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Re: Unnecessary Boolean Warning
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Re: Unnecessary Boolean Warning

  • Subject: Re: Unnecessary Boolean Warning
  • From: Charles Srstka <email@hidden>
  • Date: Wed, 03 Aug 2011 01:43:19 -0500
On Aug 2, 2011, at 7:57 PM, Dale Miller wrote:

> It is disconcerting that if A = 0x'0110' and B = '1001' then A & B returns true but A && B returns 0, so "if (A && B)' is executed, the 'true' leg is not taken

Don’t you have that backwards? Assuming B was supposed to be hex, i.e. 0x1001, then A & B would be 0, whereas A && B would be true.

Charles_______________________________________________

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