampersand for. says "use address of variable".

printf, on other hand, wants values themselves. doesn't need address. (though, technically, strings passed in pointers. not distinction need worry right now.)


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