Re: Unsignedness of std::size_t
On Apr 17, 6:53 pm, Kaba <REkalleMOunderscoreVErutane...@hotmail.com>
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'?
That's easy, it is 'b.size() - a.size()'.
If you know beforehand that b.size() >= a.size(). In this case, yes.
- How much bigger is 'a' than 'b'?
'a.size() - b.size()' ? 2^n - 8?
Well, but this makes no sense. What excuses you from testing a.size()
= b.size() in the first place before performing the subtraction?
The modular integer is a wrong model for the 'size', yet it is
implicated by the 'unsigned'.
No. What is wrong is that the second question makes no sense still
you are trying to find out an answer for it :-)
Probably, you can get away with such a simple function to calculate
distance between two size_t values:
size_t dist(size_t a, size_t b)
{
return a >= b ? a - b : b - a;
}
Here, both dist(a.size(), b.size()) and dist(b.size(), a.size()) are
perfectly valid and meaningful.
--
Regards,
Alex Shulgin
--
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]