Thread: passing to a function y should be 4 but its zero

here is a simplfied verstion of the function i am passing x and y to but it doesnt work y should equate to 4 when passed to the function but for some reason its zero.

Code:

#include <stdio.h>
#include <stdlib.h>

int func(int x, int y);

int main()
{
    int coordinate_x=2, coordinate_y=6;
    int multiply_x = 1, multiply_y = -1;

    if (func(coordinate_x + (2 * multiply_x), coordinate_y + (2 * multiply_y) == 0)) //check in case more than one jump
    {
        printf("func worked\n");
    }

    return 0;
}

int func(int x, int y)
{
    printf("x = %d and y = %d\n", x, y);
    if (x ==4 && y ==4)
    {
        printf("this bit worked\n");
        return 0;
    }
    else
    {
        printf("it didnt\n");
        return -1;
    }
}

i been staring at this for 2 hours and i cant see what is wrong. the same method works 20 times in other places.

many thanks
coop

Line 11 reads:

Code:

if ( func(coordinate_x + (2 * multiply_x), 
          coordinate_y + (2 * multiply_y) == 0) )

Notice the second argument is cordinate_y + (2 * multiply_y) == 0.
It should be:

Code:

if ( func(coordinate_x + (2 * multiply_x), 
          coordinate_y + (2 * multiply_y)) == 0)

If your text editor or IDE has a "highlight matching parenthesis" feature, it's a really useful tool for wrangling nested parentheses like this.