“未定义参考”到静态字段模板专业化 [英] "undefined reference" to static field template specialization

问题描述

我有一个模板类的头文件，它只有静态函数和字段。

I have and header with a template class, which has only static functions and fields.

template<typename T> class Luaproxy {

    static std::map<std::string, fieldproxy> fields;
    static const char * const CLASS_NAME;
    static void addfields();
    static int __newindex(lua_State * l){
        //implemented stuff, references to fields...
    }
    //etc
}

正如你所看到的，一些函数只能被声明，因为我打算用模板专门化来实现它们。

As you can see some of the functions are only declared, because I intend to implement them with template specialization.

在.ccp文件中，我有：

In a .ccp file I have:

struct test { int a; }
template<> map<string, fieldproxy> Luaproxy<test>::fields;
template<> const char * const Luaproxy<test>::CLASS_NAME=typeid(test).name();
template<> void Luaproxy<test>::addfields(){
    //stuff, references to fields...
}

从标题中实现的两个函数中获取未定义的引用错误到 Luaproxy< test> :: fields 在.cpp中。请注意，在链接中找到 Luaproxy< test> :: CLASS_NAME 和 Luaproxy< test> :: addfields

I get undefined reference errors to Luaproxy<test>::fields from both functions that are implemented in the header and those that are only specialized in the .cpp. Note that Luaproxy<test>::CLASS_NAME and Luaproxy<test>::addfields seem to be found in linking.

是什么使得映射如此特别？

推荐答案

我终于得到它的工作，但我不能真正说为什么我的编译器（gcc 4.6.1）需要这样： template< std :: map< std :: string，fieldproxy> Luaproxy< test> :: fields = std :: map< std :: string，fieldproxy>（）;

I finally managed to get it working, but I couldn't really say why my compiler (gcc 4.6.1) needs it this way: template<> std::map<std::string, fieldproxy> Luaproxy<test>::fields=std::map<std::string, fieldproxy>();

显式构造函数似乎说服gcc有效地发射变量。我要求#gcc澄清，但不幸的是，频道总是保持沉默。

The explicit constructor seem to convince gcc to effectively emit the variable. I asked #gcc for clarification, but unfortunately that channel is always silent.

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