Thread: Pointers and Arrays.

Hello to all. I have a single question. I have read the C faq .

The Standard says that if I have an array for example

Code:

int arr[3]={1,2,3}

the expression arr yiels a pointer .... except when arr appears in a sizeof like sizeof(arr) which gives 12 and & operator.....

Why sizeof(&arr) gives 8? The same with a pointer size..... According to prototype &arr is not a pointer?

thank you in advance.

In addition, what sizeof returns depends on the platform and also glibc installed on the system (processor X86 or x86_64). Here is what I get when I run the following on a Linux system (Fedora Core 17 X86_64; ldd (GNU libc) 2.15)

Code:

#include <stdio.h>

int main(int argc, char *argv[])
{
    int array[3] = {1,2,3};
    printf("sizeof(array) = %lu\n", sizeof(array));
    printf("sizeof(array[0]) = %lu\n", sizeof(array[0]));
    printf("sizeof(int) = %lu\n", sizeof(int));
    printf("sizeof(int *) = %lu\n", sizeof(int *));
    printf("size_t = %lu\n", sizeof(size_t));
    
    return 0;
}

Which returns the following (depending on your system/platform you may get different result)

Code:

$ gcc -Wall testscript.c -o testscript
$ ./testscript
sizeof(array) = 12
sizeof(array[0]) = 4
sizeof(int) = 4
sizeof(int *) = 8
size_t = 8
$

As far as I remember according to C99, size_of is at least 16 bits (therefore 8 is returned by sizeof)

Regards,
Dariyoosh

Thank you guys.

dariyoosh my problem is not the difference in the value due to implemention issues from a particular platform.

I have read that whenever array appears in an expression it will be converted to pointer on the first its element.

THe exceptions are when we have sizeof , & and literal initializer..... And I can't understand what the exception is for &array
because we get a pointer size when we apply sizeof... I hope to make my query understood :-/

thank you for your time.

Originally Posted by Mr.Lnx

THe exceptions are when we have sizeof , & and literal initializer..... And I can't understand what the exception is for &array
because we get a pointer size when we apply sizeof... I hope to make my query understood :-/

I'm not sure whether I understood your point

An array, let's an int array, is a pointer to its first cell. Therefore it includes the address of its first cell, and that's why the sizeof operator returns 8 (sizeof(array) = 8). That's why I said it depends on architecture, because sizeof (int) an sizeof(an address of int) are not necessarily equal.
array is a pointer (in the case of this example int *), &array is the address of the pointer, in other words, a pointer of pointer (int **), so again sizeof return 8 because again, here we are talking about the sizeof an address.

Is it what you were asking for?

Regards,
Dariyoosh

Yes. But I do not agree with the term pointer to pointer for an &array.

An adress is not always equivalent with a pointer

Check 2D arrays for example.... is there any double pointer???

Originally Posted by dariyoosh

There are for sure double pointers. Each row is an array, therefore a pointer to the address of its first element. And each element is also an array therefore again a pointer to the address of its first element (which is not an array).

Conclusion: a 2D array variable's name = a pointer to a pointer

Regards,
Dariyoosh

So you're saying that this always works?

Code:

void function (int **foo);

int main(void) {
    int foo[5][5];
    function(foo);
    return 0;
}

Try and compile that, you'll see how wrong it is. Confusion arises because pointers can simulate arrays, but the C standard has specific requirements for how real array types decay. An array is indeed converted to a pointer to the first element in many contexts. This conversion applies only to the first level of indirection, though, so in a 2d-array the row dimension will become a pointer, but the column dimension will stay an array type. The subsequent array types for other dimensions is the reason why you are required to supply a dimension size when making a parameter for a 2d-array in a function.

Now, the reason that & (address of) is an exception to the "array decays to a pointer to the first element" rule of thumb is because of the type of the expression: &array is a pointer to an array, not unlike any other expression involving &. So for example, that might look like:

Code:

 char (*ptr2ar)[50];
char ar[50] = "Goodbye cruel world!";
ptr2arr = &ar;

For more information, consult the C FAQ. Of particular interest is Question 6.12.

@whiteflags

Many thanks for this clarification.

Regards,
Dariyoosh

Code:

#include<stdio.h>

int main(void)
{
	int arr[3]={1,2,3};
	int (*p)[3];
	
	p = &arr;
	
	printf("%d" , **p);

return 0;
}

what about this?