Does a pointer extend the lifetime of an automatic-storage variable?

Question

int main() 
{
    float* ptr;

    {
        float f{10.f};
        ptr = &f;
    }

    *ptr = 13.f;
    // Do more stuff with `*ptr`...
}

It it valid or undefined behavior to use/access *ptr?

I tested situations similar to the above example and everything seems to work as if the lifetime of the variable in the nested block was extended thanks to the pointer.

I know that const& (const references) will extend the lifetime of a temporary. Is this the same for pointers?

Where did you get the information about `const &` extending the lifetime of the referenced temporary? I think, it's not true. — Tilman Vogel, Dec 15 '13 at 13:50
@TilmanVogel http://herbsutter.com/2008/01/01/gotw-88-a-candidate-for-the-most-important-const/ — Vittorio Romeo, Dec 15 '13 at 13:51
"Does a pointer extend the lifetime of an automatic-storage variable?" - no. — , Dec 15 '13 at 13:51
@TilmanVogel It **is** true. A pointer, however, is **not** a reference. — , Dec 15 '13 at 13:52
The connection with references is tenuous at best. Your question is not about temporary values. — Kerrek SB, Dec 15 '13 at 14:01

Shoe · Accepted Answer · 2013-12-15T14:06:06.717

6

It's undefined behavior because you are accessing an object that has been deallocated.

The variable f is declared within that specific block of scope. When the execution flow reaches:

*ptr = 13.f;

the object has been deallocated and ptr points to the old address of f.

Therefore no, the lifetime of f has not been extended.

edited Dec 15 '13 at 14:06

answered Dec 15 '13 at 13:48

Shoe

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score 5 · Answer 2 · answered Dec 15 '13 at 13:49

5

The float will go out of scope and your pointer will reference a non-allocated memory region -> using it will lead to UB.

answered Dec 15 '13 at 13:49

rems4e

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Does a pointer extend the lifetime of an automatic-storage variable?

2 Answers2