Pointer allocation and Memory Location-CodePudding

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;
   }

CodePudding user response：

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.

CodePudding user response：

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.

CodePudding user response：

This is a simple code to show you if pointer is dangling

#include <stdio.h>
 
struct Books {
   int   book_id;
};

struct Books* func1 () {
    struct Books Book1;
    Book1.book_id = 10;
    struct Books* b = &Book1;
    return b;
}

void printBook( struct Books *book ) {
   printf( "book_id : %d\n", book->book_id);
}

int main( ) {
    struct Books* Book1 = func1();
    printf("comment me\n");
    printBook(Book1);

    return 0;
}