Question

Why the two printf statements are outputting different values?

int main()
{
    int n=10;

    printf("%d\n",(n&0xAAAAAAAA)>>1 + n&0x55555555  ); //prints 0

    printf("%d\n", n&0x55555555 + (n&0xAAAAAAAA)>>1 ); //prints 10

    return 0;
}

http://ideone.com/B33YB

Answer 1

Because of the operator precedence.

+ is executed earlier than >>.

When you change

(n&0xAAAAAAAA)>>1 + n&0x55555555)

to

n&0x55555555 + (n&0xAAAAAAAA)>>1)

you are actually changing the order in which the operations are executed.

(n&0xAAAAAAAA)>>1 + n&0x55555555 can be rewritten as (n&0xAAAAAAAA)>>(1 + n&0x55555555) which is different compared to ((n&0xAAAAAAAA)>>1) + n&0x55555555 (which is what the second line states)

The same goes for the + and the & operator.

So to make the output their outputs similar you need additional parenthesis:

int main()
{
    int n=10;

    printf("%d\n",((n&0xAAAAAAAA)>>1) + (n&0x55555555) ); // prints 5
    printf("%d\n",(n&0x55555555) + ((n&0xAAAAAAAA)>>1) ); // prints 5

    return 0;
}

See http://ideone.com/d3mHT