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Re: [Scheme-reports] procedure identity
John Cowan <cowan@x> writes:
> Taylan Ulrich B. scripsit:
>> eq? is a (typically) more efficient eqv?,
> I have yet to be convinced that `eq?` actually is more efficient
> than `eqv?` on the safe types.
I suppose type-inference can reduce many uses of eqv? to eq?, and when
inference fails, the difference is merely a dispatch on the type-tag.
It still cannot be as efficient in absolutely all cases though, can it?
Perhaps the total cost of globally substituting eqv? for eq? is not as
big as I would've imagined, given a sufficiently intelligent compiler.
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