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Subject: Re: [boost] [dynamic_bitset] Comparison operators without assertions
From: Edward Diener (eldiener_at_[hidden])
Date: 2017-05-11 05:11:30

On 5/11/2017 12:24 AM, degski via Boost wrote:
> On 11 May 2017 at 04:15, Edward Diener via Boost <boost_at_[hidden]>
> wrote:
>> On 5/10/2017 8:28 PM, Peter Bartlett via Boost wrote:
>> Or do you mean to suggest that your second choice would compare the last
>> min(N,M) bits from left to right to determine ordering rather than the
>> current right to left ?
> Seems to me that comparing bitsets of different sizes makes little 'sense',
> but, having said that, having an order (any) in bitsets of different size
> is the right thing to have, as it allows for putting those bitsets in
> std::(multi_)map/std::(multi_)set. The choice of Left/Right or Right/Left
> should I think be guided by what is the cheapest way to compare them, as
> the 'most significant (to the user) bits' might be on the left, on the
> right, in the middle or even on both ends. I would go for memcmp().

There is no reason to change the right-to-left order as the most
significant bit is on the right. Furthermore it would be foolish to
break backward compatibility on a whim. I will probably keep the same
order but just add that the longer size, the rest being equal, will
always be considered greater than the shorter size. This will eliminate
the assert and, hopefully, will not break any backward compatibility as
the assert itself should never have been expected in normal code. The
only other possibility that I can imagine, if the assert is eliminated,
is to throw some exception if the sizes are not equal, which at least
has the possibility of keeping the program running if the exception is

> degski

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