No implicit conversion from std::string to std::string_view in C++17 (was in std::experimental::basic_string_view)

Question

My question is regarding C++17: http://en.cppreference.com/w/cpp/string/basic_string_view/basic_string_view

What's the caveat of implicit conversion from std::basic_string to std::basic_string_view that it wasn't included in the interface of the latter?

I believe it would greatly improve this class. Especially the family of comparison operators that, also do not accept std::string as neither lhs nor rhs.

There is such conversion in library fundamentals TS specification: http://en.cppreference.com/w/cpp/experimental/basic_string_view/basic_string_view

This question is about why it was removed. Or rather not adopted.

C++17 conversion goes in the opposite direction, from string_vew to string. — Cubbi
– Cubbi, Commented Aug 29, 2016 at 13:55
@Cubbi For one, that conversion is explicit. For the second, conversion to std::string would be a rather expensive way to compare std::string_view to std::string. — GreenScape
– GreenScape, Commented Aug 29, 2016 at 13:58
It was decided that this should be basic_string's responsibility. en.cppreference.com/w/cpp/string/basic_string/… — T.C.
– T.C., Commented Aug 29, 2016 at 14:02
@T.C. That pretty much makes a full answer to my question. Or rather my complete oversight. Can you make it so? I will accept. — GreenScape
– GreenScape, Commented Aug 29, 2016 at 14:04

Nicol Bolas · Accepted Answer · 2016-08-29 14:28:41Z

18

basic_string_view is considered the lower level class. It's the providers of string containers who have the responsibility of providing implicit conversions to string_view. If you have your own string type, then you would give it a possibly explicit operator string_view() overload to perform implicit conversion.

As such, it was decided (PDF) that basic_string would provide the conversion to basic_string_view. The original Library Fundamentals version put the implicit conversion on basic_string_view, because a TS is usually an extension. It can't affect an existing type without effectively forking that type.

answered Aug 29, 2016 at 14:28

Nicol Bolas

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