我有一个vector2d类。我在以下代码中收到错误“初始化类型没有匹配的构造函数”:
vector2d vector2d::operator+(const vector2d& vector)
{
return vector2d((this->x + vector.x), (this->y + vector.y));
}
vector2d vector2d::operator-(const vector2d& vector)
{
return vector2d(this->x - vector.x, this->y - vector.y);
}
我的向量类声明和定义是:
#ifndef __VECTOR2D_H__
#define __VECTOR2D_H__
class vector2d
{
public:
float x, y , w;
vector2d(const float x, const float y) ;
vector2d(vector2d& v) ;
vector2d operator+(const vector2d& rhs);
vector2d operator-(const vector2d& rhs);
vector2d& operator+=(const vector2d& rhs);
vector2d& operator-=(const vector2d& rhs);
float operator*(const vector2d& rhs);
float crossProduct(const vector2d& vec);
vector2d normalize();
float magnitude();
};
#endif
vector2d.cpp:
#include "vector2d.h"
#include <cmath>
vector2d::vector2d(const float x,const float y) :x(x),y(y),w(1)
{
}
vector2d::vector2d(vector2d& vector) : x(vector.x), y(vector.y), w(1)
{
}
vector2d vector2d::operator+(const vector2d& vector)
{
return vector2d((this->x + vector.x), (this->y + vector.y));
}
vector2d vector2d::operator-(const vector2d& vector)
{
return vector2d(this->x - vector.x, this->y - vector.y);
}
vector2d& vector2d::operator+=(const vector2d& vector)
{
this->x += vector.x;
this->y += vector.y;
return *this;
}
vector2d& vector2d::operator-=(const vector2d& vector)
{
this->x -= vector.x;
this->y -= vector.y;
return *this;
}
float vector2d::magnitude()
{
return sqrt(this->x * this->x + this->y * this->y);
}
//Make Unit Vector
vector2d vector2d::normalize()
{
float magnitude = this->magnitude();
float nx = 0.0f;
float ny = 0.0f;
nx = this->x / magnitude;
ny = this->y / magnitude;
return vector2d(nx,ny);
}
float vector2d::operator*(const vector2d& rhs)
{
return ( (this->x * rhs.x) + (this->y * rhs.y) );
}
float vector2d::crossProduct(const vector2d& vec)
{
return (x * vec.y - y * vec.x);
}
我没有使用默认构造函数参数创建对象,那么该错误的原因是什么?请注意,代码在Visual Studio上运行完全正常。在Xcode上我得到了错误。
答案 0 :(得分:1)
你的问题将是这个构造函数:
vector2d(vector2d& v) ;
标准的复制构造函数如下所示:
vector2d(const vector2d& v) ;
因为在标准c ++中你不能将临时绑定到可变的左值引用
不幸的是,微软凭借他们的智慧释放了许多扩展&#34; (即,与标准的恶意偏差)进入他们的编译器和仅在MSVC中,临时将绑定到可变的l值引用。
在 real 标准c ++中,临时可以绑定到:
vector2d(vector2d v) ;
vector2d(const vector2d& v) ;
vector2d(vector2d&& v) ;
`