如何在C ++中显式实例化模板构造函数？ [英] How to explicit instantiate template constructor in C++?

问题描述

当前我有以下课程

//AABB.h
class AABB {
   public:
    template <class T>
    AABB(const std::vector<T>& verties);
    template <class T>
    AABB(const std::vector<int>& indicies, const std::vector<T>& verties);
    template <class T>
    AABB(const std::vector<int>::const_iterator begin,
         const std::vector<int>::const_iterator end,
         const std::vector<T>& verties);
    AABB();
    AABB(const vec3 min, const vec3 max);
    AABB(const AABB& other);

    const vec3& min_p() const { return m_min; }
    const vec3& max_p() const { return m_max; }

    vec3 center();
    float volume();
    float surface_area();
    bool inside(vec3 point);
    float intersect(const Ray& ray);

   private:
    vec3 m_min;
    vec3 m_max;
};

有三个 AABB 构造函数的模板。有没有一种方法可以声明使用显式实例化？像下面这样？

There are three AABB constructors that are templated. Is there a possible way I could declare use explicit instantiation? Like the following?

// AABB.cpp
template AABB::AABB<Triangle>(const idx_t, const idx_t, const vector<Triangle>&);

给予

// AABB.cpp
using idx_t = std::vector<int>::const_iterator;

当前来自编译器的错误是

Currently the error from the compiler is

/Users/DarwinSenior/Desktop/cs419/tracer2/src/AABB.cpp:7:16: error: qualified
      reference to 'AABB' is a constructor name rather than a type wherever a
      constructor can be declared
template AABB::AABB<Triangle>(const idx_t, const idx_t,
               ^

推荐答案

您快到了，问题在于最后一个参数必须是模板声明中的引用：

You're almost there, the problem is the last argument must be a reference as it's in the template declaration:

template AABB::AABB<Triangle>(const idx_t, const idx_t, const vector<Triangle>&);
//                                                                           ^^^

此外，正如TC指出的那样，您可以删除< ; Triangle> ，因为它可以从参数的类型中推断出来：

Also, as T.C pointed out, you can remove <Triangle> because it can be inferred from the type of the argument:

template AABB::AABB(const idx_t, const idx_t, const vector<Triangle>&);

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