Re: Unsignedness of std::size_t

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++.moderated

Date:

Wed, 18 Apr 2007 14:22:53 CST

Message-ID:

<f02rka$dpb$1@news.datemas.de>

Kaba wrote:

To give you an example of why unsigned integers should not represent
normal integers:

Let us have:
std::vector<int> a(5);
std::vector<int> b(8);

std::vector<int>::size() returns a std::size_t.

Now conceptually test if an unsigned integer, or a modular integer, is
really a good choice for representing the 'size', by asking the
following questions:

- How much bigger is 'b' than 'a'?

This question is meaningless *unless* 'b' is actually bigger. So,
first you need to check

b.size() > a.size()

and only then use the difference.

That's easy, it is 'b.size() - a.size()'.

Not so easy, from where I'm standing.

- How much bigger is 'a' than 'b'?

It's _not_. You cannot calculate the value of something that isn't.

'a.size() - b.size()' ? 2^n - 8?

The modular integer is a wrong model for the 'size', yet it is
implicated by the 'unsigned'.

The impropriety of using an unsigned for "how much bigger" does not
infer the impropriety of using unsigned for size.

V
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