c++ - 在 C++ 中使用模板进行矩阵乘法

Question

我正在尝试将类模板用于矩阵。但是我遇到了矩阵乘法的问题。

template<typename T, unsigned int N, unsigned int M>
class Matrix : public MatrixBase<Matrix<T, N, M>, T, N, M> {

    template<unsigned int K>
    friend Matrix<T, N, K> operator*(const Matrix<T, N, M>& m1, const Matrix<T, M, K>& m2) {
        Matrix<T, N, K> ret;
        for (unsigned int n = 0; n != N; n++) {
            for (unsigned int k = 0; k != K; k++) {
                ret.i[n][k] = 0;
                for (unsigned int m = 0; m != M; m++) {
                    ret.i[n][k] += m1.i[n][m]*m2.i[m][k];
                }
            }
        }
        return ret;
    }
};

当涉及到将两个 mat4(4x4 矩阵)相乘时，如下所示:

m_model = (m_view*m_model);

它给出错误 Invalid operands to binary expression ('mat4' (aka 'Matrix<float, 4, 4>') and 'mat4') .在线查看后，我发现这不是函数模板的预期用途，因为您必须在调用模板参数时进行分配。有没有一种类似于我最初打算的解决方法，即根据函数的第二个参数自动分配模板参数？

这里分别是 MatrixBase 和 Matrix(aka mat4) 的定义:

矩阵基

template<typename T , unsigned int M>
struct ComponentColumn{
    T& operator[](int m) {
        return i[m];
    }

    const T& operator[](int m) const {
        return i[m];
    }


    T i[M];
};


//-----------MATRIXBASE-----------
template <typename ChildT, typename T, unsigned int N, unsigned int M>
class MatrixBase {
public:
    MatrixBase() {}

    MatrixBase<ChildT, T, N, M> operator*=(const MatrixBase<ChildT, T, N, M>& m1) {
        MatrixBase<ChildT, T, N, M> ret;
        for (unsigned int n = 0; n != N; n++) {
            for (int k = 0; k != M; k++) {
                ret.i[n][k] = 0;
                for (unsigned int m = 0; m != M; m++) {
                    ret.i[n][k] += (*this).i[n][m]*m1.i[m][k];
                }
            }
        }

        *this = ret;

        return ret;
    }

    MatrixBase<ChildT, T, N, M> operator+(const MatrixBase<ChildT, T, N, M>& m1) {
        MatrixBase<ChildT, T, N, M> ret;
        for (int n = 0; n != N; n++) {
            for (int m = 0; m != M; m++) {
                ret.i[n][m] = i[n][m];
            }
        }
        return ret;
    }

    ComponentColumn<T, M>& operator[](int n) {
        return this->i[n];
    }


    const ComponentColumn<T, M>& operator[](int n) const {
        return this->i[n];
    }

    explicit operator T*() {
        return &(*this)[0][0];
    }

protected:
    ComponentColumn<T, M> i[N];
};

垫4

template<typename T>
class Matrix<T, 4, 4> : public MatrixBase<Matrix<T, 4, 4>, T, 4, 4> {
public:
    Matrix<T, 4, 4>() {
        for (unsigned int n = 0; n != 4; n++) {
            for (unsigned int m = 0; m != 4; m++) {
                if (n == m) {
                    (*this)[n][m] = 1;
                } else {
                    (*this)[n][m] = 0;
                }
            }
        }
    }

    Matrix<T, 4, 4>(const Matrix<T, 3, 3>& m) {
        (*this)[0][0] = m[0][0]; (*this)[1][0] = m[1][0]; (*this)[2][0] = m[2][0]; (*this)[3][0] = 0;
        (*this)[0][1] = m[0][1]; (*this)[1][1] = m[1][1]; (*this)[2][1] = m[2][1]; (*this)[3][1] = 0;
        (*this)[0][2] = m[0][2]; (*this)[1][2] = m[1][2]; (*this)[2][2] = m[2][2]; (*this)[3][2] = 0;
        (*this)[0][3] = 0; (*this)[1][3] = 0; (*this)[2][3] = 0; (*this)[3][3] = 1;
    }

    static Matrix<T, 4, 4> Translate(T x, T y, T z);
    static Matrix<T, 4, 4> Translate(const vec3& v);
    static Matrix<T, 4, 4> Scale(T s);
    static Matrix<T, 4, 4> Rotate(T degrees);
    static Matrix<T, 4, 4> Frustum(T left, T right, T bottom, T top, T near, T far);

    explicit operator Matrix<T, 3, 3>() {
        Matrix<T, 3, 3> ret;
        for (int n = 0; n != 3; n++) {
            for (int m = 0; m != 3; m++) {
                ret[n][m] = (*this)[n][m];
            }
        }

        return ret;
    }

    Matrix<T, 4, 4> Transpose() {
        Matrix<T, 4, 4> ret = Matrix<T, 4, 4>();
        for (unsigned int n = 0; n != 4; n++) {
            for (unsigned int m = 0; m != 4; m++) {
                ret.i[n][m] = this->i[m][n];
            }
        }
        *this = ret;
        return ret;
    }

    Matrix<T, 4, 4> Inverse();
};

Answer 1

除非你这样做是为了练习，这将是一个很好的练习，否则我只会使用一个现有的线性代数库来实现矩阵 vector 运算。如 Armadillo :http://arma.sourceforge.net/