Inequivalence of templated types using aliased template names [duplicate]

Question

We have observed a strange behaviour in the compilation of the follwing source code:

template<template<class> class TT> struct X { }; template<class> struct Y { }; template<class T> using Z = Y<T>;  int main() {   X<Y> y;   X<Z> z;   z = y; // it fails here } 

This is a slightly modified example taken from the c++11 standard proposal for template aliases: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2258.pdf (See page 4) Also note that the proposal "declares y and z to be of the same type." In our interpretation it should therefore be possible to assign (or copy construct) z from y.

However, this code doesn't compile with gcc 4.8.1 nor with clang 3.3. Is this an error in the compiler or did we misunderstand the standard?

Thanks in advance, craffael et al ;)

P.S. The Clang error message is:

error: no viable overloaded '='  note: candidate function (the implicit copy assignment operator) not viable: no known conversion from 'X<template Y>' to 'const X<template Z>' for 1st argument template<template<class> class TT> struct X { };  note: candidate function (the implicit move assignment operator) not viable: no known conversion from 'X<template Y>' to 'X<template Z>' for 1st argument template<template<class> class TT> struct X { }; 

Answer 1

The current standard doesn't say so, but the intention is that y and z have the same type. There is an open Core Working Group issue for this: http://wg21.cmeerw.net/cwg/issue1286

Answer 2

I think you are confusing a type and a template (or a template alias). You have Y, which is one template and Z, which is another one. If you think that Y == Z, you are wrong. Only if you turn them into types, those types are the same, e.g. Y<int> is the same type as Z<int>. In your example:

template<class T> struct X { };  template<class> struct Y { }; template<class T> using Z = Y<T>;  int main() {   X<Y<int>> y;   X<Z<int>> z;   z = y; // works } 

In your original code you referred to them with X<Y> and X<Z>, but as Y is not the same as Z, so are X<Y> and X<Z> different types.