std::string(count, ch) not called when using uniform initialization

Question

I just got caught by this seemingly innocent attempt to call the std::string constructor that takes a size and a character value, using uniform initialization syntax:

std::string s{ 10, '\0' };

I thought this would create a string with the length of 10, initialized with \0.

Though it actually calls the constructor that takes an initializer list and thus creates a string with the length of 2, initialized with {'\n', '\0'}!

See demo at Coliru.

Is there a way to avoid this pitfall when using uniform initialization? Or do I just have to be careful?

Note: Similar question has been asked before, but no answer was given on how to avoid that pitfall.

I think the wikipedia entry on uniform initialization describes it nicely: "Uniform initialization does not replace constructor syntax, which is still needed at times. If a class has an initializer list constructor (TypeName(initializer_list<SomeType>);), then it takes priority over other forms of construction, provided that the initializer list conforms to the sequence constructor's type." — clcto
– clcto, Commented May 10, 2017 at 22:03
@clcto Compiler seems to be eager to use initializer list constructor, even if the list consist of non-conforming types (Coliru example). To be fair there is a compiler warning. — zett42
– zett42, Commented May 10, 2017 at 22:09
@zett42: For the record, that behaviour is guaranteed by [over.match.list]/1; basically; constructors are looked up in two phases and if any initializer-list constructors exist at all then that's what you get, for better or for worse. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented May 10, 2017 at 22:14

Lightness Races in Orbit · Accepted Answer · 2017-05-10 22:00:12Z

9

You just have to be careful.

In general, when people tell you to "use uniform initialisation so that everything is consistent and, well, uniform", ignore them. I mean, by all means use uniform initialisation, but don't believe that fairytale.

Also, ignore the name of the feature.

answered May 10, 2017 at 22:00

Lightness Races in Orbit

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2 Comments

T.C. Over a year ago

Well, the feature isn't officially named "uniform initialization" in the standard (or unicorn, for that matter).

Nicol Bolas Over a year ago

To be fair, list initialization didn't have to work out this way. It could have subsumed all forms of initialization, if the committee had not make two mistakes: the preference for initializer_list constructors and forbidding narrowing conversions. Remove both of those, and you could legitimately replace any case of () initialization with {}.

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