Single-Element-Vector Initialization in a Function Call

Question

Consider the following example code:

Example:

void print(int n) {
    cout << \"element print\\n\";
}

void print(vector

        

Answer 1

Because the 1st overload wins in the overload resolution for print({2});.

In both cases copy list initialization applies, for the 1st overload taking int,

(emphasis mine)

Otherwise (if T is not a class type), if the braced-init-list has only one element and either T isn't a reference type or is a reference type that is compatible with the type of the element, T is direct-initialized (in direct-list-initialization) or copy-initialized (in copy-list-initialization), except that narrowing conversions are not allowed.

{2} has only one element, it could be used to initialize an int as the argument directly; this is an exact match.

For the 2nd overload taking std::vector,

Otherwise, the constructors of T are considered, in two phases:

• All constructors that take std::initializer_list as the only argument, or as the first argument if the remaining arguments have default values, are examined, and matched by overload resolution against a single argument of type std::initializer_list

That means an std::initializer_list is constructed and used as the constructor's argument of std::vector (to construct the argument for print). One user-defined conversion (via the constructor of std::vector taking one std::initializer_list) is required, then it's worse match than the 1st overload.