具有参考模板参数的功能模板 [英] Function template with reference template parameter

问题描述

有此代码:

#include <iostream>

template<const double& f>
void fun5(){
    std::cout << f << std::endl;
} 

int main() 
{ 
    const double dddd = 5.0;
    fun5<dddd>();
    return 0;
} 

编译期间的编译器错误:

Compiler error during compilation:

$ g++ klasa.cpp -o klasa
klasa.cpp: In function ‘int main()’:
klasa.cpp:11:10: error: ‘dddd’ cannot appear in a constant-expression
klasa.cpp:11:16: error: no matching function for call to ‘fun5()’
klasa.cpp:11:16: note: candidate is:
klasa.cpp:4:6: note: template<const double& f> void fun5()

为什么不能将'dddd'用作模板参数，应该怎么做才能使其起作用?

Why placing 'dddd' as template parameter doesn't work and what should be done to make it work?

推荐答案

模板参数的引用和指针必须具有外部链接(对于C ++ 11，则为内部链接，但需要静态存储持续时间).因此，如果必须使用 dddd 作为模板参数，则需要将其移至全局范围并使其成为 extern :

References and pointers for template arguments must have external linkage (or internal linkage, for C++11, but static storage duration is required). So if you have to use dddd as a template argument, you need to move it to global scope and make it extern:

extern const double dddd = 5.0;
int main()
{
    fun5<dddd>();
    return 0;
}

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