C++14 combining generic lambdas and variable templates

Question

I know about generic lambdas, and I know about variable templates, but, what does this do? Is it even allowed?

template<typename T>
auto f = [](auto a, T b){ /**/ };

If it's allowed, can it be used as expected? That is, as f<type>(var_a, var_b)?

Did you find this code somewhere, or are you just making stuff up? — sehe, Oct 06 '13 at 21:49
http://stackoverflow.com/questions/17233547/how-does-generic-lambda-work-in-c14 — Brandon, Oct 06 '13 at 22:16
@Simple I don't know, I thought it was a good question - I'd like some explanation too. — LB--, Oct 07 '13 at 13:59

Simple · Accepted Answer · 2013-10-07T10:56:18.377

A variable template must be declared constexpr. A lambda cannot occur in a constant-expression, so the initialisation is not allowed, and its operator() is not declared constexpr, so calling it isn't allowed.

In summary, this is ill-formed in the current C++14 draft.

Note: curiously, even though a lambda-expression cannot occur in a constant-expression, it seems that the closure type of a lambda may have a constexpr copy/move constructor.

score 1 · Answer 2 · answered May 20 '14 at 14:20

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This code is legal in the current draft of C++14 now and it compiles fine with clang 3.5 trunk

answered May 20 '14 at 14:20

Chandra Shekhar Kumar

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C++14 combining generic lambdas and variable templates

2 Answers2