Why does braced-init-list behave differently in a function call versus a constructor invocation?

Question

Compiling the following code with clang 3.5.0 and gcc 4.9.1 yields an error at the last statement.

#include <iostream>

struct Foo { Foo(int x, int y) { std::cout << "Foo(int = " << x << ", int = " << y << ")" << std::endl; } };

void bar(int x, int y) { std::cout << "bar(int = " << x << ", int = " << y << ")" << std::endl; }

int main()
{
   Foo({}, {});   // Foo(int = 0, int = 0)
   Foo({1}, {2}); // Foo(int = 1, int = 2)
   Foo({1, 2});   // Foo(int = 1, int = 2)

   bar({}, {});   // bar(int = 0, int = 0)
   bar({1}, {2}); // bar(int = 1, int = 2)
   bar({1, 2});   // error: no matching function for call to 'bar'  <<< Why? <<<
}

Why is Foo({1, 2}) okay while bar({1, 2}) is not?

Particularly, it would be great to learn about the rationale.

Answer 1

Foo({1,2}) creates a temporary Foo object and calls the copy constructor.

See this modified example with copy constructor delete: http://coliru.stacked-crooked.com/a/6cb80746a8479799

It errors with:

main.cpp:6:5: note: candidate constructor has been explicitly deleted
    Foo(const Foo& f) = delete;

Answer 2

The line

bar({1, 2});

actually passes in the bar function, a temporary object of type

<brace-enclosed initializer list> // it's made clear in the comments that brace initializers have no type

and there is no way to convert this temporary object to the type of the first argument which is an int. Thus the error

cannot convert <brace-enclosed initializer list> to int for argument 1 to

void bar(int, int)