Re: equality of non-type template arguments
matthias.neubauer@gmail.com wrote:
14.4 Type equivalence [temp.type] states, that ...
"Two template-ids refer to the same class or function if [...] their
non-type template-arguments of integral
or enumeration type have identical values ... [..]."
And 14.3.2 Template non-type arguments, para 5 [temp.arg.nontype] says
that
"for a non-type template-parameter of integral or enumeration type,
integral promotions (4.5) and integral
conversions (4.7) are applied."
However, a recent g++ gives me an error (error: could not convert
template argument ?1? to ?E?) when compiling the following translation
unit. After reading the cited paragraphes of the standard I reckon
this would be legal!?!
Regards,
Matthias
enum E {
lab = 1
};
template <E t>
struct C {
void foo();
};
template <>
void C<lab>::foo() {}
void bar () {
C<1> c;
c.foo();
}
There are no integral conversions or promotions *to* an enumeration type
*from* an int. There are, however, conversions the other way, which
should make the types of 'foo' and 'boo' the same:
template<int a> class T;
enum { zero = 0 };
...
T<0> foo;
T<zero> boo;
In this case 'zero' is converted to 'int' and yields the value 0.
V
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