Return address of local variable in C

Question

Say I have the following two functions:

1

int * foo()
{
  int b=8;
  int * temp=&b;
  return temp;
}

2

int * foo()
{
   int b=8;
   return &b;
}

I don't get any warning for the first one (e.g. function returns address of a local variable) but I know this is illegal since b disappears from the stack and we are left with a pointer to undefined memory.

So when do I need to be careful about returning the address of a temporary value?

The canonical question may or may not be Can a local variable's memory be accessed outside its scope?. Though there must be one from 2008 or 2009 somewhere. — Peter Mortensen
– Peter Mortensen, Commented Dec 10, 2023 at 22:18
From 2009: Is returning a pointer to a static local variable safe? — Peter Mortensen
– Peter Mortensen, Commented Dec 10, 2023 at 22:20
"illegal" is the wrong word. If that were so it would not compile. "Unsafe" and semantically erroneous it may be, but the compiler will not stop you doing such things - in either case. — Clifford
– Clifford, Commented Dec 10, 2023 at 22:54

Community · Accepted Answer · 2017-05-23 12:00:21Z

22

The reason you are not getting a warning in the first snippet is because you aren't (from the compiler's perspective) returning an address to a local variable.

You are returning the value of int * temp. Even though this variable might be (and in this example is) containing a value which is an address of a local variable, the compiler will not go up the code execution stack to see whether this is the case.

^{Note: Both of the snippets are equally bad, even though your compiler doesn't warn you about the former. Do not use this approach.}

You should always be careful when returning addresses to local variables; as a rule, you could say that you never should.

static variables are a whole different case though, which is being discussed in this thread.

edited May 23, 2017 at 12:00

CommunityBot

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answered Jan 5, 2012 at 13:40

Filip Roséen

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2 Comments

mary Over a year ago

thanks! and the fact that When I print the return value in main it gets me the right result it actually means that it prints junk?

Filip Roséen Over a year ago

It means that you are invoking what we normally refer to as undefined behavior, ie. behavior that is unspecified in the standard.

Peter Mortensen · Accepted Answer · 2023-12-10 23:25:42Z

2

I would say that you can return the local variable or rather pointers to such if it was dynamically allocated by malloc.

In that case, the memory used for storing variable want to be on the stack, but on the heap and won't be cleared or reused after exiting the function as it takes place in the case of automatic local variables (created without malloc). Am I right?

edited Dec 10, 2023 at 23:25

Peter Mortensen

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answered Jan 5, 2012 at 14:38

user989583

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Comments

Peter Mortensen · Accepted Answer · 2023-12-10 22:33:54Z

1

This question is probably one of the most discussed on Stack Overflow. Below, are two replies from similar questions on Stack Overflow which I found interesting, and probably made me leave this bad practice of de-referencing local variables despite the chances of me getting a wrong (undefined behavior) being very slim.

Here is a simple answer

I especially liked the this reply.

Here is an example where this bad practice caused some real hardware damage

edited Dec 10, 2023 at 22:33

Peter Mortensen

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answered Apr 24, 2012 at 18:54

Abhinav

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1 Comment

Peter Mortensen Over a year ago

This is essentially a link-only answer. The second link has been deleted.

Lindydancer · Accepted Answer · 2012-01-05 13:39:20Z

0

Both examples are equally incorrect. My guess is that your compiler doesn't see the danger and thus don't issue a warning when you store the address in a temporary variable.

answered Jan 5, 2012 at 13:39

Lindydancer

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Comments

Peter Mortensen · Accepted Answer · 2023-12-10 23:35:28Z

-1

These are both bad, and a good compiler should be able to detect and warn about both situations. You may get better results by turning the warning level to maximum (which you always should do anyway), and turn on optimization.

edited Dec 10, 2023 at 23:35

Peter Mortensen

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answered Jan 5, 2012 at 13:38

harald

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4 Comments

Rob Over a year ago

The specific case, maybe. But there are lots of similar cases involving shuffling the address to and from other variables (or via other functions) that a compiler simply will not be able to detect. Also, optimisation rarely increases chances of detecting such an problem - it just increases the chance of the problem having observable effects.

harald Over a year ago

Of course there are situations the compiler can't detect, but these specific cases should be easily detectable. Optimization works because it reduces the first example to the second.

Rob Over a year ago

Sorry, not true. There are few "shoulds" when it comes to quality of implementation of a compiler - these are vendor decisions based on their own cost/benefit analysis, which may not meet developer preconceptions. It is also very easy to construct examples (e.g. a function call that may change the pointer) that means the optimisation you claim may not be feasible - which also makes it a vendor decision on whether or not to optimise such things, not a given.

harald Over a year ago

Of course, optimization is not part of the standard, and is therefore entirely up to the vendor. Any compiler not able to optimize away the uneccesary temporary in this example though, is worthless. Also if you read my answer it says "You may get better results..." Have a nice day.

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