pointer getting reference to a return of a function

Question

Imagine we have a scope where a variable is defined and a pointer with larger scope gets this variable's reference. What happens when we leave the scope knowing that the pointer is defined in the outer scope, a class member for example.

   {
       int a = 6;
       pointa = &a; //defined out of this scope
   }
   //what happens here, pointa is still defined in this scope

EDIT: My question was concerning a more specific case which I cannot find an answer to (it feels like the answer lies within the dangling pointer explanation though).

Suppose foo() is a function that returns a double. I tried this:

    int* p;
    p = &foo();

But then p becomes empty. Can you please explain how this relates to a dangling pointer?

You could answer this question with a tiny amount of basic knowledge of the language and/or a tiny amount of common sense; either alone should be sufficient. — underscore_d, Nov 07 '17 at 16:27
UB => https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a — underscore_d, Nov 07 '17 at 16:28
Hi again guys, do you know why I got downvoted? I am new here and now it seems I cannot ask other questions because of the downvote. What should I do in this situation? Hope someone will see this comment — hamza keurti, Nov 09 '17 at 14:00

score 0 · Accepted Answer · edited Nov 07 '17 at 16:31

0

What happens when we leave the scope knowing that the pointer is defined in the outer scope, a class member for example.

The pointer becomes a dangling pointer.

Dereferencing it to access the object's members, member variables as well as member functions, will lead to undefined behavior.

edited Nov 07 '17 at 16:31

Chris Dodd

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answered Nov 07 '17 at 16:28

R Sahu

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pointer getting reference to a return of a function

1 Answers1