Forcing std::vector overload instead of int overload on list with one element

Question

Consider the code below:

#include 
#include 

void f(std::vector v) {std::cout << __PRETTY_FUNCTION__ <<         


        

Answer 1

Forcing std::vector overload

int main()
{
    f(std::vector<int>{42}); // the vector overload is being picked up now
}

Why isn't the vector(initializer_list) constructor being picked up?

Assume that another header declares a void f(std::set<int> v).

How would you like the compiler to react when faced with f({1}): construct a vector or construct a set?

Answer 2

Braced initializer has no type, we can't say {42} is an int or std::initializer_list<int>. When it's used as an argument, special rules for overload resolution will be applied for overloaded function call.

(emphasis mine)

• Otherwise, if the parameter type is not a class and the initializer list has one element, the implicit conversion sequence is the one required to convert the element to the parameter type

{42} has only one element with type int, then it's exact match for the overload void f(int). While for void f(std::vector<int>) a user-defined conversion is needed. So void f(int) will be picked up here.

Is there any way of forcing the compiler to pick the std::vector overload (without explicitly constructing std::vector<int>{42}) even on 1-element lists?

As a wordaround, you can put additional braces to force the compiler construct a std::initializer_list<int> and then pick up void f(std::vector<int>):

f({{42}});

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