Bitwise AND adds bits together, and if two 1's line up, it's a 1, otherwise, it's a 0, like so, yes?

So, in an if statement:Code:`0011`

& 1010

= 0010

So that means if just one of the resulting bits is 1, the condition is true?Code:`int y=2; //0010`

if (y & 3)//0011

{

cout<< "This is true.";

}

So saying if (y & 3) is the same as saying if (y==1 || y==2 || y==3), yes?