Question

I have a quick question.

I’ve been playing around with bit manipulation in c/c++ for a while and I recently discovered that when I compare 2UL and 10UL to a regular unsigned int they seem to return the same bit.

For example,

#define JUMP 2UL
#define FALL 10UL

unsigned int flags = 0UL;
this->flags |= FALL;

//this returns true
this->is(JUMP);

bool Player::is(const unsigned long &isThis)
{
return ((this->flags & isThis) == isThis);
}

Please confirm if 2U equals 10U and if so, how would I go around it if I need more than 8(?) flags in a single unsigned integer.

Kind regards,

-Markus

Answer 1

Of course. 10ul is 1010 in binary and 2 is 10. Therefore, doing x |= 10 sets the second bit too.

You probably wanted to use 0x10 and 0x2 as your flags. These would work as you expect.

As an aside: a single digit in the hex notation represent 4 bits, not 8.