Pointer Arrays

[joezamboni]

When I run this code:

#include <stdio.h>

int main() {
	int x[] = {1,2,3,4,5,6,7,8,9,0};
	printf("Address of x : %p\n", &x);
	printf("Contents of x: %p\n", x);
	return 0;
}

It outputs the same address twice. Why?
It looks like x is pointing to itself.

[tabstop]

The name of an array, used by itself, "decays" to a pointer to its first element. So &x and x have the same value, but different types (the first is the address of an array, the second is the address of an int).

[joezamboni]

How is this held in memory?

[tabstop]

How is what held in memory? The array is just enough contiguous space for 10 ints.

[joezamboni]

x is a pointer to x[0], am I right?

[tabstop]

x is a pointer to x[0], am I right?

Sure. (If you mean is there extra space set aside for that pointer, then no, there's no extra space there.)

[joezamboni]

I dont understand how x can have a different value than x[0]. Where is x in relation to the data in the array?

[laserlight]

I dont understand how x can have a different value than x[0]. Where is x in relation to the data in the array?

Since x is an array, its value (or should I say values?) is all its elements.

[joezamboni]

I think I'm starting to understand. But, why does printf("Contents of x: %p\n", x); give me an address.

[laserlight]

But, why does printf("Contents of x: %p\n", x); give me an address.

Because in this context (and many others), an array is converted to a pointer to its first element.

By the way, to be pedantically correct, you should write:

printf("Address of x : %p\n", (void*)&x);
printf("Contents of x: %p\n", (void*)x);

Since %p is supposed to correspond to a void pointer. However, since pointers to objects are convertible to pointers to void, this probably does not matter in practice.

[joezamboni]

But my question is, why does x contain it's own address?

[tabstop]

x doesn't contain it's own address. None of x[0] through x[9] contain the address of x. The system has to store pointers to all the variables for its own use, and since you are not asking for the contents of x, but its location, the system is happy to tell you where that location is.

[laserlight]

But my question is, why does x contain it's own address?

It does not. The conversion to a pointer is just the way it works.

[joezamboni]

OK, thanks.

One more thing: How would I print the value at x?

[laserlight]

How would I print the value at x?

For example:

printf("%d\n", x[0]);

EDIT:
Actually, this is not quite correct. Put it in another way: suppose I told you: show me your hand! Which of your hands am I referring to (assuming that you have both of them)? Likewise, printing the value of x... which of the values of x?