Forms of list initialization

问题

See the following code:

std::vector<int> v1{1, 2, 3};
std::vector<int> v2 = {1, 2, 3};

My questions are:

1. Is there a difference between the two? I know the first one must be list initialization, but how about the second?

2. Because there is a assign sign for the second, it makes me think that the compiler will use the std::initializer_list to create a temporary vector first, then it use copy constructor to copy the temp vector to v2. Is this the fact?

回答1:

The two (direct-list-initialization vs copy-list-initialization) are exactly the same in this case. No temporary std::vector is constructed and there's no std::vector::operator= called. The equals sign is part of the initialization syntax.

There would be a difference if std::vector's constructor overload no. 7 was marked explicit, in which case any copy-initialization fails, but that would be a flaw in the design of the standard library.

回答2:

They're both list initialization (since C++11). The first one is direct-list-initialization, and the second one is copy-list-initialization.

1. Is there any difference between the two?

For direct-list-initialization both explicit and non-explicit constructors will be considered, while copy-list-initialization only non-explicit constructors might be called. For this case (i.e. std::vector) the actual result is same.

2. Because there is a assign sign for the second, it makes me think that the compiler will use the initializer_list to create an temp vector first, then it use copy constructor to copy the temp vector to v2. Is this the fact?

No. Even for the copy-list-initialization, the object will be constructed by the appropriate constructor directly. For this case, a temporary std::initializer_list<int> will be created and then std::vector::vector(std::initializer_list<T>, const Allocator& = Allocator()) will be called to construct v2.

来源：https://stackoverflow.com/questions/38582981/forms-of-list-initialization