c++ question

[Dave Smith]

Levi Pearson wrote:
> On Sep 23, 2006, at 1:49 PM, Dave Smith wrote:
>
>>
>> Wrong. In both cases (char* and char[]), the variable itself stores 
>> the address. You do not need the & operator in either case.
>>
>> In fact, using the & operator is incorrect, because it will give you 
>> the address of the variable that was already holding the address of 
>> the character array, instead of the address of the character array.
>
> No, you are wrong, at least partially.  Both a char * and char[] 
> variable will evaluate to an address, but in the case of char[] that 
> address is both the address of the first element /and/ the address of 
> the variable itself.  With a char *, the address it is pointing to is 
> not the same as its own address, so the & operator acts differently on 
> char * than char[].
>

At least we agree that the & operator is not needed to print the address 
of a character array, which was the original poster's question.

--Dave