I personally prefer std::enable_if<>. For e.g.
template<typename T, class = typename std::enable_if<std::is_same<T,
int>::value>
Class Foo { }
I don’t like that something inside the body of a class / function would cause a
template to be enabled or not.
--
Chris Dumez
> On Aug 22, 2017, at 8:34 PM, Keith Miller <[email protected]> wrote:
>
> Hello fellow WebKittens,
>
> I’ve noticed over time that we don’t have standard way that we enable
> versions of template functions/classes (flasses?). For the most part it seems
> that people use std::enable_if, although, it seems like it is attached to
> every possible place in the function/class.
>
> I propose that we choose a standard way to conditionally enable a template.
>
> There are a ton of options; my personal favorite is to add the following
> macro:
>
> #define ENABLE_TEMPLATE_IF(condition) static_assert(condition, “template
> disabled”)
>
> Then have every function do:
>
> template<typename T>
> void foo(…)
> {
> ENABLE_TEMPLATE_IF(std::is_same<T, int>::value);
> …
> }
>
> And classes:
>
> template<typename T>
> class Foo {
> ENABLE_TEMPLATE_IF(std::is_same<T, int>::value);
> };
>
> I like this proposal because it doesn’t obstruct the signature/declaration of
> the function/class but it’s still obvious when the class is enabled.
> Obviously, I think we should require that this macro is the first line of the
> function or class for visibility. Does anyone else have thoughts or ideas?
>
> Cheers,
> Keith
>
> P.S. in case you are wondering why this macro works (ugh C++), it’s because
> if there is any compile time error in a template it cannot be selected as the
> final candidate. In my examples, if you provided a type other than int
> foo/Foo could not be selected because the static_assert condition would be
> false, which is a compile error.
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