Is This Actually Ambiguous?

问题

So I am aware that braces in code can mean more than just an initializer_list: What Is a Curly-Brace Enclosed List If Not an intializer_list?

But what should they default to?

For example, say that I define an overloaded function:

void foo(const initializer_list<int>& row_vector) { cout << size(row_vector) << "x1 - FIRST\n"; }
void foo(const initializer_list<initializer_list<int>>& matrix) { cout << size(matrix) << 'x' << size(*begin(matrix)) << " - SECOND\n"; }

If I call foo({ 1, 2, 3 }) the 1^st will obviously be called. And if I call foo({ { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }) the 2^nd will obviously be called.

But what if I call:

foo({ { 1 }, { 2 }, { 3 } })

Are those nested braces int-initializers or initializer_list<int> intializers? gcc says it's ambiguous But if you take that code and run it on http://webcompiler.cloudapp.net/ Visual Studio says it's just constructing an initializer_list<int>. Who's right? Should there be a default here?

回答1:

The rule is in [over.ics.list]:

Otherwise, if the parameter type is std::initializer_list<X> and all the elements of the initializer list can be implicitly converted to X, the implicit conversion sequence is the worst conversion necessary to convert an element of the list to X, or if the initializer list has no elements, the identity conversion.

In both your overloads, the worst conversion necessary is the identity conversion. So we have two implicit conversion sequences with rank identity. There is a rule in [over.match.best] which prefers a list-initialization of a std::initializer_list<X> over alternatives (so std::initializer_list<int> is preferred to int for {1}), but there nothing to suggest that this rule should apply recursively.

Since there's nothing to disambiguate the two conversion sequences, the call is ambiguous. gcc and clang are correct to reject.

来源：https://stackoverflow.com/questions/38795374/is-this-actually-ambiguous