使用template初始化头文件中的静态变量

Question

使用模板在头文件初始化静态变量(使用c++11，不支持'inline'方法)

猫base_class.hpp

#include <string>
#ifndef PROGRAM_BASE
#define PROGRAM_BASE
template <typename T>
struct S
{
static std::string s_elfName;
};
template <typename T>
std::string S<T>::s_elfName; //static initialization
class program_base : public S<void>{
public:
static std::string GetElfName() { return S<std::string>::s_elfName; }
bool update_string();
};
#endif

猫base_class.cpp

#include "base_class.hpp"
bool program_base::update_string(){
S<std::string>::s_elfName =  "UpdateString";
}

猫read_string.cpp

#include <iostream>
#include "base_class.hpp"
using namespace std;
int main () {
program_base pb; pb.update_string();
cout << program_base::GetElfName() << endl;
}

上面的工作很好，当我尝试在类program_base">

中添加模板时猫base_class.hpp

#include <string>
#ifndef PROGRAM_BASE
#define PROGRAM_BASE
class program_base {
public:
template <typename T>
struct S
{
static std::string s_elfName;
};
template <typename T>
std::string S<T>::s_elfName; //static initialization
static std::string GetElfName() { return S<std::string>::s_elfName; }
bool update_string();
};
#endif

它给出一个错误"错误:成员's_elfName'声明为模板">

为什么我不能在类中声明模板而不是继承它?

Answer 1

如c++标准嵌套类声明中所述:

Member functions and static data members of a nested class can be defined in a namespace scope enclosingthe definition of their class.
[Example:
struct enclose {
struct inner {
static int x;
void f(int i);
};
};
int enclose::inner::x = 1;
void enclose::inner::f(int i) { /* ... */ }
— end example]

如果您将定义从外部类中取出，那么应该没问题:

template <typename T>
std::string program_base::S<T>::s_elfName;

从c++ 17开始，您还可以进行内联静态初始化:

class program_base {
public:
template <typename T>
struct S
{
inline static std::string s_elfName;
};
static std::string GetElfName() { return S<std::string>::s_elfName; }
bool update_string();
};