Ok, that's not exactly true. Pointers and arrays are not exactly the same entity, but they're very close. In fact, an array is for all intents and purposes a constant pointer.
What?! How can an array be a pointer, and how can a pointer be an array? Before we delve into that question, we first need to discuss pointer arithmetic.
If you have an integer variable, you can add the number 1 to it and the contents f that variable will increase by 1. You could add an 'a' to a character variable and it would increase by the value 'a'. With pointers, arithmetic is also possible, but a little quirky. Once we understand how pointer arithmetic works, however, it is an invaluable aid. In fact, as we'll see, arrays work properly because of pointer arithmetic.
As we've seen, pointers store an address in memory. If we have an integer pointer, it points to a memory location that can hold an integer. If we have a character pointer, it points to an address in memory that can hold a character. And so forth. So all a pointer really holds is a big number, say, for example, 0x4b14 (or in binary 0b0100101100010100). We might expect that if we added a number to this, say the number 1, that the pointer would then hold the number 0x4b15. Fortunately, that is not always the case.
When you add numbers to pointers, the address stored in the pointer isn't necessarily increased by that many bytes. That would cause trouble, for example, with an integer pointer. Let's say we had a series of integers in a row located at addresses 0x4b14, 0x4b18, 0x4b1b, and 0x4b1f (remember that integers on most modern machines, and the example machine we're dealing with here, are a 4-byte data type, meaning they take up 4 bytes). Let's say that we have a pointer ptr that currently holds the address 0x4b14. If we executed the instruction: ptr = ptr + 1; without special arithmetic, if ptr were just an integer variable, we would end up with the value 0x4b15. But this number doesn't make any sense in terms of memory addresses. A single integer resides at the memory addresses 0x4b14 through 0x4b17, so accessing the memory at address 0x4b15 would be accessing into the middle of an integer. Fortunately, this is not how pointer arithmetic works.
When you add a number to a pointer, the computer knows what type of data the pointer points to, and multiplies the number you're adding by the size of the pointer's type before adding it to the pointer. For example, taking the case from above, if we have the following code:
and if ptr is a pointer to an integer and originally contained the value 0x4b14, then the computer really does the math: ptr = 0x4b14 + 4*2 = 0x4b1c meaning that it adds 4 bytes to the pointer for every 1 unit being added. If we were dealing with a character, normally a one byte data structure, then the math would be: ptr = 0x4b14 + 1*2 = 0x4b16 and if we were dealing with some big data structure that took up 200 bytes for every structure, then the math would be: ptr = 0x4b14 + 200*2 = 0x4ca4ptr = ptr + 2