Re: Why is this
Den onsdag den 1. oktober 2014 22.05.42 UTC+2 skrev Bint:
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, toge=
ther
with all the nice misfeatures like legalized wrap-around and type tain=
ting.
Better to avoid mixing signed and unsigned in such expressions. And lo=
ok 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::allocato=
r
and it is unsigned.
/Flibble
Boy, that makes no sense to me. I mean, I understand that an unsigned va=
lue
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. Let the left
side of the equation determine whether the result can be negative or
positive. I can't believe that's how C works, that it forces the whole
expression to be unsigned! That's crazy, man!
No it's not. Some rule (inherited from C, so it is not going to change) has=
to be there. C++ does not overload based on "what is on the left side". Fo=
r one thing, "what is on the left side" does not always exist - e.g.
void f(unsigned);
void f(int);
unsigned i;
int j;
f(i + j);
Which f do you call?
You have learnt a lesson: do not mix signed and unsigned integral values. I=
f I were you, I would avoid the unsigned types. I use them mostly for low-l=
evel stuff - e.g. working with "raw" data and bits.
It is an open debate whether the standard library should have used signed o=
r unsigned types e.g. when indexing a std::vector. Today I believe that sig=
ned types would be a better choice, but back then with the segmented archit=
ecture used for the popular x86 processors then, I would also have chosen a=
n unsigned value.
/Peter
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