ychen added inline comments.
================
Comment at: clang/lib/Sema/SemaTemplateDeduction.cpp:5179-5231
if (!Better1 && !Better2) // Neither is better than the other
return JudgeByConstraints();
// C++ [temp.deduct.partial]p11:
// ... and if G has a trailing function parameter pack for which F does not
// have a corresponding parameter, and if F does not have a trailing
// function parameter pack, then F is more specialized than G.
----------------
mizvekov wrote:
> One thing I have been thinking about all of these "more constrained" checks,
> is whether we should actually be performing them over the results of the
> deductions for the least constrained checks.
>
> Ie especially with regards to this new check and the one for
> `[temp.deduct.partial]p11:`, it seems that we should be instead considering
> as more constrained the deduction which deduced a lesser amount of template
> parameters, including ones in packs. Or something along this line.
>
> This way, we do one sweeping general rule to handle all these cases, instead
> of just appending these ad-hoc tie-breaking rules which look more like
> workarounds.
>
> What do you think about that?
>
> But this is a big change and I don't want to block you on that. Because
> otherwise, this new change is keeping in line with other workarounds that
> have already been added here, so this is within existing practice.
>
> Since this patch was submitted quite recently and it's a speculative fix, I
> would let this patch marinate a little more and wait for any other reviewers.
> Ie especially with regards to this new check and the one for
> [temp.deduct.partial]p11:, it seems that we should be instead considering as
> more constrained the deduction which deduced a lesser amount of template
> parameters, including ones in packs. Or something along this line.
>
>This way, we do one sweeping general rule to handle all these cases, instead
>of just appending these ad-hoc tie-breaking rules which look more like
>workarounds.
>
> What do you think about that?
I think you meant the cases that for a `P`/`A` deduction pair, either `P`/`A`
is pack and the other is not empty, right? If so, it already works that way.
https://eel.is/c++draft/temp.deduct#type-9
If neither `P`/`A` is pack and either is empty, then they wouldn't get into the
stage of partial ordering at all because the function arguments list can only
have one length, `P`/`A` would not be in the same candidate sets.
Here handles the case that either `P`/`A` is pack and the other is empty. Then
the deduction rule could not kick in because one of the `P`/`A`pair is none. So
we have to tie-break after the deduction.
Repository:
rG LLVM Github Monorepo
CHANGES SINCE LAST ACTION
https://reviews.llvm.org/D133683/new/
https://reviews.llvm.org/D133683
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