Do objects of built-in types have special static initialisation order precedence?

Question

I\'d have expected the following code to yield a segmentation fault (or otherwise UB):

struct T {
   T();
};

T t;
char const* str = \"Test string\";

T::T()         


        

Answer 1

I think I found it; what's happening here is not so much about the built-in type, but about the constant initialiser:

[C++11: 3.6.2/2]: Variables with static storage duration (3.7.1) or thread storage duration (3.7.2) shall be zero-initialized (8.5) before any other initialization takes place.

Constant initialization is performed:

• if each full-expression (including implicit conversions) that appears in the initializer of a reference with static or thread storage duration is a constant expression (5.19) and the reference is bound to an lvalue designating an object with static storage duration or to a temporary (see 12.2);
• if an object with static or thread storage duration is initialized by a constructor call, if the constructor is a constexpr constructor, if all constructor arguments are constant expressions (including conversions), and if, after function invocation substitution (7.1.5), every constructor call and full-expression in the mem-initializers and in the brace-or-equal-initializers for non-static data members is a constant expression;
• if an object with static or thread storage duration is not initialized by a constructor call and if every full-expression that appears in its initializer is a constant expression.

Together, zero-initialization and constant initialization are called static initialization; all other initialization is dynamic initialization. Static initialization shall be performed before any dynamic initialization takes place. [..]

That final sentence would seem to override subsequent sequencing rules, making this ordering apply across Translation Units.