C programming passing function in if condition

Question

I'm writing simple function which returns maskable array of zeros and ones to array of numbers.

If number is a power of two maskable index will be one if not then zero.

It works fine without passing my simple function isPower() , but when I do pass it. weird things happen.

I want my code works with

if(isPower(nums[i]))

Without or condition

#include<stdio.h>

int *isPowerOf2(int num, int*nums,int *result);
int isPower(int num);
int main(void)
{
int array[]= {1,2,2,4,4,3,4,3,3,4,4,4,4,4};
int size=0;
int *result=NULL;
result=isPowerOf2(14,array,&size);

for(int i=0; i<size; i++)
{
    printf("%i : %i\n",array[i],result[i]);
}
free(result);
printf("%i",size);
return 0 ;
}

int *isPowerOf2(int num, int*nums,int *result_num)
{
int *result=(int *)malloc(num*sizeof(int));

*result_num=num;
for(int i=0; i<num; ++i)
{
    if(isPower(nums[i])||!(nums[i]&nums[i]-1)) //this won't work  but when I use !(nums[i]&nums[i]-1) it works fine
    {
        result[i]=1;
    }
    else
    {
        result[i]=0;
    }

}
return result;
}

 int isPower(int var)
 {
 return ((var)&(var-1)==0)?1:0;
 }

You need to learn basics of C language and OR operator. Second operand of OR statement is NOT executed if first one evaluates to non-zero value. — unalignedmemoryaccess
– unalignedmemoryaccess, Commented Mar 22, 2018 at 8:42
I think you did not get my point I added second operand because without it won't work. I did add it to clarify my question, but I want code work without it. it supposed not be there. — IDEN
– IDEN, Commented Mar 22, 2018 at 8:45
You second operand nums[i]&nums[i]-1. Do you know the sequence of execution here? — unalignedmemoryaccess
– unalignedmemoryaccess, Commented Mar 22, 2018 at 8:47
Focus on isPower function: return ((var)&(var-1)==0)?1:0; should be return (((var)&(var-1))==0)?1:0; — unalignedmemoryaccess
– unalignedmemoryaccess, Commented Mar 22, 2018 at 8:50
Also, clean it up. If I can't immediately identify by eye the start/end of functions, I close-voe as 'Unclear' and move on to the next question. — Martin James
– Martin James, Commented Mar 22, 2018 at 9:00

kabanus · Accepted Answer · 2018-03-22 08:56:03Z

3

At times like this it pays to check what you're actually computing:

#include<stdio.h>

int main() {
    printf("%d %d\n",!(7&6),((7&6==0) ? 1 : 0)); //Not a power of 2
    printf("%d %d\n",!(8&7),((8&7==0) ? 1 : 0)); //Power of 2
    return 0;
}

Returns:

0 0
1 0

If you think about it a bit, should make sense. Or does it? Operator precedence is killing you here. Why would the bottom right be 0?

edited Mar 22, 2018 at 8:56

answered Mar 22, 2018 at 8:50

kabanus

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1 Comment

IDEN Over a year ago

Yeah, hope I give more attention next time :)

maze · Accepted Answer · 2018-03-22 08:59:09Z

1

Befare that (1 & 2) is a bitwise operand while (1 && 2) is the logical operand.

So to be clear: (1 & 2) == 0 is something completely different than !(1 & 2).

answered Mar 22, 2018 at 8:59

maze

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