🚀 go-pugleaf

In article  <65409@lanl.gov> jlg@lanl.gov (Jim Giles) writes:
]
]A pointer, on the other hand, is a variable whose _value_ is an
]address...  But, addresses
]aren't particularly important to me.  I almost never need to know
]or manipulate them in any way.  So, why do I need a data type who's
]values are addresses?

An int, on the other hand, is a variable whose _value_ is a
bit-string.  But, bit-strings aren't particularly important to me.  I
almost never need to know or manipulate them in any way.  So why do I
need a data type who's values are bit-strings?

The above probably won't make the point (though it should) so I will
expand on it.  A pointer might have an address as its concrete value,
but you should be thinking of a pointer as an abstraction, not by its
concrete representation.  A pointer is a data type with certain
operations on it:

for pointers declared "TYPE *p, *q", the following

	*p returns an l-value referencing an object of type TYPE (as
	   for any other data object, if p is not initialized the
	   outcome of any operation is undefined)

	p + i returns a pointer to the ith element of type TYPE
	   following *p.  This is only valid if there is a contiguous
	   sequence of at least i elements  of type TYPE following *p.

	p < q returns TRUE iff *p preceeds *q in a sequence of
	   elements of type TYPE, assuming that both *p and *q are
	   members of the same contiguous sequence.

etc.  There is no need to refer to addresses.  Pointers are no less
abstract than arrays.  There is no guarantee that "p + i" will return
an address "i*sizeof(TYPE)" greater than that of *p, anymore than it
is guaranteed that the address of A[i] is "(i-j)*sizeof(TYPE)" more
than the address of A[j].  I know implementations of both arrays and
pointers where these conditions do _not_ hold.
-- 
					David Gudeman
Department of Computer Science
The University of Arizona        gudeman@cs.arizona.edu
Tucson, AZ 85721                 noao!arizona!gudeman