Thread: pointer int math wrong

ive been having alot of fishy problems lately

if i run this it adds the numbers wrong. i tried adding a * to it but it didnt help any.

Code:

int main() {
	int x = 1;
	
	test(x);
	
	return 0;
}

int test(int *x) {
	printf("X = %i\n", x);
	++x;
	printf("X Now = %i", x);
	
	return 0;
}

i get back
X = 1
X Now = 5
i want
X Now = 2

You have a few problems with your code. Here is what it should look like.

Code:

#include <stdio.h>

int test( int *x );

int main() {
	int x = 1;
	
	test(&x);
	
	return 0;
}

int test(int *x) {
	printf("X = %i\n", *x);
	++(*x);
	printf("X Now = %i", *x);
	
	return 0;
}

When you pass a regular variable to a function as a pointer you need to pass the address of it. Also you have to dereference the pointer to get the value located there. Hope this helps.

EDIT: I don't know how it even compiled for you. There are several errors that should occur. If you just print out x in the test function without dereferencing you will get the address not the value.

i didnt get why it compiled either, it didnt seem right. but how can 1 be the adress? i thought they took a completly diff form(i was thinking hex) and wouldnt my kernel take up alot of the begining?

thanks

i just compiled your code, seg fault.

now im really confused.

What compiler are you using??

i copy and pasted it into vi, but instead of typing :q i had caps on and did :Q. sorry