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Subject: Re: [boost] [Review:Algorithms] Order of args to clamp
From: Jeffrey Lee Hellrung, Jr. (jeffrey.hellrung_at_[hidden])
Date: 2011-09-25 19:33:47
On Sun, Sep 25, 2011 at 3:55 PM, Steven Watanabe <watanabesj_at_[hidden]>wrote:
> AMDG
>
> On 09/25/2011 03:26 PM, Jeffrey Lee Hellrung, Jr. wrote:
> > On Sun, Sep 25, 2011 at 3:15 PM, Steven Watanabe <watanabesj_at_[hidden]
> >wrote:
> >
> >> Precondition: operator< defines an ordering on objects of type U. For
> >> any
> >>> 2 objects a and b of types A and B, respectively, where A and B are
> each
> >> one
> >>> of {T,L,H}, a < b is equivalent to (U)a < (U)b. !(hi < lo).
> >>
> >> This should be !(U(hi) < U(lo)).
> >>
> >
> > Equivalent, given the preconditions, no?
> >
> > Your restrictions on the comparison are
> >> too strict. You don't want to constrain
> >> (lo < lo), (hi < hi), (hi < lo), or (lo < hi),
> >> because that would exclude T = std::string,
> >> L = H = const char*.
> >>
> >
> > I don't follow.
> > - (lo < lo) and (hi < hi) should both return false, so that clamp(x, x,
> hi)
> > and clamp(x, lo, x) both return x. I think this is desirable.
>
> Sorry. I was actually thinking in terms of the types,
> rather than the actual values.
>
> > - I think !(hi < lo) is a reasonable precondition so that the result is
> > independent of whether you first compare to lo or hi.
>
> I agree, the only question is how to express it.
>
> > How do these preconditions exclude T == std::string and L == H == char
> const
> > *? Doesn't the STL define comparison operators between these types that
> do
> > the "obvious" thing?
> >
>
> Comparing const char*'s compares the pointers,
> not the contents.
>
Ah, there's the rub. I missed that.
Now I'm troubled that certain mixes of std::string and char const * would be
bad if operator< were used by default:
clamp("x", "L", std::string("H")) // bad
clamp(std::string("x"), "L", "H") // okay, assuming current
implementation...and no asssertion that !(hi < lo)
- Jeff
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