Question

我正在尝试实现一个非常基本的Vector3（Vec3）类。我正在努力解决一个特殊情况：Vec3<size_t>添加Vec3<int>。

如何为此案例制作模板专业化？

任何帮助将不胜感激。本

#include <array>
#include <ostream>
#include <vector>

// #define Vec3f std::array<float, 3>
// #define Vec3u std::array<size_t, 3>

#define Vec3f Vec3<float>
#define Vec3u Vec3<size_t>
#define Vec3i Vec3<int>

template <typename T>
class Vec3
{
    public:
        Vec3(): _a() {}
        Vec3(T x, T y, T z): _a({x, y, z}) {}
        Vec3(const Vec3<T> & a): _a({a[0], a[1], a[2]}) {}

        ~Vec3() {}

        /// Print Vec3.
        friend std::ostream & operator<<(std::ostream & os, const Vec3<T> & v)
        {
            os << "(" << v[0] << ", " << v[1] << ", " << v[2] << ")";
            return os;
        }

        inline typename std::array<T, 3>::reference operator[](size_t i)
        {
            return _a[i];
        }

        inline typename std::array<T, 3>::const_reference operator[](size_t i) const
        {
            return _a[i];
        }


        /// Test equality.
        inline bool operator==(const Vec3<T> & other) const
        {
            bool a = abs(_a[0] - other._a[0]) < 1e-6;
            bool b = abs(_a[1] - other._a[1]) < 1e-6;
            bool c = abs(_a[2] - other._a[2]) < 1e-6;
            return (a and b and c);
        }

        /// Test non-equality.
        inline bool operator!=(const Vec3<T> & other) const
        {
            return not (*this == other);
        }


        /// Vec3 same type addition.
        inline Vec3<T> operator+(const Vec3<T> & other) const
        {
            return {_a[0] + other[0], _a[1] + other[1], _a[2] + other[2]};
        }


    protected:
        std::array<T, 3> _a;
};

Answer 1

您的问题是找到size_t和int之间的公共类型，它是结果的模板参数。这是一个可能的解决方案：

/// Vec3 addition between vectors of different base type.
template <class U>
Vec3<typename std::common_type<T, U>::type> operator+(const Vec3<U> & other) const
{
    return{ _a[0] + other[0], _a[1] + other[1], _a[2] + other[2] };
}

成员函数的C ++模板专业化

1 个答案: