Question

When I’m reading the C++ standard, it seems that the following code is perfectly fine according to the standard.

int main() {
   goto lol;
   {
      int x;
lol:
      cout << x << endl;
   }
}

// OK

[n3290: 6.7/3]: It is possible to transfer into a block, but not in a
way that bypasses declarations with initialization. A program that
jumps from a point where a variable with automatic storage duration is
not in scope to a point where it is in scope is ill-formed unless the
variable has scalar type, class type with a trivial default
constructor and a trivial destructor, a cv-qualified version of one of
these types, or an array of one of the preceding types and is declared
without an initializer.

Why should it work? Isn’t it still dangerous to jump over its definition and use undefined x? And why should the existence of an initializer make any difference?

Answer 1

You’d use an uninitialized x anyways, since int x; is as uninitialized as it’s going to get. The existence of an initializer makes of course a difference, because you’d be skipping it. int x = 5; for example initializes x, so it would make a difference if you jumped over it or not.