OpenCV源码阅读(2)---matx.h---函数的内联实现

外部矩阵计算函数

namespace internal
{

template<typename _Tp, int m> struct Matx_DetOp
{
    double operator ()(const Matx<_Tp, m, m>& a) const
    {
        Matx<_Tp, m, m> temp = a;
        double p = LU(temp.val, m*sizeof(_Tp), m, 0, 0, 0);
        if( p == 0 )
            return p;
        for( int i = 0; i < m; i++ )
            p *= temp(i, i);
        return 1./p;
    }
};

template<typename _Tp> struct Matx_DetOp<_Tp, 1>
{
    double operator ()(const Matx<_Tp, 1, 1>& a) const
    {
        return a(0,0);
    }
};

template<typename _Tp> struct Matx_DetOp<_Tp, 2>
{
    double operator ()(const Matx<_Tp, 2, 2>& a) const
    {
        return a(0,0)*a(1,1) - a(0,1)*a(1,0);
    }
};

template<typename _Tp> struct Matx_DetOp<_Tp, 3>
{
    double operator ()(const Matx<_Tp, 3, 3>& a) const
    {
        return a(0,0)*(a(1,1)*a(2,2) - a(2,1)*a(1,2)) -
            a(0,1)*(a(1,0)*a(2,2) - a(2,0)*a(1,2)) +
            a(0,2)*(a(1,0)*a(2,1) - a(2,0)*a(1,1));
    }
};

上面的函数定义了矩阵行列式计算的计算。高于3阶的矩阵使用LU分解算法，低于3阶矩阵对Matx_DetOp进行了重载，使用直接计算行列式的方式来计算。这里使用的是在结构体里定义计算的方式。这样做的目的是什么呢？需要继续看类是如何调用这些操作的

template<typename _Tp> Vec<_Tp, 2> inline conjugate(const Vec<_Tp, 2>& v)
{
    return Vec<_Tp, 2>(v[0], -v[1]);
}

template<typename _Tp> Vec<_Tp, 4> inline conjugate(const Vec<_Tp, 4>& v)
{
    return Vec<_Tp, 4>(v[0], -v[1], -v[2], -v[3]);
}

这里使用内联的方式来实现向量共轭的计算。。。但是向量类中并没有定义共轭函数conjugate，只有一个conj。这是错误吗？

矩阵构造函数与基本运算

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx()
{
    for(int i = 0; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0)
{
    val[0] = v0;
    for(int i = 1; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1)
{
    CV_StaticAssert(channels >= 2, "Matx should have at least 2 elaments.");
    val[0] = v0; val[1] = v1;
    for(int i = 2; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2)
{
    CV_StaticAssert(channels >= 3, "Matx should have at least 3 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2;
    for(int i = 3; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3)
{
    CV_StaticAssert(channels >= 4, "Matx should have at least 4 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    for(int i = 4; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4)
{
    CV_StaticAssert(channels >= 5, "Matx should have at least 5 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3; val[4] = v4;
    for(int i = 5; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5)
{
    CV_StaticAssert(channels >= 6, "Matx should have at least 6 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5;
    for(int i = 6; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5, _Tp v6)
{
    CV_StaticAssert(channels >= 7, "Matx should have at least 7 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5; val[6] = v6;
    for(int i = 7; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5, _Tp v6, _Tp v7)
{
    CV_StaticAssert(channels >= 8, "Matx should have at least 8 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5; val[6] = v6; val[7] = v7;
    for(int i = 8; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5, _Tp v6, _Tp v7, _Tp v8)
{
    CV_StaticAssert(channels >= 9, "Matx should have at least 9 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5; val[6] = v6; val[7] = v7;
    val[8] = v8;
    for(int i = 9; i < channels; i++) val[i] = _Tp(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5, _Tp v6, _Tp v7, _Tp v8, _Tp v9)
{
    CV_StaticAssert(channels >= 10, "Matx should have at least 10 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5; val[6] = v6; val[7] = v7;
    val[8] = v8; val[9] = v9;
    for(int i = 10; i < channels; i++) val[i] = _Tp(0);
}


