How does this "&" work in this statement?

Question

I know how bitwise AND works,but I don't understand how does (sourceDragMask & NSDragOperationGeneric) work here,I don't get the point.Is there anyone can explain to me?Thanks a lot.

- (NSDragOperation)draggingEntered:(id <NSDraggingInfo>)sender
{
    NSPasteboard *pboard;
    NSDragOperation sourceDragMask;

    sourceDragMask = [sender draggingSourceOperationMask];
    pboard = [sender draggingPasteboard];

    if ( [[pboard types] containsObject:NSColorPboardType] )
    {
        if (sourceDragMask & NSDragOperationGeneric)
        {
            return NSDragOperationGeneric;
        }
    }

    return NSDragOperationNone;
}

Answer 1

sourceDragMask is what's called a bitmask: an integer that represents a a number of boolean flags. It's declared as a variable of type NSDragOperation, which is actually typedefed from NSUInteger—it really is just that.

First off, the & operator is what's known as a "bitwise AND", which simply takes two numbers ANDs each of the bits in them to create a new value. For example, 0011 1001 & 0101 1010 == 0001 0001—57 & 90 = 17. Similarly, the | operator does a bitwise OR. Using these operators, we can use each of those 1s and 0s as a boolean flag separately.

So, if, for example, sourceDragMask is the value 12, what does this tell us? In binary, 12 is represented as 0000 1100, or 8 | 4, which according to the NSDraggingInfo Protocol Reference, is NSDragOperationPrivate | NSDragOperationGeneric. So if we & this with NSDragOperationGeneric, we get 4 out, which is non-zero and therefore true. In this way, we can check whether a bitmask contains a certain flag.

Answer 2

I know how bitwise AND works,but I don't understand how does (sourceDragMask & NSDragOperationGeneric) work here,I don't get the point.

NSDragOperationGeneric is most likely a power of two, which means it has only one bit set. This is deliberate: Bit masks are almost all defined as powers of two (single bits) to enable bit-mask operations like this one.

The bitwise-AND operation, as you know, evaluates to only those bits that are set in both sides. If one side has only one bit (NSDragOperationGeneric) set, then the operation effectively tests whether that bit is set in the other side.

That's the point of the operation: To test whether the NSDragOperationGeneric bit is set.

There is one gotcha: As you know, a successful bitwise AND test will evaluate to the tested-for bit mask, not 1. So, for example, if you test for a bit mask that's defined as 0x100 (1 followed by 8 clear bits), then assign that result to a BOOL (which is a signed char) variable, you'll assign zero to the variable! This is why you sometimes see code like this:

BOOL supportsCopyOperation = ((dragOperations & NSDragOperationCopy) == NSDragOperationCopy);

or this:

BOOL supportsCopyOperation = ((dragOperations & NSDragOperationCopy) != 0);

or this:

BOOL supportsCopyOperation = !!(dragOperations & NSDragOperationCopy);

Other bit-mask operations include bitwise-OR (|) to set bits in a value (return NSDragOperationCopy | NSDragOperationMove;, for example) and bitwise-NOT (~, a.k.a. two's complement) to invert the bits of a value, usually for “anything but” tests.