g++ and clang++ different behaviour with recursive initialization of a static member

Question

Given the following code:

#include <iostream>  template <std::size_t N> struct foo  { static std::size_t value; };  template <> std::size_t foo<0>::value = 0u;  template <size_t N> std::size_t foo<N>::value = 1u + foo<N - 1u>::value;  int main()  {    std::cout       << foo<3u>::value << ' '       << foo<2u>::value << ' '       << foo<1u>::value << ' '       << foo<0u>::value << std::endl;  } 

where the static member value of the template struct foo is recursively initialized, I get different outputs from g++:

3 2 1 0 

and from clang++:

1 1 1 0 

So seem that g++ initializes foo<N>::value recursively using the initialized value of foo<N-1u>::value where clang++ uses zero for foo<N-1u>::value.

Two questions:

1. is the preceding code legit or is it Undefined Behavior in some way?
2. if the preceding code is legit, who's right: g++ or clang++?

Answer 1

It is unspecified. Both compilers are right.

Here are the relevant pieces from cppreference "initialization".

Static initialization

For all other non-local static and thread-local variables, Zero initialization takes place

So for all these variables, they are zero when the program loads. Then:

Dynamic initialization

After all static initialization is completed, dynamic initialization of non-local variables occurs in the following situations:

1) Unordered dynamic initialization, which applies only to (static/thread-local) class template static data members and ... that aren't explicitly specialized.

And these variables match the criteria. And then it says:

Initialization of such static variables is indeterminately sequenced with respect to all other dynamic initialization ....

Which means that any sequence of initialization is fine. Both compilers are correct.

To avoid the issue use constexpr to force a "constant initialization" instead.