template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5, _Tp v6, _Tp v7, _Tp v8, _Tp v9, _Tp v10, _Tp v11)
{
    CV_StaticAssert(channels == 12, "Matx should have at least 12 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5; val[6] = v6; val[7] = v7;
    val[8] = v8; val[9] = v9; val[10] = v10; val[11] = v11;
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(_Tp v0, _Tp v1, _Tp v2, _Tp v3, _Tp v4, _Tp v5, _Tp v6, _Tp v7, _Tp v8, _Tp v9, _Tp v10, _Tp v11, _Tp v12, _Tp v13, _Tp v14, _Tp v15)
{
    CV_StaticAssert(channels == 16, "Matx should have at least 16 elaments.");
    val[0] = v0; val[1] = v1; val[2] = v2; val[3] = v3;
    val[4] = v4; val[5] = v5; val[6] = v6; val[7] = v7;
    val[8] = v8; val[9] = v9; val[10] = v10; val[11] = v11;
    val[12] = v12; val[13] = v13; val[14] = v14; val[15] = v15;
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n>::Matx(const _Tp* values)
{
    for( int i = 0; i < channels; i++ ) val[i] = values[i];
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n> Matx<_Tp, m, n>::all(_Tp alpha)
{
    Matx<_Tp, m, n> M;
    for( int i = 0; i < m*n; i++ ) M.val[i] = alpha;
    return M;
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n> Matx<_Tp,m,n>::zeros()
{
    return all(0);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n> Matx<_Tp,m,n>::ones()
{
    return all(1);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n> Matx<_Tp,m,n>::eye()
{
    Matx<_Tp,m,n> M;
    for(int i = 0; i < shortdim; i++)
        M(i,i) = 1;
    return M;
}

template<typename _Tp, int m, int n> inline
_Tp Matx<_Tp, m, n>::dot(const Matx<_Tp, m, n>& M) const
{
    _Tp s = 0;
    for( int i = 0; i < channels; i++ ) s += val[i]*M.val[i];
    return s;
}

template<typename _Tp, int m, int n> inline
double Matx<_Tp, m, n>::ddot(const Matx<_Tp, m, n>& M) const
{
    double s = 0;
    for( int i = 0; i < channels; i++ ) s += (double)val[i]*M.val[i];
    return s;
}

/** @cond IGNORED */
template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n> Matx<_Tp,m,n>::diag(const typename Matx<_Tp,m,n>::diag_type& d)
{
    Matx<_Tp,m,n> M;
    for(int i = 0; i < shortdim; i++)
        M(i,i) = d(i, 0);
    return M;
}

这一大段使用内联函数实现了矩阵的定义和加，减，点乘，点除等基础操作，使用内联的作用是提高效率。可以看出，对于低阶矩阵，opencv的做法十分粗暴，直接访问数组数据成员，然后赋值。不过在赋值和构造之前使用了CV_staticAssert来验证是否会溢出，这是c++的断言功能，不知opencv是如何重新利用的。

template<typename _Tp, int m, int n> template<typename T2>
inline Matx<_Tp, m, n>::operator Matx<T2, m, n>() const
{
    Matx<T2, m, n> M;
    for( int i = 0; i < m*n; i++ ) M.val[i] = saturate_cast<T2>(val[i]);
    return M;
}

这个函数作用是操作符的重载，重载了操作符()，作用是复制一个矩阵。其中使用了saturate_cast<T2> 模板函数，作用是防止内存溢出，但是这个函数不在这个文件中，我猜在bufferpool.h里。

template<typename _Tp, int m, int n> template<int m1, int n1> inline
Matx<_Tp, m1, n1> Matx<_Tp, m, n>::reshape() const
{
    CV_StaticAssert(m1*n1 == m*n, "Input and destnarion matrices must have the same number of elements");
    return (const Matx<_Tp, m1, n1>&)*this;
}

reshape函数，作用是将具有相同元素数目的矩阵转换形式。例如9*1–>3*3

template<typename _Tp, int m, int n>
template<int m1, int n1> inline
Matx<_Tp, m1, n1> Matx<_Tp, m, n>::get_minor(int i, int j) const
{
    CV_DbgAssert(0 <= i && i+m1 <= m && 0 <= j && j+n1 <= n);
    Matx<_Tp, m1, n1> s;
    for( int di = 0; di < m1; di++ )
        for( int dj = 0; dj < n1; dj++ )
            s(di, dj) = (*this)(i+di, j+dj);
    return s;
}

