Comments on: How to Make SFINAE Pretty – Part 1: What SFINAE Brings to Code

By: Dusan Jovanovic

Dusan Jovanovic — Tue, 24 Jul 2018 07:41:00 +0000

Good and sober text on SFINAE … Although, one might ask if SFINAE was such a good concept why do we need (c++20) concepts, at all 🙂

By: Nick Harris

Nick Harris — Sun, 27 May 2018 00:12:00 +0000

In reply to Jonathan Boccara. I believe so!

By: Jonathan Boccara

Jonathan Boccara — Fri, 25 May 2018 11:12:00 +0000

In reply to Nick Harris. I hadn't considered this approach! So do you mean that the template function is SFINAEd away when it is a redefinition of the non-template one?

By: Jonathan Boccara

Jonathan Boccara — Fri, 25 May 2018 11:04:00 +0000

In reply to Liquidify. The result of enable_if is void when the condition is true. Otherwise it fails to be instantiated at all, because it falls back to the definition of enable_if that doesn't have an underlying type. Does that make sense?

By: Jonathan Boccara

Jonathan Boccara — Fri, 25 May 2018 11:02:00 +0000

In reply to Michael Ellery. You are absolutely right Michael. I fixed this error, thanks.

By: Nick Harris

Nick Harris — Tue, 22 May 2018 18:23:00 +0000

So what am I missing here? Simply forcing the rvalue version of f to be a template
allows the compiler to distinguish properly between f and f as
the previous function prevents f from being created. So
wouldn’t all that is needed be:

template
class MyClass
{
public:

void f(T const &t) { cout << "lvalue" << endl; }
template
void f(T &&t) { cout << "rvalue" << endl; }
};

Using this template:

int i = 0;
MyClass{}.f(i); //prints lvalue
MyClass{}.f(move(i)); //prints rvalue
MyClass{}.f(i); //print lvalue
MyClass{}.f(move(i)); //compiler error due to T=const int& requires lvalue

By: Liquidify

Liquidify — Sat, 19 May 2018 05:26:00 +0000

“If that Condition is false, enable_if has no underling type.”

If the condition is false then doesn’t the result of enable_if has a type of void?

By: Michael Ellery

Michael Ellery — Fri, 18 May 2018 18:18:00 +0000

great summary/intro. BTW, your first two examples with `std::is_reference` have it as a positive check instead of a negative, and then after that you switch to negating the check (which I expected all along). I wasn’t sure if I was following incorrectly or probably I just missed something about the first two examples. Thanks again.

By: duibudui

duibudui — Fri, 18 May 2018 13:05:00 +0000

In reply to duibudui. Thanks!

By: Jonathan Boccara

Jonathan Boccara — Fri, 18 May 2018 11:41:00 +0000

In reply to duibudui. Yes, that's what I meant indeed!