Print the address or pointer for value in C

I want to do something that seems fairly simple. I get results but the problem is, I have no way to know if the results are correct.

I'm working in C and I have two pointers; I want to print the contents of the pointer. I don't want to dereference the pointer to get the value pointed at, I just want the address that the pointer has stored.

I wrote the following code and what I need to know is if it is right and if not, how can I correct it.

/* item one is a parameter and it comes in as: const void* item1   */
const Emp* emp1 = (const Emp*) item1; 

printf("\n comp1-> emp1 = %p; item1 = %p \n", emp1, item1 );

While I'm posting this (and the reason it is important that it is correct) is that I eventually need to do this for a pointer-to-a-pointer. That is:

const Emp** emp1 = (const Emp**) item1; 

Solution 1:

To print address in pointer to pointer:

printf("%p",emp1)

to dereference once and print the second address:

printf("%p",*emp1)

You can always verify with debugger, if you are on linux use ddd and display memory, or just plain gdb, you will see the memory address so you can compare with the values in your pointers.

Solution 2:

What you have is correct. Of course, you'll see that emp1 and item1 have the same pointer value.

Solution 3:

I believe this would be most correct.

printf("%p", (void *)emp1);
printf("%p", (void *)*emp1);

printf() is a variadic function and must be passed arguments of the right types. The standard says %p takes void *.

Solution 4:

Since you already seem to have solved the basic pointer address display, here's how you would check the address of a double pointer:

char **a;
char *b;
char c = 'H';

b = &c;
a = &b;

You would be able to access the address of the double pointer a by doing:

printf("a points at this memory location: %p", a);
printf("which points at this other memory location: %p", *a);