由矩阵的第i行，第j列开始，抽取一个较小的矩阵。

这里使用了this指针的方式，把this指针式只想类所定义的对象的指针，也就是说，如果定义Matx m（3，3）, m.get_minor(2,2) 那么this指针是只想m的，所以可以用(*this)的方式表示提取m中的数据。

template<typename _Tp, int m, int n> inline
Matx<_Tp, 1, n> Matx<_Tp, m, n>::row(int i) const
{
    CV_DbgAssert((unsigned)i < (unsigned)m);
    return Matx<_Tp, 1, n>(&val[i*n]);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, 1> Matx<_Tp, m, n>::col(int j) const
{
    CV_DbgAssert((unsigned)j < (unsigned)n);
    Matx<_Tp, m, 1> v;
    for( int i = 0; i < m; i++ )
        v.val[i] = val[i*n + j];
    return v;
}

提取第i行第j列的函数，对于提取行，可以直接使用val[i*n]的方式，这种方式表达的是，因为val[m*n] 是一个m*n的数组，是一个一维数组。如果是一个行矩阵的话，可以用这个数组的首地址进行初始化，将母矩阵该行首地址作为子矩阵数据首地址即可。并且这些函数都是const类型，保护了原本矩阵的数据。

template<typename _Tp, int m, int n> inline
typename Matx<_Tp, m, n>::diag_type Matx<_Tp, m, n>::diag() const
{
    diag_type d;
    for( int i = 0; i < shortdim; i++ )
        d.val[i] = val[i*n + i];
    return d;
}

提取对角元素。

template<typename _Tp, int m, int n> inline
const _Tp& Matx<_Tp, m, n>::operator()(int i, int j) const
{
    CV_DbgAssert( (unsigned)i < (unsigned)m && (unsigned)j < (unsigned)n );
    return this->val[i*n + j];
}

template<typename _Tp, int m, int n> inline
_Tp& Matx<_Tp, m, n>::operator ()(int i, int j)
{
    CV_DbgAssert( (unsigned)i < (unsigned)m && (unsigned)j < (unsigned)n );
    return val[i*n + j];
}

template<typename _Tp, int m, int n> inline
const _Tp& Matx<_Tp, m, n>::operator ()(int i) const
{
    CV_StaticAssert(m == 1 || n == 1, "Single index indexation requires matrix to be a column or a row");
    CV_DbgAssert( (unsigned)i < (unsigned)(m+n-1) );
    return val[i];
}

template<typename _Tp, int m, int n> inline
_Tp& Matx<_Tp, m, n>::operator ()(int i)
{
    CV_StaticAssert(m == 1 || n == 1, "Single index indexation requires matrix to be a column or a row");
    CV_DbgAssert( (unsigned)i < (unsigned)(m+n-1) );
    return val[i];
}

重载操作符（）,用于访问矩阵内元素，操作符重载是c++中非常重要的操作。这里使用了_Tp & Matx<_Tp,m,n>::operator ()(int i,int j){} 是把操作符重载为成员函数的做法。其中,i是第一个操作数，j是第二个操作数。

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, m, n>& a, const Matx<_Tp, m, n>& b, Matx_AddOp)
{
    for( int i = 0; i < channels; i++ )
        val[i] = saturate_cast<_Tp>(a.val[i] + b.val[i]);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, m, n>& a, const Matx<_Tp, m, n>& b, Matx_SubOp)
{
    for( int i = 0; i < channels; i++ )
        val[i] = saturate_cast<_Tp>(a.val[i] - b.val[i]);
}

template<typename _Tp, int m, int n> template<typename _T2> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, m, n>& a, _T2 alpha, Matx_ScaleOp)
{
    for( int i = 0; i < channels; i++ )
        val[i] = saturate_cast<_Tp>(a.val[i] * alpha);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, m, n>& a, const Matx<_Tp, m, n>& b, Matx_MulOp)
{
    for( int i = 0; i < channels; i++ )
        val[i] = saturate_cast<_Tp>(a.val[i] * b.val[i]);
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, m, n>& a, const Matx<_Tp, m, n>& b, Matx_DivOp)
{
    for( int i = 0; i < channels; i++ )
        val[i] = saturate_cast<_Tp>(a.val[i] / b.val[i]);
}

