From:	horst DOT kraemer AT gmx DOT de (Horst Kraemer)
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: error flaging I can't reason with..
Date:	Sat, 10 Jun 2000 08:58:00 GMT
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On Fri, 09 Jun 2000 23:03:46 -0700, Uriel Weizmann <uw1 AT cam DOT org>
wrote:

> The present is a c++ source program containing two lines commented
> respectively CASE1 and CASE2.
 
> Until this moment I can't find an explanation why CASE1 does not work.
> I thing that POINTR1 and ARRAY1 are exactly the same type variables with
> the same data.
 
> The puzzle even grows by the fact that if we eliminate the & from the
> parameter ARRAY in the prototype of FILLARRAY (changing ARRAY from
> POINTER REFERENCE  to the older style POINTER TO INTEGER,  CASE1  works
> fine.

Yes. That's what you should do if you want to use your function for
true arrays and for pointers pointing to the initial element of some
array. Your declaration makes it unusable for entities declared as
arrays.

The parameter declaration

> void fillarray (int* &array)

expects a reference to a non-const int*. Now you are calling

	int arr[10];
	int *ptr = arr;

	f(ptr);
	f(arr);

In the first case ptr is a variable of type (non-const) int*. This
matches exactly a reference to a non-const int*. In the second case
arr is converted implicitly to a pointer to int pointing to the
initial element of the array arr. Lets call this temporary pointer
parr. The difference is that ptr is a variable of type int* and aptr
is only a _value_ - like a number - and not a variable.

According to the C++ Standard a reference to a non-const 'foo' is
expecting a non-const variable of type 'foo' may not be initialized
with a simple value if type 'foo'. Values of type 'foo' are treated in
this context as if they where const variables of type 'foo'.

If you change your declaration to

	void fillarray (int* & const array)

it will be compiled because the type of the parameter is now 

	reference to  const (int*)

which matches a _value_ of type int* _and_ a variable of type int*,
too. But there is no point in declaring it

	void fillarray (int* & const array)

because there is no practical difference to

	void fillarray (int* array)

On the contrary. The compiler may create more code for the reference
definition because it may create a pointer variable of its own where
it stores the temporary pointer. The only theoretical difference is
that you cannot change array itself like

	array = 0;

_in_ the function which you would never to anyway. And still you could
achieve this by a simple

	void fillarray (int* const array)

in order to protect yourself against an accidental change of 'array'
itself in the function. Thus there is no point in using the reference
version at all.

Regards
Horst

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