What is &= and |=

Question

I was going through some VC++ code in a large code-base and came across this:

    if (nState & TOOL_TIPS_VISIBLE)
        nState &= ~TOOL_TIPS_VISI         


        

Answer 1

& and | are similar to && and ||, only they works in bitwise fashion. So now you could imagine &= and |= works similar to +=. That is x &= y; ==> x = x & y;

Answer 2

What hasn't been mentioned is that both &= and |= operators can be overloaded. Therefore, the code you posted depends on the type of nState (although it's quite clearly an int, so most probably this doesn't apply here). Overloading &= does not implicitly overload &, so in this case

x &= y might not be the same as x = x & y

It might also depend on what TOOL_TIPS_VISIBLE is.

struct s{
   int x;
};

void operator &= (int& x, s y)
{
   x = 0;
}

Now whenever you do:

s TOOL_TIPS_VISIBLE;
x &= TOOL_TIPS_VISIBLE;

x will become 0. Again, highly unlikely, but good to know nevertheless.

All other answer probably apply here, but this is worth considering.

Answer 3

Its bitwise and or

in the first case the flag (bit) is turned off

nState &= ~TOOL_TIPS_VISIBLE

in the second case the flag is turned on

nState |= TOOL_TIPS_VISIBLE

Answer 4

x &= y is the same as x = x & y
x |= y is the same as x = x | y

Answer 5

Yes. &= is to & what += is to +

Answer 6

x &= y means x = x & y. So yes you're right.