Implicit conversion in array initializer list

Question

Why implicit conversion from const char* to std::string does not work in the latter case? Link a reference to C++ standard if possible, please.

Variant 1:

struct Foo {
    Foo(const char* a) {}
};

int main() {
   // works well for a "const char*" accepting constructor
   Foo* foo = new Foo[1] { "a" };
}

Variant 2:

struct Foo {
    Foo(std::string a) {}
};

int main() {
   // could not convert from "const char*" to "Foo"
   Foo* foo = new Foo[1] { "a" };
}

There's no such thing as an implicit cast. There is a cast (explicit type conversion) and type coalescing (implicit type conversion), decide which one. — user529758
– user529758, Commented Apr 12, 2013 at 10:21
check how's the page named: en.cppreference.com/w/cpp/language/implicit_cast — Hertz
– Hertz, Commented Feb 19, 2014 at 5:53

ecatmur · Accepted Answer · 2013-04-12 11:13:23Z

6

At most one user-defined conversion is allowed in a user-defined conversion sequence (12.3p4).

You can use an extra level of braces to make it work:

   Foo* foo = new Foo[1] { {"a"} };

Note that because of a bug in clang it requires Foo to have a default constructor Foo::Foo() even though it will not actually be called.

edited Apr 12, 2013 at 11:13

answered Apr 12, 2013 at 10:26

ecatmur

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2 Comments

awesoon Over a year ago

Hmm... Looks good, works well with g++, but not with clang(LWS example)

ecatmur Over a year ago

@balki clang is broken; there's a simple workaround (add a default constructor). Reporting the bug now.

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