Point taken, si I made a simple mistake it should have been
"an array name parameter is a pointer"  K&R page 99



Bob Ammerman wrote:

> Ok,
>
> I've put on my language lawyer hat because I really have to clear up a
> misconception that is dogging this thread:
>
> NOTE: An array name is _not_ a pointer in C.
>
> An array name is an identifier for an area of memory (the array).
>
> It is just that in _most_ contexts you can treat the array name as a pointer
> to the first element of the array. The terminology typically used is that it
> 'decays' to the pointer.
>
> There are at least two places where treating the array name as a pointer
> breaks down:
>
> 1) You cannot change this 'pointer' to point to anything else, ever.
> For example, given:
>
> char buf[20];
> char buf2[10];
>
> buf = buf2; // This is illegal. You are trying to change the array name buf
> to point to something else (in this case to be equal to the pointer to which
> 'buf2' decays).
>
> 2) When using the array name in a sizeof operation, you get the size of the
> entire array, _not_ the size of a pointer to the elements. Thus, given:
>
> char buf[20];
>
> sizeof(buf) will be 20.
>
> 2) [Optional for advanced language-law students] There is actually an
> exception to the rules described in 1 and 2, above. When an array is
> declared as argument to a function:
>
> void foo(char buf[20])
> {
> }
>
> Then within the function 'foo', buf really _is_ treated as a pointer to
> char. You can freely adjust it or assign things to it:
>
>     char buf2[20];
>     ++buf;
>     buf = buf2;
>
> And sizeof(buf) will return the size of the pointer, not the buffer!
>
> Bob Ammerman
> RAm Systems
> (contract development of high performance, high function, low-level
> software)
>
> --
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