What happens when you deallocate a pointer twice or more in C++?

Question

int main(){
Employee *e = new Employee();

delete e;
delete e;
...
delete e;
return 0;

}

Answer 1

e's not a reference, it's a pointer. You get undefined behaviour if you try to delete an object through a pointer more that once.

This means that pretty much anything can happen from 'appearing to work' to 'crashing' or something completely random.

Answer 2

It's not safe, and it's undefined what might actually happen:

http://www.parashift.com/c++-faq-lite/freestore-mgmt.html#faq-16.2

Answer 3

You are likely venturing into 'undefined behavior' territory.

On many systems this will cause a crash; for example, on my Linux machine:

*** glibc detected *** ./cctest: double free or corruption (fasttop): 0x0000000000d59900 ***
======= Backtrace: =========
/lib/libc.so.6[0x7f399f4cbdd6]
/lib/libc.so.6(cfree+0x6c)[0x7f399f4d074c]
./cctest[0x400a7a]
/lib/libc.so.6(__libc_start_main+0xfd)[0x7f399f474abd]
./cctest[0x400959]

Answer 4

Aside from the old saw about "undefined behavior" meaning anything could happen from nothing to a gateway to the seventh circle of the inferno opening up in main memory, in practice what will usually happen in most implementations is that the program will continue to run past the deletes, and then mysteriously crash sometime later in some unrelated memory allocation.

Answer 5

It's undefined behavior, so anything can happen.

What's likely to happen is bad. Typically, the free store is a carefully managed system of free and allocated blocks, and new and delete do bookkeeping to keep everything in a consistent state. If you delete again, the system is likely to do the same bookkeeping on invalid data, and suddenly the free store is in an inconsistent state. This is known as "heap corruption".

Once that happens, anything you do with new or delete may have unpredictable results, which can include attempting to write outside the application's memory area, silently corrupting data, erroneously thinking there's no more memory, or double or overlapping allocation. If you're lucky, the program will crash soon, although you'll still have problems figuring out why. If you're unlucky, it will continue to run with bad results.

Answer 6

If you're really lucky it will crash. What normally happens is it stores up karma until your CEO is demonstrating the code to your most important new customer when it will corrupt/destroy all of their data.

In checked or debug builds often this kind of thing is caught, but it can go completely undetected and cause havoc later. This is especially profound when multiple threads get involved.

Answer 7

I assume this question is related to garbage collection. Like what happens if one function deletes, and another one later on deletes it? It might create hard to find bugs or other problems.

I can imagine a situation where you do something like:

void doSomething(int *p)
{
 //.... do something....
 delete p;
}

int main()
{
 int *p = new int;
 //...
 doSomething(p);
 delete p;
 return 0;
}

Answer 8

Yeah, it's 'undefined' like everyone says, but if you want to be a real jerk about it, i'd say it depends on what happens inside the destructor of the object your are deleting. What if it calls exit()? I'd say calling delete on it multiple times is a-ok!

Answer 9

If you are worried this might happen in your apps, either stop using raw pointers completely, so that you don't need delete (eg switch over to shared_ptr) or always set pointers to NULL (or 0, or better still nullptr) after you delete them. Calling delete on a null pointer is guaranteed to do nothing.