C++ try/catch after deleting a pointer

Question

After I searched about how to verify if a pointer is deleted in C++, i found out that there is no certain way, some really ugly work-around or smart pointers (I first want to understand how normal pointers works). My question is why C++ crashes when I try in a try/catch to show the value of a deleted pointer? shouldn't it handle the crash and just print the exception ??

void main()
{
    int *a = new int[5];

    for (int i = 0; i < 5; i++)
    {
        *(a + i) = i;
    }

    for (int i = 0; i < 5; i++)
    {
        if (*(a + i) != NULL) // useless verify, cuz if it would be NULL, it 
                              //would crash imediately
        {
            cout << (a + i) << ", " << *(a + i) << endl;
        }
    }

    delete a;
    cout << a << ", ";// << *a << endl;

    for (int i = 0; i < 5; i++)
    {
        try{
            cout << (a + i) << ", " << *(a + i) << endl;
        }
        catch (int e)
        {
            cout << "Error: " << e << endl;
        }
    }

    cin.get();
}

Does it throw exception, if not then what are you trying to catch? — Killzone Kid, May 07 '18 at 13:07
Messing up with pointers doesn't throw an exception. It just does whatever. — NathanOliver, May 07 '18 at 13:07
maybe this will help: https://stackoverflow.com/questions/23621677/what-happens-to-the-pointer-itself-after-delete — k.v., May 07 '18 at 13:08
`*(a + i) != NULL` Is useless to verify, since the value of `*(a + i)` is not a pointer, hence, you are effectively comparing such a value with a value of `0`. — Algirdas Preidžius, May 07 '18 at 13:08
Your program invokes undefined behavior. It should **not** _handle the crash and just print the exception_. Making parallels between the C++ and other languages is often counter-productive. — Ron, May 07 '18 at 13:08

score 4 · Answer 1 · answered May 07 '18 at 13:09

4

try{
        cout << (a + i) << ", " << *(a + i) << endl;
    }
    catch (int e)
    {
        cout << "Error: " << e << endl;
    }

You can catch an exception only if one is thrown. Accessing freed memory is Undefined Beahvior and does not throw an exception, so there is nothing to catch.

answered May 07 '18 at 13:09

Gaurav Sehgal

7,422
2
18
34

Thanks! easy and understandable answer :) – Andrei Bratu May 07 '18 at 21:33

Abhijit Pritam Dutta · Answer 2 · 2018-05-07T13:12:47.403

0

First for array deletion the syntax is delete [] a; secondly your try catch block is not for handling corrupted address, and unfortunate there is not exception handling for memory access violation in C++.

edited May 07 '18 at 13:12

answered May 07 '18 at 13:07

Abhijit Pritam Dutta

5,521
2
11
17

1

What about `void main`? – Killzone Kid May 07 '18 at 13:09
1

@KillzoneKid, you are true but his question is specifically regarding try/catch. – Abhijit Pritam Dutta May 07 '18 at 13:13

Kostas · Answer 3 · 2018-05-07T13:17:12.310

-1

There are no exception thrown in this case, simply undefined behavior. Exceptions are made to explicitly catch exceptions that are being thrown.

Also, use delete [] a; instead of delete a; to delete an array.

Also, when we delete memory, the memory does not actually get deleted but is free to be overwritten from now on.

What cout would print does not say much on whether the pointer was deleted.

edited May 07 '18 at 13:17

answered May 07 '18 at 13:14

Kostas

4,061
1
14
32

1

*The former only deletes the first element.* I think it is UB. – Gaurav Sehgal May 07 '18 at 13:16

C++ try/catch after deleting a pointer

3 Answers3