Hi,

I am a bit confused here. I have an example of bit operators verse logical operators from http://cplus.about.com/od/advancedtu.../aa042203f.htm . I was wondering if I am on the track saying that a bit operator does the same thing as a logical operator with the only exception that it does it on bits rather then variables ect?

#include <iostream>
using namespace std;

int main()
{
unsigned short flags1;
unsigned short flags2;

flags1 = 0x00FF;
flags2 = 0xFF00;

cout << "Bitwise Operator: " << (flags1 & flags2) << endl;
cout << "Logical Operator: " <<(flags1 && flags2) << endl;

return 0;
}

Thanks for your helpl

AFAIK, with && both will be converted to bool first for integral/POD types, then the result returned will also be bool. For &, each bit is treated as its own bool and like positions are compared:If either operand is a class object then the operation will only be evaluated if there is an overloaded/global operator that will take both operands as arguments in the order that they are given. In that case, it's up to the implementor to decide how the operator will behave.
ta0kira

just as a side note: If you want to combine your bitfields then you can 'or' them:

0000 0000 1111 1111 & 1010 1010 0000 0000 = 1010 1010 1111 1111

You mean:

0000 0000 1111 1111 | 1010 1010 0000 0000 = 1010 1010 1111 1111

Narg.. Yes. I meant | instead of &. Sorry bout that.