Re: [boost] RFC Expression Validity Asserts

2015-01-23 10:11 GMT+08:00 Matt Calabrese :

On Thu, Jan 22, 2015 at 5:24 PM, TONGARI J wrote:

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2015-01-23 8:26 GMT+08:00 Matt Calabrese :

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On Thu, Jan 22, 2015 at 4:12 PM, Matt Calabrese wrote:

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On Thu, Jan 22, 2015 at 4:09 PM, Matt Calabrese wrote:

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For instance, you can use it to easily do:

template <class T> struct has_preincrement_operator : std::integral_constant {};

Hmm, actually this use doesn't seem to work, though I'm not sure why... I should have tested it in the browser before posting the sample. I'm pretty sure it fails due to a compile-bug but I'm not entirely certain.

I'm sure that lambda cannot be used in unevaluated context.

Right, but the magic of this trick is that it's not actually using the lambda in an unevaluated context (at least that's my understanding, and I've searched the standard for all instances of the term unevaluated)! A lambda itself never actually appears in a typeid/decltype/sizeof/alignof/noexcept. I do the tricks indirectly, using the lambdas just as a means to introduce declarations inside a nested scope of the expression and to create nested templates via generic lambdas for SFINAE. No lambda itself is actually invoked or directly used in an unevaluated context. I deduce information that the lambdas contain indirectly.

Anyway, what is interesting is that the following *does* compile, as I thought it would:

////////// using T = std::integral_constant; //////////

The above uses the lambda internally in the same way as the previous email, but without failure. I just pass the result of the macro invocation as a compile-time bool template argument, with the main difference being that I use "int" here instead of a dependent type. I don't immediately see why that difference should cause failure. As far as I can tell, they should both compile, or if I'm wrong somewhere, it seems that NEITHER should compile.

You're right, I misunderstood the situation. Your code doesn't work for clang, it seems to be more strict than g++: "error: a lambda expression may not appear inside of a constant expression" I suspect it's also a bug, since a hand-written functor does compile.