template<typename _Tp, int m, int n> template<int l> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, m, l>& a, const Matx<_Tp, l, n>& b, Matx_MatMulOp)
{
    for( int i = 0; i < m; i++ )
        for( int j = 0; j < n; j++ )
        {
            _Tp s = 0;
            for( int k = 0; k < l; k++ )
                s += a(i, k) * b(k, j);
            val[i*n + j] = s;
        }
}

template<typename _Tp, int m, int n> inline
Matx<_Tp,m,n>::Matx(const Matx<_Tp, n, m>& a, Matx_TOp)
{
    for( int i = 0; i < m; i++ )
        for( int j = 0; j < n; j++ )
            val[i*n + j] = a(j, i);
}

特殊的矩阵构造函数
定义了矩阵的基本操作，包括加减乘除缩放，这些操作作为矩阵的构造函数，可以生成一个新的矩阵，也就是支持由两个矩阵生成新的矩阵。

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n> Matx<_Tp, m, n>::mul(const Matx<_Tp, m, n>& a) const
{
    return Matx<_Tp, m, n>(*this, a, Matx_MulOp());
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n> Matx<_Tp, m, n>::div(const Matx<_Tp, m, n>& a) const
{
    return Matx<_Tp, m, n>(*this, a, Matx_DivOp());
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, n, m> Matx<_Tp, m, n>::t() const
{
    return Matx<_Tp, n, m>(*this, Matx_TOp());
    template<typename _Tp, int m, int n> inline
}
Vec<_Tp, n> Matx<_Tp, m, n>::solve(const Vec<_Tp, m>& rhs, int method) const
{
    Matx<_Tp, n, 1> x = solve((const Matx<_Tp, m, 1>&)(rhs), method);
    return (Vec<_Tp, n>&)(x);
}

template<typename _Tp, int m> static inline
double determinant(const Matx<_Tp, m, m>& a)
{
    return internal::Matx_DetOp<_Tp, m>()(a);
}

template<typename _Tp, int m, int n> static inline
double trace(const Matx<_Tp, m, n>& a)
{
    _Tp s = 0;
    for( int i = 0; i < std::min(m, n); i++ )
        s += a(i,i);
    return s;
}

template<typename _Tp, int m, int n> static inline
double norm(const Matx<_Tp, m, n>& M)
{
    return std::sqrt(normL2Sqr<_Tp, double>(M.val, m*n));
}

template<typename _Tp, int m, int n> static inline
double norm(const Matx<_Tp, m, n>& M, int normType)
{
    return normType == NORM_INF ? (double)normInf<_Tp, typename DataType<_Tp>::work_type>(M.val, m*n) :
        normType == NORM_L1 ? (double)normL1<_Tp, typename DataType<_Tp>::work_type>(M.val, m*n) :
        std::sqrt((double)normL2Sqr<_Tp, typename DataType<_Tp>::work_type>(M.val, m*n));
}

实现矩阵本身的乘，除，转置操作。如果已经有了一个两个矩阵，可以以成员函数的方式来生成结果。例如

Matx a(####);
Matx b(####);
Matx c;
c=a.mul(b);
Matx d(a,b,Mul_OP)

template<typename _Tp, typename _T2, int m, int n> static inline
MatxCommaInitializer<_Tp, m, n> operator << (const Matx<_Tp, m, n>& mtx, _T2 val)
{
    MatxCommaInitializer<_Tp, m, n> commaInitializer((Matx<_Tp, m, n>*)&mtx);
    return (commaInitializer, val);
}

template<typename _Tp, int m, int n> inline
MatxCommaInitializer<_Tp, m, n>::MatxCommaInitializer(Matx<_Tp, m, n>* _mtx)
    : dst(_mtx), idx(0)
{}

template<typename _Tp, int m, int n> template<typename _T2> inline
MatxCommaInitializer<_Tp, m, n>& MatxCommaInitializer<_Tp, m, n>::operator , (_T2 value)
{
    CV_DbgAssert( idx < m*n );
    dst->val[idx++] = saturate_cast<_Tp>(value);
    return *this;
}

template<typename _Tp, int m, int n> inline
Matx<_Tp, m, n> MatxCommaInitializer<_Tp, m, n>::operator *() const
{
    CV_DbgAssert( idx == n*m );
    return *dst;
}

