Gotta watch out for precedence when dealing with the bitwise ops, but there's no precedence issue in the code you posted. 

a & b == c     # XXX, means a & (b == c)
(a & b) == c   # ok

[download]

Just trying to guess at your problem since the code you posted does not exhibit the problem you describe. (It exhibits a strict error.)

In reply to Re: bitmask check by ikegami
in thread bitmask check by wwe

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