Why is the lifetime of an object on the free store bound to scope?

Question

If I have some object pointing to NULL and i want to assign it using new in the scope of a function, why is the lifetime of this object limited to the scope?

SomeObject *o = nullptr;

void f(SomeObject *o)
{
    o = new SomeObject; //Not null
}

o->doWhatever(); //Null again

And: how can i assign someObject else then (without returning)?

Ages-old question -- you passed `o` by value into the function, so the original `o` is unaffected. — Quentin, Feb 08 '17 at 13:13

score 0 · Answer 1 · answered Feb 08 '17 at 13:14

0

The input argument that you are passing to function f has the same name as the global variable.

So the assignment inside function f applies for the local variable instead of the global variable.

answered Feb 08 '17 at 13:14

barak manos

29,648
10
62
114

I don't think `o` is in global scope. No more than the executable statement `o->doWhatever();` is. – StoryTeller - Unslander Monica Feb 08 '17 at 13:16
@StoryTeller: I see a global declaration `SomeObject *o = nullptr;` in the code. – barak manos Feb 08 '17 at 13:17
And I see it followed by a global executable statement. So either the code is ill-formed, or those aren't really globals. – StoryTeller - Unslander Monica Feb 08 '17 at 13:18
1

@StoryTeller: Yep, I understand your point. I've taken it for granted that the "global statement" was just "cut off" from some function, since the code wouldn't even compile otherwise (in opposed to the global declaration). But you are right that the code in its present state is ill-formed. – barak manos Feb 08 '17 at 13:22

Why is the lifetime of an object on the free store bound to scope?

1 Answers1