一种往已有矩阵添加数的方法，具体情况不清楚。

///////////////////////////// Matx out-of-class operators ////////////////////////////////

template<typename _Tp1, typename _Tp2, int m, int n> static inline
Matx<_Tp1, m, n>& operator += (Matx<_Tp1, m, n>& a, const Matx<_Tp2, m, n>& b)
{
    for( int i = 0; i < m*n; i++ )
        a.val[i] = saturate_cast<_Tp1>(a.val[i] + b.val[i]);
    return a;
}

template<typename _Tp1, typename _Tp2, int m, int n> static inline
Matx<_Tp1, m, n>& operator -= (Matx<_Tp1, m, n>& a, const Matx<_Tp2, m, n>& b)
{
    for( int i = 0; i < m*n; i++ )
        a.val[i] = saturate_cast<_Tp1>(a.val[i] - b.val[i]);
    return a;
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator + (const Matx<_Tp, m, n>& a, const Matx<_Tp, m, n>& b)
{
    return Matx<_Tp, m, n>(a, b, Matx_AddOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator - (const Matx<_Tp, m, n>& a, const Matx<_Tp, m, n>& b)
{
    return Matx<_Tp, m, n>(a, b, Matx_SubOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n>& operator *= (Matx<_Tp, m, n>& a, int alpha)
{
    for( int i = 0; i < m*n; i++ )
        a.val[i] = saturate_cast<_Tp>(a.val[i] * alpha);
    return a;
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n>& operator *= (Matx<_Tp, m, n>& a, float alpha)
{
    for( int i = 0; i < m*n; i++ )
        a.val[i] = saturate_cast<_Tp>(a.val[i] * alpha);
    return a;
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n>& operator *= (Matx<_Tp, m, n>& a, double alpha)
{
    for( int i = 0; i < m*n; i++ )
        a.val[i] = saturate_cast<_Tp>(a.val[i] * alpha);
    return a;
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator * (const Matx<_Tp, m, n>& a, int alpha)
{
    return Matx<_Tp, m, n>(a, alpha, Matx_ScaleOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator * (const Matx<_Tp, m, n>& a, float alpha)
{
    return Matx<_Tp, m, n>(a, alpha, Matx_ScaleOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator * (const Matx<_Tp, m, n>& a, double alpha)
{
    return Matx<_Tp, m, n>(a, alpha, Matx_ScaleOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator * (int alpha, const Matx<_Tp, m, n>& a)
{
    return Matx<_Tp, m, n>(a, alpha, Matx_ScaleOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator * (float alpha, const Matx<_Tp, m, n>& a)
{
    return Matx<_Tp, m, n>(a, alpha, Matx_ScaleOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator * (double alpha, const Matx<_Tp, m, n>& a)
{
    return Matx<_Tp, m, n>(a, alpha, Matx_ScaleOp());
}

template<typename _Tp, int m, int n> static inline
Matx<_Tp, m, n> operator - (const Matx<_Tp, m, n>& a)
{
    return Matx<_Tp, m, n>(a, -1, Matx_ScaleOp());
}

template<typename _Tp, int m, int n, int l> static inline
Matx<_Tp, m, n> operator * (const Matx<_Tp, m, l>& a, const Matx<_Tp, l, n>& b)
{
    return Matx<_Tp, m, n>(a, b, Matx_MatMulOp());
}

template<typename _Tp, int m, int n> static inline
Vec<_Tp, m> operator * (const Matx<_Tp, m, n>& a, const Vec<_Tp, n>& b)
{
    Matx<_Tp, m, 1> c(a, b, Matx_MatMulOp());
    return (const Vec<_Tp, m>&)(c);
}

非成员函数的运算符重载，将重载后的操作定义在类外。

_Tp& operator *(const _Tp &a,const _Tp &b){a=a+b;return a;}
返回引用的意思是返回返回值的引用。
比如上面代码的意思就是返回a的引用
执行c=a*b那么c就是a的引用，而a又是a+b并且这个函数没有定义变量，也就是说没有额外的内存开销，所有使用的变量都是前面程序里面已经有的。这里千万不能有const 因为一旦加上数据保护，那么这个数据就不能再更改了，a就还是a 最后c返回a 的引用没有什么意义

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