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Subject: Re: [boost] function convert() -- do you find it useful?
From: Krzysztof Czainski (1czajnik_at_[hidden])
Date: 2014-06-28 17:52:39


2014-06-28 22:58 GMT+02:00 Andrzej Krzemienski <akrzemi1_at_[hidden]>:

> Hi,
> This is the second time that I have found the following function useful
> when playing with variants of Optional library:
>
> template <typename T, typename U>
> typename std::enable_if<std::is_convertible<U, T>::value, T>::type
> convert(U && u)
> { return std::forward<U>(u); }
>
> It performs an implicit conversion from U to T, but you can call it
> explicitly.
>

+1

I even have such a function in my tools, and use it often.

I chose a shorter name "as", because since this only forces an implicit
conversion, I find verbosity unwanted here.

Consider preventing dangerous uses like this (not tested):
auto& x = convert<T const&>( U() );

You can prevent that by adding a static_assert in this function:
// Prevent returning an lvalue reference to a temporary
BOOST_STATIC_ASSERT( !is_lvalue_reference<T>::value ||
is_reference<U>::value );

BTW, why not get rid of the enable_if< is_convertible > part? Won't a
message "U isn't convertible to T" pointing into the function be more
readable, than a message "There's no 'convert' function"?

Regards,
Kris

PS. Look, how nice this looks (it even sounds English):
cout << boost::as<double>(2) / 3 << endl;

Compared to:
cout << boost::convert<double>(2) / 3 << endl;

Not to mention:
cout << static_cast<double>(2) / 3 << endl;
which (like Andrzej explained) brings in potentially unwanted explicit
conversions.


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