What is the operator "" in C++?

Question

I fell on this page where the author talks about the standardisation of the operator "":

The decision of the C++ standards committee to standardise operator "" was [...]

What is he/she talking about? I can't find any information about this, and I don't understand what it could imply (overload for constant strings? Or something more conceptual, that doesn't affect the final use of the language?)

score 68 · Accepted Answer · edited Mar 16 '17 at 17:33

Those are user-defined literals. They allow you to create stuff like std::string, std::chrono::durations or any user defined type (you can make your own literals) in place:

auto str = "Hello"s; // str is std::string("Hello")
auto sec = 5s;       // sec is 5 std::chrono::seconds

A list of the literal-operators provided by the standard library and their documentation can be found at the bottom of the documentation page I linked.

score 3 · Answer 2 · edited Mar 16 '17 at 17:34

3

It's the user-defined literal operator which will allow the introduction of new literal syntax based on existing literals.

For more information, show this reference link.

edited Mar 16 '17 at 17:34

Peter Mortensen

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answered Mar 16 '17 at 12:35

msc

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@BaummitAugen I think `36.0_mi` is a user defined literal, isn't it? – Peter - Reinstate Monica Mar 16 '17 at 16:25
@PeterA.Schneider: The answer's examples were `128u`, `128l`, `128.0f`, and `0xBAD`. – user2357112 Mar 16 '17 at 16:52
@PeterA.Schneider Yeah, the examples I was talking about were removed in the meantime. – Baum mit Augen Mar 16 '17 at 17:53

What is the operator "" in C++?

2 Answers2

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