1st element of "an array of CHAR strings"

[Rolf Magnus]

On Apr 2, 11:55 pm, "Daniel T." <[email protected]> wrote:

Given 'arr' is declared as "const char* arr[]" then:

arr and &arr[0] are effectively the same thing.

&arr is the address of arr.

So. arr[0] is a const char*, arr and &arr[0] is the address of the const
char*, and &arr is the address of the address of a const char*.

Now check this out:

arr[0][0] is a const char...

i got it, say:

const char* arr[] = { "Daniel", "Tartaglia" }

it is an array of ararys,

No, it's not. It's an array of pointers.

because elements, e.g. /Daniel/ itself is a character array.

Yes, that's right. Still, arr is an array of _pointers_. The first element
(arr[0]) points to the first character of the character array "Daniel", the
second element of arr points to the first character of "Tartaglia".

so

1.) /arr/ and /&arr[0]/ give the address of 1st element of arr[]
Yes.

2.) &arr, will give the address of "D" of /Daniel/ (array inside an
array)

No, it won't. &arr gives the address of the array. *arr would give you the
address of the 'D'.

 

[Old Wolf]

Given 'arr' is declared as "const char* arr[]" then:

arr and &arr[0] are effectively the same thing.

Only if 'arr' is used in certain contexts...

&arr is the address of arr.

....not including this one &arr is different to &(&arr[0])
In fact the latter is an error because &arr[0] is not an lvalue.

So. arr[0] is a const char*, arr and &arr[0] is the address of the const
char*, and &arr is the address of the address of a const char*.

No, &arr is the address of an array of const char* .
Have a read of http://c-faq.com/aryptr/ .

arr[0][0] is a const char...

To clarify, the char need not be const; it is only the lvalue
expression arr[0][0] that designates it const. For example, this
is legal code:

char ch;
const char *arr[] = { &ch };

const_cast<char &>(arr[0][0]) = 'x';