C++98 curly brace const scalar initialization

Question

I stumbled upon the code which I do not understand. Here's a simplified version of it:

template <int> struct A {};

int const i = { 42 };
typedef A<i> Ai;

int const j = 42;
typedef A<j> Aj;

This code compiles with GCC in C++98 mode, but not in Clang. Clang produces the following error:

$ clang -Wall -Wextra -std=c++98 -c test.cpp

test.cpp:4:11: error: non-type template argument of type 'int' is not an integral constant expression
typedef A<i> Ai;
          ^
test.cpp:4:11: note: initializer of 'i' is not a constant expression
test.cpp:3:11: note: declared here
int const i = { 42 };
          ^

As far as I understand initialization of int with and without curly braces should be equivalent. Clang initializes i correctly to 42, just doesn't think it's a compile time constant.

This code compiles well in C++11 mode.

Is there a reason j is treated as a compile time constant and i is not? Or is it simply a bug in Clang?

Update: I opened a ticket in LLVM bug tracker with this issue.

Answer 1

Yes, both declarations are equivalent, per C++98 8.5/13:

If T is a scalar type, then a declaration of the form

T x = { a };

is equivalent to

T x = a;

So both variables are constant, and initialised from a constant expression, so (as far as I can see) should both be usable as constant expressions.