fis*_*ood 2 c++ operator-overloading
我正在用c ++编写矩阵3x3类.
glm :: mat3通过[][] operator语法提供对矩阵数据的访问.
例如,myMatrix[0][0] = 1.0f; 将第一行,第一列输入设置为1.0f.
我想提供类似的访问权限.怎么能超载[][] operators?
我尝试了以下内容,但是我收到了错误:
必须将运算符名称声明为函数
const real operator[][](int row, int col) const
{
// should really throw an exception for out of bounds indices
return ((row >= 0 && row <= 2) && (col >= 0 && col <= 2)) ? _data[row][col] : 0.0f;
}
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重载此运算符的正确方法是什么?
没有运算符[][],因此您需要重载[]运算符两次:一次在矩阵上,返回行的代理对象,并返回一次返回的代理行:
// Matrix's operator[]
const row_proxy operator[](int row) const
{
return row_proxy(this, row);
}
// Proxy's operator[]
const real operator[](int col) const
{
// Proxy stores a pointer to matrix and the row passed into the first [] operator
return ((this->row >= 0 && this->row <= 2) && (col >= 0 && col <= 2)) ? this->matrix->_data[this->row][col] : 0.0f;
}
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通常,您想要使用多个参数operator(),而不是operator[].
嗯,如果不是很明显,operator[][]在C++中就没有了.它只是operator[]应用了两次.这意味着如果你想要那个符号,那么你必须让第一个返回第二个可以应用的结果(可索引的东西或代理).
下面的代码草拟了一些方法,选择你喜欢的:
#include <iostream>
#include <vector>
template< int n >
int& dummy() { static int elem = n; return elem; }
struct Mat1
{
int operator() ( int const x, int const y ) const
{ return dummy<1>(); }
int& operator() ( int const x, int const y )
{ return dummy<1>(); }
Mat1( int, int ) {}
};
struct Mat2
{
int at( int const x, int const y ) const
{ return dummy<2>(); }
int& at( int const x, int const y )
{ return dummy<2>(); }
Mat2( int, int ) {}
};
struct Mat3
{
struct At { At( int x, int y ) {} };
int operator[]( At const i ) const
{ return dummy<3>(); }
int& operator[]( At const i )
{ return dummy<3>(); }
Mat3( int, int ) {}
};
class Mat4
{
protected:
int get( int const x, int const y ) const
{ return dummy<4>(); }
void set( int const x, int const y, int const v ) {}
class AssignmentProxy
{
private:
Mat4* pMat_;
int x_;
int y_;
public:
void operator=( int const v ) const
{ pMat_->set( x_, y_, v ); }
int value() const { return pMat_->get( x_, y_ ); }
operator int () const { return value(); }
AssignmentProxy( Mat4& mat, int const x, int const y )
: pMat_( &mat ), x_( x ), y_( y )
{}
};
public:
int operator()( int const x, int const y ) const
{ return get( x, y ); }
AssignmentProxy operator()( int const x, int const y )
{ return AssignmentProxy( *this, x, y ); }
Mat4( int, int ) {}
};
class Mat5
{
protected:
int at( int const x, int const y ) const
{ return dummy<4>(); }
int& at( int const x, int const y )
{ return dummy<5>(); }
class RowReadAccess
{
private:
Mat5 const* pMat_;
int y_;
public:
int operator[]( int const x ) const
{
return pMat_->at( x, y_ );
}
RowReadAccess( Mat5 const& m, int const y )
: pMat_( &m ), y_( y )
{}
};
class RowRWAccess
{
private:
Mat5* pMat_;
int y_;
public:
int operator[]( int const x ) const
{
return pMat_->at( x, y_ );
}
int& operator[]( int const x )
{
return pMat_->at( x, y_ );
}
RowRWAccess( Mat5& m, int const y )
: pMat_( &m ), y_( y )
{}
};
public:
RowReadAccess operator[]( int const y ) const
{ return RowReadAccess( *this, y ); }
RowRWAccess operator[]( int const y )
{ return RowRWAccess( *this, y ); }
Mat5( int, int ) {}
};
struct Mat6
{
private:
std::vector<int> elems_;
int width_;
int height_;
int indexFor( int const x, int const y ) const
{
return y*width_ + x;
}
public:
int const* operator[]( int const y ) const
{
return &elems_[indexFor( 0, y )];
}
int* operator[]( int const y )
{
return &elems_[indexFor( 0, y )];
}
Mat6( int const w, int const h )
: elems_( w*h, 6 ), width_( w ), height_( h )
{}
};
int main()
{
using namespace std;
enum{ w = 1024, h = 1024 };
typedef Mat3::At At;
Mat1 m1( w, h );
Mat2 m2( w, h );
Mat3 m3( w, h );
Mat4 m4( w, h );
Mat5 m5( w, h );
Mat6 m6( w, h );
wcout
<< m1( 100, 200 ) // No fuss simple, but exposes element ref.
<< m2.at( 100, 200 ) // For those who don't like operators.
<< m3[At( 100, 200)] // If you really want square brackets mnemonic.
<< m4( 100, 200 ) // Hides element ref by using assignment proxy.
<< m5[200][100] // Ditto but with square brackets (more complex).
<< m6[200][100] // The minimum fuss square brackets, exposes elem ref.
<< endl;
}
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哦,我在发布代码后发现我还没有完全隐藏内部存储Mat5:它需要一个额外的代理级别,如Mat4.所以这种方法非常复杂.我不会这样做(Mat1我觉得很好很容易),但有些人认为代理很酷,数据隐藏得更酷......
总而言之,没有"正确的"超载方式operator[].有很多方法(如上面的代码所示),每种方式都有一些权衡.通常你最好使用operator(),因为相反operator[]它可以采取任何数量的论点.