Doubts on 2D array

**sanddune008** · 03-31-2014

Hi All,

Following code :

Code:

int main()
{
	int ar[][5]= {{1,2,3,4,5},{6,7,8,9,10}};
	
	printf("Val       : %d\n",ar[1][2]);
	printf("Val       : %d\n",*(ar[1]+2));	
	printf("*(ar)     : %x\n",*(ar));
	printf("*(*(ar))  : %x\n",*(*(ar)));
	printf("ar 		  : %x\n",ar);
	printf("&ar 		  : %x\n",&ar);
	printf("ar[0]     : %x\n",ar[0]);
	printf("&ar[0][0] : %x\n",&ar[0][0]);
	printf("ar[0][0]  : %x\n",ar[0][0]);
	printf("(ar[0]+1)  : %x\n",(ar[0]+1));
	printf("(ar[1])  : %x\n",(ar[1]));
	return 0;
}

Code:

Output:
Val       : 8
Val       : 8
*(ar)     : 6a335340
*(*(ar))  : 1
ar                : 6a335340
&ar               : 6a335340
ar[0]     : 6a335340
&ar[0][0] : 6a335340
ar[0][0]  : 1
(ar[0]+1)  : 6a335344
(ar[1])  : 6a335354

Question:
1. Why are the following highlighted values the same (for *ar,ar,&ar)? What is the logic behind such assignment wrt the execution perspective?

Thanks in advance

**laserlight** · 04-01-2014

Originally Posted by sanddune008

1. Why are the following highlighted values the same (for *ar,ar,&ar)? What is the logic behind such assignment wrt the execution perspective?

In other words, the results show that the address of the array is the same in value as the address of its first element, although these addresses differ in type (i.e., they are not merely numeric addresses but pointers). This makes sense since it saves space.

By the way, to print a pointer with printf, use %p and cast the pointer to void*

**Hodor** · 04-03-2014

Originally Posted by laserlight

(i.e., they are not merely numeric addresses but pointers).

Nitpick... they behave like pointers (they're not "real" pointers because they're not lvalues). Or is that what you meant by "although these addresses differ in type"?

**laserlight** · 04-03-2014

Originally Posted by Hodor

Nitpick... they behave like pointers (they're not "real" pointers because they're not lvalues).

They (or rather the result of the conversion, in the case of ar and *ar) are of pointer type, hence they are pointers. Otherwise, you better be prepared to argue that 123 is not an integer, because although it is of integer type, it is not an lvalue. Likewise, you better be prepared to argue that malloc does not return a pointer, since the return value of a function is not an lvalue.

Originally Posted by Hodor

Or is that what you meant by "although these addresses differ in type"?

The thing is, "address" can refer to a numeric address, without any type information. In the context of C, such addresses can come with type information, i.e., they can be of a pointer type. The object under consideration by this conceptual bundling of a numeric address with type information is known as a pointer, though it can also be called an address, with the bundling of type implied by context.

**Hodor** · 04-03-2014

Originally Posted by laserlight

. Likewise, you better be prepared to argue that malloc does not return a pointer, since the return value of a function is not an lvalue.

Umm, but it does return an lvalue, or at least something that can be used as an lvalue.

**grumpy** · 04-03-2014

const pointers are not lvalues.

**Hodor** · 04-03-2014

I should explain what I meant.

int array[];

array = &whatever; /* Not ok */

int *blah = malloc(1024);

blah = &whatever; /* Ok */

So, although the array name behaves like a pointer it's not, strictly, a pointer.

**laserlight** · 04-03-2014

Originally Posted by Hodor

Umm, but it does return an lvalue, or at least something that can be used as an lvalue.

Are you sure? As far as I can recall, the return value of a function is not an lvalue (as opposed to say, not a modifiable lvalue). In C++ reference return types are possible, so my statement would not apply.

Originally Posted by Hodor

So, although the array name behaves like a pointer it's not, strictly, a pointer.

An array is not a pointer; an array is converted to a pointer to its first element in most contexts, including the context from post #1. My statements were with respect to "the address of the array" and "the address of its first element", not with respect to the array and the array's first element, so you have a strawman argument.

**Hodor** · 04-03-2014

Originally Posted by laserlight

Are you sure? As far as I can recall, the return value of a function is not an lvalue (as opposed to say, not a modifiable lvalue). In C++ reference return types are possible, so my statement would not apply.

An array is not a pointer; an array is converted to a pointer to its first element in most contexts, including the context from post #1. My statements were with respect to "the address of the array" and "the address of its first element", not with respect to the array and the array's first element, so you have a strawman argument.

I have to make a confession. The strawman from the Wizard of Oz is a great character. Much better than the tinman. Infinitely better than those stupid flying monkeys (or whatever they were), so thank you for the compliment. That little dog, Toto, went onto a successful singing career (linked below) so my hat off to him as well.

Toto - Africa - YouTube

Edit: I used to be scared of the wizard, but it turned out that it was only grumpy behind a curtain pulling levers and twisting dials that released smoke and amplified his voice. He was actually quite nice once revealed, so I'm not scared anymore.

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