Created on 1999-01-20.00:00:00 last changed 301 months ago
Rationale (10/99): The Standard is already sufficiently clear on this question.
Is this valid C++? The question is whether a member constant can be specialized. My inclination is to say no.
template <class T> struct A {
static const T i = 0;
};
template<> const int A<int>::i = 42;
int main () {
return A<int>::i;
}
John Spicer: This is ill-formed because
11.4.9.3 [class.static.data]
paragraph 4 prohibits an initializer on a definition of a static data member
for which an initializer was provided in the class.
The program would be valid if the initializer were removed from the specialization.
Daveed Vandevoorde: Or at least, the specialized member should not be allowed in constant-expressions.
Bill Gibbons: Alternatively, the use of a member constant within the definition could be treated the same as the use of "sizeof(member class)". For example:
template <class T> struct A {
static const T i = 1;
struct B { char b[100]; };
char x[sizeof(B)]; // specialization can affect array size
char y[i]; // specialization can affect array size
};
template<> const int A<int>::i = 42;
template<> struct A<int>::B { char z[200] };
int main () {
A<int> a;
return sizeof(a.x) // 200 (unspecialized value is 100)
+ sizeof(a.y); // 42 (unspecialized value is 1)
}
For the member template case, the array size "sizeof(B)" cannot
be evaluated until the template is instantiated because B might
be specialized. Similarly, the array size "i" cannot be evaluated
until the template is instantiated.
Rationale (10/99): The Standard is already sufficiently clear on this question.
| History | |||
|---|---|---|---|
| Date | User | Action | Args |
| 2000-02-23 00:00:00 | admin | set | messages: + msg292 |
| 2000-02-23 00:00:00 | admin | set | status: open -> nad |
| 1999-01-20 00:00:00 | admin | create | |