矢量类型的模板类特化 - 不同的有效语法? [英] Template class specialization for vector type - different valid syntax?

问题描述

在以下代码段中，template<> 对于专业化是可选的吗?我是否包含它有什么区别吗?我的第一直觉是将它包括在内，因为它或多或少表明它是一种专业化.它在 g++ 4.9.2 和 Intel 16 下以两种方式编译

In the following snippit, is the template<> optional for the specialization? Is there any difference whether I include it or not? My first instinct was to include it as it more-or-less signifies that it is a specialization. It compiles both ways under both g++ 4.9.2 and Intel 16

#include <vector>
#include <iostream>

template<typename T>
struct PrintMe
{
    static void Print(const T & t)
    {
        std::cout << "In general templated struct: " << t << "\n";
    }   
};      


template<> // <--- optional?
template<typename T>
struct PrintMe<std::vector<T>>
{   
    static void Print(const std::vector<T> & t)
    {   
        std::cout << "In general specialization for vector: " << t.size() << "\n";
        for(const auto & it : t)
            std::cout << "        " << it << "\n";
    }

};


int main(void)
{   
    PrintMe<int>::Print(5);   
    PrintMe<double>::Print(5);    
    PrintMe<std::vector<float>>::Print({10,20,30,40});

    return 0;
}

注意:出于好奇，我尝试添加多个 template<>.即，

Note: Out of curiosity, I tried adding multiple template<>. Ie,

template<>
template<>
template<>
template<typename T>
struct PrintMe<std::vector<T>>

这仍然可以用 Intel 编译，但不能用 g++ 编译.不确定这意味着什么，但很有趣.

This still compiles with Intel, but not with g++. Not sure what that means, but it's interesting.

注意 2:哇，这和我 5 年前的一个问题非常相似:模板类特化，其中模板参数是模板.在那里它被称为冗余语法.

Note 2: Wow, this is very similar to a question of mine from 5 years ago: Templated class specialization where template argument is a template . There it was mentioned as redundant syntax.

推荐答案

给定类模板的定义，

template<> // <--- optional?
template<typename T>
struct PrintMe<std::vector<T>> { ... };

无效.

您需要删除该行并使用:

You need to remove that line and use:

template<typename T>
struct PrintMe<std::vector<T>> { ... };

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