Re: Integer template argument warning

From:
"Ben Voigt [C++ MVP]" <rbv@newsgroups.nospam>
Newsgroups:
microsoft.public.dotnet.languages.vc
Date:
Mon, 6 Jul 2009 14:37:21 -0500
Message-ID:
<DFABDFC5-BAED-4543-B698-F48C1F0D5035@microsoft.com>
"user790" <user790@gmail.com> wrote in message
news:1f986a40-dee4-46c1-b8de-89fcc99b1a1e@t13g2000yqt.googlegroups.com...

Hello,

I am compiling the following code

<code>

template < unsigned int I >
class A {};

template < int I >
class B {};

template < int I >
B<-I> foo(A<I> const &)
{
 return B<-I>();
}

template < int I >
void bar(A<I> const &)
{
}

int main()
{
 A<1> a;
 foo(a); // warning here
 bar(a);
 return 0;
}

</code>

At the indicated line, Visual Studio 2009 emits the following warning:

warning C4146: unary minus operator applied to unsigned type, result
still unsigned

I don't really understand the warning. It seems to me that the problem
could be the unsigned int to int conversion of the template argument
-- once it has been converted to int, appling the unary minus operator
should be a problem. But I get this warning for foo and no warning for
bar. What is the reason behind this?


Ah, but it isn't converted to int until after the negation operator is
applied. Hence the problem, and the solution.

Right now you have

B<-I>

which is effectively

B<(int)(-I)>

Instead use

B<-(int)I>

More importantly, I would like to know if there would be any work-
around to avoid the warning.

Thank you,

Paul

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