Can a local variable's memory be accessed outside its scope?

2018-05-31 03:48:53

I have the following code.

int * foo()
{
    int a = 5;
    return &a;
}

int main()
{
    int* p = foo();
    cout << *p;
    *p = 8;
    cout << *p;
}

And the code is just running with no runtime exceptions!

The output was 58

How can it be? Isn't the memory of a local variable inaccessible outside its function?

You rent a hotel room. You put a book in the top drawer of the bedside table and go to sleep. You check out the next morning, but "forget" to give back your key. You steal the key!

A week later, you return to the hotel, do not check in, sneak into your old room with your stolen key, and look in the drawer. Your book is still there. Astonishing!

How can that be? Aren't the contents of a hotel room drawer inaccessible if you haven't rented the room?

Well, obviously that scenario can happen in the real world no problem. There is no mysterious force that causes your book to disappear when you are no longer authorized to be in the room. Nor is there a mysterious force that prevents you from entering a room with a stolen key.

The hotel management is not required to remove your book. You didn't make a contract with them that said that if you leave stuff behind, they'll shred it for you. If you illegally re-enter your room with a stolen key to get it back, the hotel security staff is not required to catch you sneaking in. You didn't make a contract with them that said "if I try to sneak back into my room later, you are required to stop me." Rather, you signed a contract with them that said "I promise not to sneak back into my room later", a contract which you broke.

In this situation anything can happen . The book can be there -- you got lucky. Someone else's book can be there and yours could be in the hotel's furnace. Someone could be there right when you come in, tearing your book to pieces. The hotel could have removed the table and book entirely and replaced it with a wardrobe. The entire hotel could be just about to be torn down and replaced with a football stadium, and you are going to die in an explosion while you are sneaking around.

You don't know what is going to happen; when you checked out of the hotel and stole a key to illegally use later, you gave up the right to live in a predictable, safe world because you chose to break the rules of the system.

C++ is not a safe language . It will cheerfully allow you to break the rules of the system. If you try to do something illegal and foolish like going back into a room you're not authorized to be in and rummaging through a desk that might not even be there anymore, C++ is not going to stop you. Safer languages than C++ solve this problem by restricting your power -- by having much stricter control over keys, for example.

UPDATE

Holy goodness, this answer is getting a lot of attention. (I'm not sure why -- I considered it to be just a "fun" little analogy, but whatever.)

I thought it might be germane to update this a bit with a few more technical thoughts.

Compilers are in the business of generating code which manages the storage of the data manipulated by that program. There are lots of different ways of generating code to manage memory, but over time two basic techniques have become entrenched.

The first is to have some sort of "long lived" storage area where the "lifetime" of each byte in the storage -- that is, the period of time when it is validly associated with some program variable -- cannot be easily predicted ahead of time. The compiler generates calls into a "heap manager" that knows how to dynamically allocate storage when it is needed and reclaim it when it is no longer needed.

The second is to have some sort of "short lived" storage area where the lifetime of each byte in the storage is well known, and, in particular, lifetimes of storages follow a "nesting" pattern. That is, the allocation of the longest-lived of the short-lived variables strictly overlaps the allocations of shorter-lived variables that come after it.

Local variables follow the latter pattern; when a method is entered, its local variables come alive. When that method calls another method, the new method's local variables come alive. They'll be dead before the first method's local variables are dead. The relative order of the beginnings and endings of lifetimes of storages associated with local variables can be worked out ahead of time.

For this reason, local variables are usually generated as storage on a "stack" data structure, because a stack has the property that the first thing pushed on it is going to be the last thing popped off.

It's like the hotel decides to only rent out rooms sequentially, and you can't check out until everyone with a room number higher than you has checked out.

So let's think about the stack. In many operating systems you get one stack per thread and the stack is allocated to be a certain fixed size. When you call a method, stuff is pushed onto the stack. If you then pass a pointer to the stack back out of your method, as the original poster does here, that's just a pointer to the middle of some entirely valid million-byte memory block. In our analogy, you check out of the hotel; when you do, you just checked out of the highest-numbered occupied room. If no one else checks in after you, and you go back to your room illegally, all your stuff is guaranteed to still be there in this particular hotel.

We use stacks for temporary stores because they are really cheap and easy. An implementation of C++ is not required to use a stack for storage of locals; it could use the heap. It doesn't, because that would make the program slower.

An implementation of C++ is not required to leave the garbage you left on the stack untouched so that you can come back for it later illegally; it is perfectly legal for the compiler to generate code that turns back to zero everything in the "room" that you just vacated. It doesn't because again, that would be expensive.

An implementation of C++ is not required to ensure that when the stack logically shrinks, the addresses that used to be valid are still mapped into memory. The implementation is allowed to tell the operating system "we're done using this page of stack now. Until I say otherwise, issue an exception that destroys the process if anyone touches the previously-valid stack page". Again, implementations do not actually do that because it is slow and unnecessary.

Instead, implementations let you make mistakes and get away with it. Most of the time. Until one day something truly awful goes wrong and the process explodes.

This is problematic. There are a lot of rules and it is very easy to break them accidentally. I certainly have many times. And worse, the problem often only surfaces when memory is detected to be corrupt billions of nanoseconds after the corruption happened, when it is very hard to figure out who messed it up.

More memory-safe languages solve this problem by restricting your power. In "normal" C# there simply is no way to take the address of a local and return it or store it for later. You can take the address of a local, but the language is cleverly designed so that it is impossible to use it after the lifetime of the local ends. In order to take the address of a local and pass it back, you have to put the compiler in a special "unsafe" mode, and put the word "unsafe" in your program, to call attention to the fact that you are probably doing something dangerous that could be breaking the rules.

For further reading:

What if C# did allow returning references? Coincidentally that is the subject of today's blog post:

http://blogs.msdn.com/b/ericlippert/archive/2011/06/23/ref-returns-and-ref-locals.aspx

Why do we use stacks to manage memory? Are value types in C# always stored on the stack? How does virtual memory work? And many more topics in how the C# memory manager works. Many of these articles are also germane to C++ programmers:

https://blogs.msdn.microsoft.com/ericlippert/tag/memory-management/

What you're doing here is simply reading and writing to memory that used to be the address of a . Now that you're outside of foo , it's just a pointer to some random memory area. It just so happens that in your example, that memory area does exist and nothing else is using it at the moment. You don't break anything by continuing to use it, and nothing else has overwritten it yet. Therefore, the 5 is still there. In a real program, that memory would be re-used almost immediately and you'd break something by doing this (though the symptoms may not appear until much later!)

When you return from foo , you tell the OS that you're no longer using that memory and it can be reassigned to something else. If you're lucky and it never does get reassigned, and the OS doesn't catch you using it again, then you'll get away with the lie. Chances are though you'll end up writing over whatever else ends up with that address.

Now if you're wondering why the compiler doesn't complain, it's probably because foo got eliminated by optimization. It usually will warn you about this sort of thing. C assumes you know what you're doing though, and technically you haven't violated scope here (there's no reference to a itself outside of foo ), only memory access rules, which only triggers a warning rather than an error.

In short: this won't usually work, but sometimes will by chance.

Because the storage space wasn't stomped on just yet. Don't count on that behavior.

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