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() {
   std::cout << str; // zero-initialised, only!
}

int main() {}

That’s because t is initialised before str. I’d expect str to hold the value (char const*)0 due to zero-initialisation. My interpretation of [C++11: 3.6.2/2] supports this.

However, the above snippet appears to output the string as expected (and I confirmed the behaviour by also printing the pointer’s value).

Is there some rule of static initialisation that I’m missing here, that allows str to be value-initialised before t begins construction? Where is it in the standard?

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^{This came up on static variable resolution at build time, where an answerer asserted that using char const* rather than std::string for a static global avoids the static initialisation order fiasco. I disagreed, but now I’m not so sure…}

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.