gcc 4.5.1, SuSE Linux i686
Suppose we have following code:
template<typename realT> class B
{
public:
B() {std::cout << "B()" << std::endl;}
};
template<typename realT> class A
{
public:
static B<realT> static_var;
};
template<typename realT> B<realT> A<realT>::static_var;
template<> B<float> A<float>::static_var;
template<> B<double> A<double>::static_var;
int main()
{
A<float> test;
return 0;
}
In this case we won't have any output in the stdout. Compiler won't generate code to initialize float and double specialization of class A.
But.. if we'll change initializations like this:
template<> B<float> A<float>::static_var = B<float>();
template<> B<double> A<double>::static_var = B<double>();
the compiler will generate such code and we'll have double "B()" in the output.
Can somebody help me with understanding of such behaviour?
n3337 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 ]
braced-init-list
is C++11
feature, so in C++03
you can use only
template<> X Q<int>::x = ...;
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