Pointer allocation and Memory Location

Question

In an Microprocessor it is said that the local variables are stored in stack. In my case if func1() is called by main function then the local variable (int a = 12;)will be created in stack. Once the Called Function is executed the and return back to main function the stack memory will be deleted. So the pointer address still holds (*b) the value 12. At stack if this 'a = 12' is deleted then 'b' should be a dangling pointer no?? Can anyone explain this ? If you have detailed explanation on what happens in memory when this code is being executed it would be helpful.

#include <stdio.h>
int* func1(void);
int main()
{
    int* b = func1();
    printf("%d\n",*b);
}

int* func1(void)
   {
       int a = 12;
       int* b = &a;
       return b;
   }

Is your question whether the pointer is dangling or not? – Ait-Gacem Nabil Feb 26 '22 at 08:51 — Ait-Gacem Nabil, Feb 26 '22 at 08:51

Armandas · Answer 1 · 2022-02-26T09:38:54.870

The pointer is dangling. The memory may still hold the previous value, but dereferencing the pointer invokes undefined behaviour.

GCC will give you a warning about this, if you pass -Wall option.

From the C standard (6.2.4):

The lifetime of an object is the portion of program execution during which storage is guaranteed to be reserved for it. An object exists, has a constant address,25) and retains its last-stored value throughout its lifetime.26) If an object is referred to outside of its lifetime, the behavior is undefined. The value of a pointer becomes indeterminate when the object it points to reaches the end of its lifetime.

score 0 · Answer 2 · answered Feb 26 '22 at 09:52

There are multiple layers here.

First, is the C programming language. It is a language. You say stuff in it and it has meaning. There are sentences that have meaning, but you can also construct grammatically valid sentences that are gibberish.

The code you posted, grammatically valid, is gibberish. The object a inside func1 stops existing when the function returns. *b tries to access an object that does not exists anymore. It is not defined what should happen when you access an object after its lifetime ended. You can read about undefined behavior.

Memory exists. It's not like it is disintegrated when a function returns. It's not like RAM chips are falling out of your PC. They are still there.

So your compiler will produce some machine instructions. These machine instructions will execute. Depending solely on your compiler decisions (the code is undefined behavior, compiler can do what it wants) the actual code can behavior in any way the compiler decides to. But, most probably, *b will generate machine instructions that will read the memory region where object a resided. That memory region may still hold the value 12 or it may have been overwritten somewhere between func1 returning and calling printf, resulting in reading some other value.

At stack if this 'a = 12' is deleted then 'b' should be a dangling pointer no?

Yes.

what happens in memory when this code

This depends on the actual machine instructions generated by the compiler. Compile your code and inspect the assembly.

score 0 · Answer 3 · answered Feb 27 '22 at 08:37

Strictly speaking, the behaviour is undefined, but repeat the printf or otherwise inspect (in your debugger for example) *b after the printf. It is likely that it will no longer be 12.

It is not accurate to say that the stack memory is "deleted". Rather the stack pointer is restored to the address it had before the call was invoked. The memory is otherwise untouched but becomes available for use by subsequent calls (or for passing arguments to such calls), so after calling printf it is likely to have been reused and modified.

So yes, the pointer is "dangling" since it points to memory that is no longer "valid" in the sense that it does not belong to the context in which the pointer exists.

Pointer allocation and Memory Location

3 Answers3