Local C pointer variable

Question

Suppose i define the following code:

int *func()
{
    int *p=(int *)malloc(sizeof(int)); // memory is allocated from heap
                                      // which can be accessed anytime.

    return p;    // but we created p which has a local scope and vanishes as
                 //soon as function exits.  
}

Then how does this thing work? p being a local varible (a pointer holding a address to the dynamic memory). I mean the memory itself from HEAP should definitely continue to exist, but the pointer variable has a local scope. How come i will be able to get this pointer?

btw (*int) func() should be int *func() the same goes for (*int) it should be (int *) as for ur other question, its cause ur returning a copy of the memory address returned from malloc not &p that would be returning the address of p which would have being deallocated. — Samy Vilar
– Samy Vilar, Commented Aug 27, 2012 at 8:15
@GreatCoder depending on the architecture this example would not be more efficient, depending on the sizeof(pointer) 32 bit will have 4 bytes , 64 will have 8 bytes, most likely your integer will have the same size as the pointers — Oliver
– Oliver, Commented Aug 27, 2012 at 8:28
@OliverStutz Right now I am mainly concerned with the concept of scope of the varibles/pointers. You may be very right with the efficiency concept — user1581106
– user1581106, Commented Aug 27, 2012 at 8:50
@OliverStutz Can u please elaborate on the size which u mentioned? I am not able to get it? — user1581106
– user1581106, Commented Aug 27, 2012 at 8:55

paxdiablo · Accepted Answer · 2012-08-27 09:12:46Z

4

p is simply a pointer which refers to the memory block that you've allocated.

The lifetime of the allocated block extends beyond the end of the function.

It's true that the actual pointer will go out of scope on return but you've already returned a copy of it it to the caller before then.

And, as an aside, you probably meant int* (in both cases) rather than *int and (*int). In any event, you don't want to explicitly cast the return value from malloc in C - it can hide subtle errors. C is perfectly capable of converting the void * returned by malloc into any suitable pointer type.

It would be better written as something like:

int *func (void) {
    int *p = malloc (sizeof (int));
    return p;
}

You'll also notice the use of void in the function to explicitly state the function takes no parameters. That's preferable to just having () which is subtly different (an indeterminate number of parameters).

edited Aug 27, 2012 at 9:12

answered Aug 27, 2012 at 8:18

paxdiablo

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2 Comments

user1581106 Over a year ago

What is bad in conveting/typecasting the void pointer to int type explicitly?

paxdiablo Over a year ago

In systems where an int and a void pointer are different sizes, and you forget to include the header for malloc, you'll run into problems because the assumed return value of a function is an int.

user1581106 · Accepted Answer · 2012-08-29 09:19:10Z

2

Finally after understanding... I was able to get the concept.

When we return the pointer p... definitely p, which was allocated on stack vanishes.
Since values are returned with a call by value mechanism and p was holding an address of memory allocated on the heap. That address is still valid and holds reference to a valid memory location.
Thus p itself vanishes, but the value returned by it is copied to a variable in the calling function and hence this justifies the correctness of the code.

answered Aug 29, 2012 at 9:19

user1581106

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