模板静态成员变量的实例化(Instantiation of template static member variable)

我试图弄清楚clang如何确定C ++模板静态成员变量需要实例化，并且我看到一些令我困惑的行为。

给出以下代码：

#include <stdio.h> #include <typeinfo> int report(const char *name) { printf("Reporting class: %s\n", name); return 0; } template<typename ReportedClass> class reported_class { public: reported_class() { _reported_instances++; } private: static int _reported_instances; }; template<typename ReportedClass> int reported_class<ReportedClass>::_reported_instances = report(typeid(ReportedClass).name()); class foo : reported_class<foo> { public: int baz() { return 0; } }; class bar : reported_class<bar> { public: bar() { } int baz() { return 0; } }; int main(int argc, char **argv) { return 0; }

当我运行它时，我看到以下内容：

$ c++ -v Apple LLVM version 5.0 (clang-500.0.68) (based on LLVM 3.3svn) Target: x86_64-apple-darwin13.1.0 Thread model: posix $ c++ test.cpp $ ./a.out Reporting class: 3bar $

为什么report_class的静态实例化，而不是foo的静态？唯一的区别似乎是构造函数的存在，但我希望在任何一种情况下都会调用reported_class构造函数（因此在构造函数中使用时会强制执行静态实例化）。标准中是否有这样的理由我不知道，这是否可以依赖？

gcc-4.7.3显示相同的行为，所以我认为这是我误解的事情。

I’m trying to puzzle out how clang determines what C++ template static member variables require instantiation, and I’m seeing some behavior that has me confused.

Given the following code:

#include <stdio.h> #include <typeinfo> int report(const char *name) { printf("Reporting class: %s\n", name); return 0; } template<typename ReportedClass> class reported_class { public: reported_class() { _reported_instances++; } private: static int _reported_instances; }; template<typename ReportedClass> int reported_class<ReportedClass>::_reported_instances = report(typeid(ReportedClass).name()); class foo : reported_class<foo> { public: int baz() { return 0; } }; class bar : reported_class<bar> { public: bar() { } int baz() { return 0; } }; int main(int argc, char **argv) { return 0; }

When I run it, I see the following:

$ c++ -v Apple LLVM version 5.0 (clang-500.0.68) (based on LLVM 3.3svn) Target: x86_64-apple-darwin13.1.0 Thread model: posix $ c++ test.cpp $ ./a.out Reporting class: 3bar $

Why is the static for reported_class being instantiated, but not the one for foo? The only difference appears to be the presence of a constructor, but I’d expect the reported_class constructor to be called in either case (therefore forcing static instantiation due to use in the constructor). Is there a reason for this in the standard that I’m not aware of, and is this something that can be relied upon?

gcc-4.7.3 displays the same behavior, so I assume that this is something that I"m misunderstanding.