c - passing a pointer to a function with a parameter of void* variable

Question

I want to pass a pointer to a variable. Sometimes it would be an integer and sometimes maybe a character. In the example below i pass the pointer p to CreateObject but when i try to retrieve the value of the variable the pointer points to i get an awkward result:

int i =0;
int *p = malloc(sizeof(int));
*p = i;

ObjectP object = CreateObject(p);

Say i want to cast it back to an int and display it:

void CreateObject(void *key)
{
   printf("%d\n", (int)key);
}

I get: 160637064 instead of 0. What am i getting instead of the integer i assigned previously and how do i retrieve it instead of the current value?

And, out of curiosity, how do you plan to distinguish the size of the element passed? Another parameter? — sidyll
– sidyll, Commented Dec 25, 2012 at 13:37

unwind · Accepted Answer · 2012-12-25 13:35:31Z

6

This:

(int) key

is not dereferencing the pointer to access the data it points at, it's re-interpreting the pointer value (the address) itself as the integer.

You need:

printf("%d\n", *(int *) key);

answered Dec 25, 2012 at 13:35

unwind

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Tom van der Woerdt · Accepted Answer · 2012-12-25 13:35:53Z

3

You're typecasting a pointer to an integer. This means that you simply get the address. After all, a pointer is just an address.

You probably want to dereference the pointer instead of casting it. *(int*)key would do that.

answered Dec 25, 2012 at 13:35

Tom van der Woerdt

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Comments

Martin Baulig · Accepted Answer · 2012-12-25 13:36:24Z

3

You're printing the pointer, ie. the address, not the value that it's pointing to.

Use this to print the value:

void CreateObject(void *key)
{
   printf("%d\n", *(int*)key);
}

answered Dec 25, 2012 at 13:36

Martin Baulig

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