I'm trying to figure out why this example doesn't compile. My understanding is that if a static variable is not expl开发者_运维问答icitly set then it defaults to 0. In the five examples below four of them behave as I would expect, but the one that's commented out won't compile.
#include <iostream>
class Foo
{
public:
static int i;
static int j;
};
template <int n>
class Bar
{
public:
Bar(int) { }
static int i;
};
static int i;
int Foo::i;
int Foo::j = 1;
template <> int Bar<2>::i;
template <> int Bar<3>::i = 3;
int main(int argc, char** argv)
{
std::cout << "i " << i << std::endl;
std::cout << "Foo::i " << Foo::i << std::endl;
std::cout << "Foo::j " << Foo::j << std::endl;
//std::cout << "Bar<2>::i " << Bar<2>::i << std::endl; // Doesn't compile?
std::cout << "Bar<3>::i " << Bar<3>::i << std::endl;
return 0;
}
Why doesn't int Bar<2>::i
do the same thing as int Foo::i
or static int i
?
Edit: I had forgotten to add template<> to the Bar<2> and Bar<3> declarations. (doesn't solve the problem though, still getting linker errors)
Under the rules of the current C++ standard, the specialisation template <> int Bar<2>::i;
is only a declaration and never a definition.
To become a definition, you must specify an initialiser. (See clause 14.7.3/15)
Apart from that, you were missing one very common case: the definition of a non-specialised static member of a template:
template <int n> int Bar<n>::i;
This provides a definition for Bar<N>::i
for N not equal to 2 or 3.
According to the latest draft of Standard C++ it says
14.7.3/13 An explicit specialization of a static data member of a template is a definition if the declaration includes an initializer; otherwise, it is a declaration.
[Note: the definition of a static data member of a template that requires default initialization must use a braced-init-list:
template<> X Q<int>::x; //declaration
template<> X Q<int>::x (); // error: declares a function
template<> X Q<int>::x { }; // definition
— end note ]
So what you are asking is possible, if your compiler is supporting it.
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