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C++ 如何默认初始化静态类模板变量_C++_Templates - Fatal编程技术网

C++ 如何默认初始化静态类模板变量

C++ 如何默认初始化静态类模板变量,c++,templates,C++,Templates,在下面的代码中 #include <iostream> #include <string> #include <type_traits> template<class T, typename = std::enable_if<std::is_default_constructible<T>::value>> struct E { static T var; }; int main() { std:

在下面的代码中

#include <iostream>
#include <string>
#include <type_traits>

template<class T, typename =  std::enable_if<std::is_default_constructible<T>::value>>
struct E {
    static T var;   
};

int main() {
    std::cout << E<std::string>::var << std::endl;
}

template te te::var{};
课后定义


为什么初始化时使用
{}
?@shane:在当前情况下,您可以省略它们(因为变量是静态的),但我更喜欢语法而不是语法
/home/tE1MjB/ccEXSZtn.o: In function `main':
prog.cpp:(.text.startup+0x12): undefined reference to `E<std::string, std::enable_if<true, void> >::var'
collect2: error: ld returned 1 exit status
template<class T, typename T2> T E<T, T2>::var{};