Thread: Pointing to the next character in an array?

Print the address of the variable in question, and then print the address of the variable + 1, and you will have your answer.


Quzah.

Originally Posted by quzah

Print the address of the variable in question, and then print the address of the variable + 1, and you will have your answer.


Quzah.

What's the best way to do this? Pretty much any way I try it, I get what I expected -- the compiler automatically takes into account the size of the variable. However, now I'm curious. Here's my code:

Code:

#include <stdio.h>
int main ()
{
	int myArray[] = {5, 6, 7, 8};
	int * myPointer = &myArray[0];
	printf("myPointer:\t0x%08lX\n", (unsigned long)myPointer);
	printf("myPointer+1:\t0x%08lX\n", (unsigned long)(myPointer+1));
	return 0;
}

If I remember correctly, the size of a pointer, if you will, is dependent on the architecture? A 32-bit architecture has a pointer (i.e. address) size of 4 bytes, while a 64-bit architecture will have a pointer size of 8 bytes?

In other words, the maximum possible address location on a 32-bit architecture would be 0xFF FF FF FF, and on a 64-bit architecture it would be 0xFF FF FF FF FF FF FF FF.

Is that correct?

Then, further, is it safe to always use unsigned long for the size of a pointer? In other words, is the size of the unsigned long variable type interpreted differently on different machines with different architectures? Is this size determined at compilation or at runtime?

Thank you in advance.

Edit: I just discovered %p as a printf() specifier, so I'll look into that. But my previous curiosity still holds! Any advice on either part is appreciated.

Use %p and type cast to (void *) to print addresses. The address of the next member in an array of type will be current address + sizeof( type ).


Quzah.

Edit: nevermind. I couldn't get the %x version to work the same as the %p version, but now I have:

Code:

#include <stdio.h>
int main ()
{
	int myArray[] = {5, 6, 7, 8};
	int * myPointer = &myArray[0];

	printf("myPointer:\t0x%08lx\n", (unsigned long)myPointer);
	printf("myPointer:\t%p\n", (void*)&myArray[0]);
	return 0;
}

Notice the difference between printing the value of a pointer (in this case converted to unsigned long) and printing the value of the address of a pointer. It is similiar to the difference between printing the value of an int variable and printing the address of the int variable.

Yes, I made a mistake, I thought the %p took care of that for me (I thought I was supposed to pass the pointer's address to it for some reason, which makes no sense in retrospect).
I edited my above post, thanks.

Edit: any recommendations or advice?

Code:

#include <stdio.h>
int main ()
{
	int myArray[] = {5, 6, 7, 8};
	int * myPointer = &myArray[0];

	printf("Print address of beginning of array of 4-byte integers:\n");

	// print address of myArray[0]
	printf("&myArray[0]:\t0x%08lx\n", (unsigned long)myPointer);
	printf("&myArray[0]:\t%p\n", (void*)&myArray[0]);
	printf("&myArray[0]:\t%p\n", (void*)myPointer);

	// print address of next element of the array
	printf("&myArray[1]:\t0x%08lx\n", (unsigned long)(myPointer+1));
	printf("&myArray[1]:\t%p\n", ((void*)&myArray[0])+sizeof(int));
	printf("&myArray[1]:\t%p\n", ((void*)myPointer)+sizeof(int));
	return 0;
}