Re: Why is this

From:

Victor Bazarov <v.bazarov@comcast.invalid>

Newsgroups:

comp.lang.c++

Date:

Wed, 01 Oct 2014 16:22:29 -0400

Message-ID:

<m0hnq6$6v1$1@dont-email.me>

On 10/1/2014 4:05 PM, Bint wrote:

On 01/10/2014 18:51, Paavo Helde wrote:

Bint <bint@ign.com> wrote in news:D051A43E.F5B8%bint@ign.com:

Can someone explain why I get this result? In the 'if' statement
below, I'm evaluating (0 - (384 % 64)) and it is passing the
greater-than-zero test.

int A = 0;
int B = 64;
unsigned long C = 500;

if ((A - (C % B)) > 0)
C += (A - (C % B)); // it gets to here -- why?

500 % 64 is 52, and 0 - 52 is not greater than zero.
Is it somehow because of the an unsigned long?

Yes, an unsigned type will turn the whole subexpression unsigned, together
with all the nice misfeatures like legalized wrap-around and type tainting.
Better to avoid mixing signed and unsigned in such expressions. And look up
boost::numeric_cast<>.

Beware that there are two parties in this group, ones who think that
unsigned types are the Right Thing, and others who think they are Real
Evil. I predict another 500 posts in this thread ;-(

If a value can never be negative as it represents a "size" or a "count"
for example then using an unsigned type to represent it is perfectly
fine. std::size_t pervades the Standard Libray thanks to std::allocator
and it is unsigned.

/Flibble

Boy, that makes no sense to me.

Which part?

> I mean, I understand that an unsigned value

can never be negative, but if you throw it into a calculation with signed
values then it should just be treated as a positive value.

And it is.

> Let the left

side of the equation determine whether the result can be negative or
positive.

And in your case what is the "left side"?

> I can't believe that's how C works, that it forces the whole

expression to be unsigned! That's crazy, man!

Are you in jest? I can't believe you can't believe it.

There are old rules about promotions and conversions of arithmetic types
(integral and floating point) in both C and C++, they are essentially
the same. There is nothing to believe, really, you just need to learn
them. And as you learn them you might just understand the logic behind
them. Eventually.

V
--
I do not respond to top-posted replies, please don't ask