Finding Difficulty in Understanding pointers in C

Question

Not able to understand the code. The correction option is d but i m not understanding how it comes out to be d.

#include <stdio.h>
int *m();
void main()
{
    int k = m();
    printf("%d",k);
}
int *m()
{
    int a[2]={5,8};
    return a;
}

options are:

a) 5
b) 8
c) nothing
d) varies

The compiler is showing following warnings:

1)warning: return makes integer from pointer without a cast
 return a;
2)warning: function returns address of local variable [-Wreturn-local-addr]

and I am not able to interprete both.

Answer 1

It's no surprise you have trouble understanding that code, because it is very poorly written.

The main problem is that the a array in the m() function ceases to exist when m returns, so any pointer to that array is no longer valid¹, and attempting to read or write anything through an invalid pointer leads to undefined behavior.

"Undefined behavior" simply means that the code is erroneous, but the compiler isn't required to handle it in any specific way. Your program may crash outright, it may be put into a bad state that doesn't manifest until later, it may corrupt data, or it may appear to run without any issues.

So yes, the result may vary, but that's understating the case by a lot. The right answer is "E) The behavior is undefined".

But there are other problems in the code. m() returns a pointer to int, but the program assigns the result to a plain int, hence the first warning. Whoever wrote that code either doesn't understand types, or assumes that pointer values and int values are interchangeable (not necessarily a valid assumption).

Either that, or the original code is actually

int k = *m();

And, finally, main() returns int, not void, as in:

int main( void )  // or int main( int argc, char **argv ) if your program takes
                  // command line arguments.

---

1. Obviously, the memory location that a occupied still exists, but after m() exits, that memory may be overwritten or used by something else.

Answer 2

Thanks for making updates to your question, I think I understand what's going on, now. My C is a bit rusty, but here goes:

1. The method m returns an int* (pointer/address to int)

2. When m is called, it creates a as a local array on the stack. Then, it tries to return the address for a.

3. When m is completed and returns control back to main, the memory on allocated by m on the stack goes out of scope. a is among the items on that stack that has gone out of scope.

4. Because the data at the address a is out of scope, it's considered "undefined behavior" to access/dereference it, which is what main (almost) does with the printf call. This is what 2)warning: function returns address of local variable [-Wreturn-local-addr] is warning you of.

The answer, "varies", would've been more accurate as "undefined behavior".

A separate issue highlighted by the error:

1)warning: return makes integer from pointer without a cast

Is that m() returns a pointer to an int, but it's being assigned to int k.

Answer 3

The first warning (MSVC compiler)

int differs in levels of indirection from int *

is because you are trying to supply a pointer for int k in main, which is not a pointer.

The second warning

returning address of local variable or temporary: a

is because a function's local variable goes out of life after the function returns.

I would suggest you use malloc in your function to obtain memory for the array, and return that to a pointer variable in main. Having done that, you can print the first element. But I won't post the code for that, since you "copied the code as it is from the website".

Answer 4

The reason that the behavior "varies" (D) is because m() is returning a, which is a memory address. Each time you run your program, that address could change because your OS can load the program anywhere in memory that it wants.

Now if your program were doing what I think you actually want it to - to print the contents at the memory address returned by m() - that would be:

int *k = m();
printf("%d\n", *k);

But this would still result in behavior "varies" though, for the local variable issue described by the other answers.

Your compiler warning: 2)warning: function returns address of local variable [-Wreturn-local-addr] is occurring because m() is returning a, which is a local variable. Look up "local variables" on the googles for more information. This is useful: Difference between static, auto, global and local variable in the context of c and c++

Your compiler warning: 1)warning: return makes integer from pointer without a cast is because you're storing the return value of m(), which is of type int* into variable k, which is of type int. You need to make these the same type in order to get rid of the compiler warning. Look up "C pointers" on the googles for more information. This tutorial seems reasonable: http://www.programiz.com/c-programming/c-pointers