vsoftco vsoftco - 1 month ago 9
C++ Question

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

Consider the code below:

#include <iostream>
#include <vector>

void f(std::vector<int> v) {std::cout << __PRETTY_FUNCTION__ << std::endl;}
void f(int n) {std::cout << __PRETTY_FUNCTION__ << std::endl;}

int main()
{
f({42}); // the int overload is being picked up
}


Live on Coliru

I was a bit surprised to realize that in this case the int overload is being picked up, i.e. the output of the program is:


void f(int)


with the warning


warning: braces around scalar initializer [-Wbraced-scalar-init] f({42});


Of course this happens only when I pass a 1-element list as an argument, otherwise the
std::vector
overload is being picked up.

Why is
{42}
treated like a scalar and not like a init-list? 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?


PS: The
std::vector
has an init-list constructor

vector(std::initializer_list<T> init, const Allocator& alloc = Allocator());


see (7) from cppreference.

Answer

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.