🚀 go-pugleaf

>>>>> "JB" == John Barton <jjb@watson.ibm.com> writes:
JB> In article <47qnbu$f48@oznet03.ozemail.com.au>, John Max Skaller <maxtal@su
phys.physics.su.oz.au> writes:
JB> |> jjb@watson.ibm.com (John Barton) wrote:
JB> |> >In article <47dn87$a17@sunsystem5.informatik.tu-muenchen.de>,
JB> |> >|> You can extend this idea:
JB> |> >|>
JB> |> >|> template<class T> T  operator- (T&a, T&b) { return a + (-b); }
JB> |> >|> template<class T> T  operator- (T&a)      { return 0 - a; }
JB> |> >|> template<class T> T& operator+ (T&a)      { return a; }
JB> |> >|> template<class T> T& operator++(T&a)      { return a += 1; }
JB> |> >|> template<class T> T& operator++(T&a,int)  { return ++a; }
JB> |> >
JB> |> >This will not work: it prevent you from writing any templates
JB> |> >for these operators over types that do not have "1" or "0"
JB> |> >or that need a definition other than the one above.
JB> |>
JB> |> How does it prevent it? [Quote WP ??]

JB> template<class T>
JB> Matrix<T> operator-(Matrix<T>& rhs) { /* doesn't matter what I code here */
 }

JB> The global operator-(T&) will always win over any more
JB> specific template like the one above.  Unless the standard has
JB> changed here as well.

AFAIK, this has never been true. My understanding is that now the
``most specialized'' (if determinable) specialization will be
selected. Before that, I think you would have an ambiguity.

A template `t<U, V, ...>' would be ``more specialized'' than a template
`t<X, Y, ...>' if there exists some subsitution for `X', `Y', ... that
makes `t<X, Y>' identical to `t<U, V>' but not vice versa.

	Daveed
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