Bitwise AND To Check Equality

Question

If you have a [Flags] enum instance called typeToAdjust, and you want to see if it's equal to clientType, is there any point doing this?

 (clientType & (int)typeToAdjust) == (int)typeToAdjust

Doesn't this do the same?:

clientType == (int)typeToAdjust

Did you mistype the second statement? Did you meant `&` instead of `==` ? — Mark Lakata, Sep 23 '15 at 02:03
possible duplicate of [How to Compare Flags in C#?](http://stackoverflow.com/questions/40211/how-to-compare-flags-in-c) — Metro Smurf, Sep 23 '15 at 02:03
not really an answer to that specific question ; but from FW4.5 you can use the [HasFlag](https://msdn.microsoft.com/en-us/library/system.enum.hasflag%28v=vs.110%29.aspx) method to simplify things ; it will do exactly that — Sehnsucht, Sep 23 '15 at 02:15

score 1 · Accepted Answer · answered Sep 23 '15 at 02:02

1

If you have more than one flag set in clientType, then you must do the first one, although you could do this also:

(clientType & (int)typeToAdjust) != 0

You seem to have missed the whole idea of [Flags] which allows one or more bits to be set.

answered Sep 23 '15 at 02:02

Mark Lakata

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score 0 · Answer 2 · answered Sep 23 '15 at 02:02

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Yes but the point of having a flags enum is so you can combine them so

   typeToAdjust = type1 
   clientType = type1 | type2
   (clientType & (int)typeToAdjust) == (int)typeToAdjust // true
   clientType == (int)typeToAdjust // false

answered Sep 23 '15 at 02:02

Mark

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score 0 · Answer 3 · answered Sep 23 '15 at 02:02

The two statements are different, but the second is the one that actually tests equality. If clientType has a superset of the bits set in typeToAdjust, then the first expression will evaluate to true since (clientType & (int)typeToAdjust) just evaluates to typeToAdjust. By superset, I mean that all of the bits set in typetoAdjust are set in clientType, but maybe some additional bits as well.

score 0 · Answer 4 · answered Sep 23 '15 at 03:35

0

(clientType & (int)typeToAdjust) == (int)typeToAdjust

and clientType == (int)typeToAdjust

Are not the same

Take clientType = 3 and typeToAdjust = 1

clientType & (int)typeToAdjust = 1 = typeToAdjust but clientType != (int)typeToAdjust

answered Sep 23 '15 at 03:35

Razack

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Bitwise AND To Check Equality

4 Answers4