Re: a DWORD with all bits set...
On Wed, 03 Jun 2009 17:44:53 -0400, r norman <r_s_norman@comcast.net>
wrote:
On Wed, 03 Jun 2009 16:36:56 -0500, "Doug Harrison [MVP]"
<dsh@mvps.org> wrote:
On Wed, 03 Jun 2009 16:55:23 -0400, r norman <r_s_norman@comcast.net>
wrote:
And what it means to have all bits set does NOT depend on arithmetic
representation, it means that all bits are set. The operator ~
inverts all bits in a value. It pays absolutely no attention to
signed/unsigned -- it just inverts bits. So if you start with a 32
bit value with all bits zero, the ~ operator will then set all 32 bits
to one.
What I said last time was:
Let me see if I can clarify this. The expression ~0 is a signed int, and
what it means to have "all bits set" depends on the representation of
signed integers. If one's complement, ~0 == 0. If sign/magnitude, ~0 ==
INT_MIN. (That's what I would expect, anyway.) Neither of those evaluate to
-1. Please read the [*] part quoted above until you see the importance of
this.
Clearly, "what it means" is referring to the value of the expression ~0,
and that depends totally on the representation of signed integers. It
doesn't matter that all bits are set in ~0 if the value is not what is
desired, i.e. -1 in this thread. I don't know how to make it any clearer.
The goal is to get all bits set, not to get a value of -1. I don't
know how to make that any clearer. It doesn't matter what arithmetic
value is represented by the "all bits set" condition if the goal is to
get all bits set.
For the purpose of assigning an unsigned integer the largest value of its
type from some value of a signed type, you are dead wrong. You are not
understanding the language rule I presented and have repeated several
times, nor do you appear to understand any of the other discussion. I don't
see any point in continuing this.
--
Doug Harrison
Visual C++ MVP
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