End of lifetime of static object at block scope versus global scope

Question

In this passage on program exit from cppreference.com

If the completion of the constructor or dynamic initialization for thread-local or static object A was sequenced-before thread-local or static object B, the completion of the destruction of B is sequenced-before the start of the destruction of A

what is the meaning of "sequenced-before"?

In particular, for this program

struct Object {
  Object() {
  }
  ~Object() {
  }
};

Object a;

void f() {
  static Object b;
}

int main() {
  f();
}

is it safe to assume that a.~Object() is called after b.~Object() because a.Object() is called before b.Object()?

see also [this answer regarding sequenced-before relations](http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points/4183735#4183735) (and also the sequence *points* of yesteryear) — jaggedSpire, Jan 19 '17 at 19:49

score 0 · Accepted Answer · answered Jan 19 '17 at 20:36

what is the meaning of "sequenced-before"?

Objects are initialized at run time by the run time environment in a sequence. If initialization of one object comes before initialization of a second object, then the construction of the first object is "sequenced-before" construction of the second object.

is it safe to assume that a.~Object() is called after b.~Object() because a.Object() is called before b.Object()?

If you can assume that a.Object() is called before b.Object(), then you can assume that a.~Object() is called after b.~Object(). However, that is not always the case. In your posted code that is true. But it is possible, in a more complex application, that f() is called before a is initialized.

End of lifetime of static object at block scope versus global scope

1 Answers1