我正在尝试为Android构建一些已经在win32上工作的C代码.我有一个问题,重载运算符.例如：

码：

Vector2 uv0 =  textures.back()->m_uv0;
Vector2 uvt =  textures.back()->m_uvt;

uv0 = m_uv0 + Vector2(uv0.x * m_uvt.x, uv0.y * m_uvt.y) + Vector2(0.01f,0.01f);

其中Vector2是上面声明的类.声明是：

class Vector2
{
public:
//Constructors
Vector2() : x(0.0f), y(0.0f){}
Vector2(GLfloat _x, GLfloat _y) : x(_x), y(_y) {}
Vector2(double _x, double _y) : x(static_cast<float>(_x)), y(static_cast<float>(_y)) {}
Vector2(int _x, double _y) : x(static_cast<float>(_x)), y(static_cast<float>(_y)) {}
Vector2(double _x, int _y) : x(static_cast<float>(_x)), y(static_cast<float>(_y)) {}
Vector2(int _x, int _y) : x(static_cast<float>(_x)), y(static_cast<float>(_y)) {}
Vector2(GLfloat * pArg) : x(pArg[0]), y(pArg[1]) {}
Vector2(const Vector2 & vector) : x(vector.x), y(vector.y) {}

//Vector's operations
GLfloat Length();
Vector2 & Normalize();
Vector2 operator + (Vector2 & vector);
Vector2 & operator += (Vector2 & vector);
Vector2 operator - ();
Vector2 operator - (Vector2 & vector);
Vector2 & operator -= (Vector2 & vector);
Vector2 operator * (GLfloat k);
Vector2 & operator *= (GLfloat k);
Vector2 operator / (GLfloat k);
Vector2 & operator /= (GLfloat k);
Vector2 & operator = (Vector2 vector);
Vector2 Modulate(Vector2 & vector);
GLfloat Dot(Vector2 & vector);
void Set(GLfloat _x, GLfloat _y);

//access to elements
GLfloat operator [] (unsigned int idx);

//data members
float x;
float y;
};

我不会在这里列出这个类的定义,因为它没有得到满足.

但不幸的是我发现了一个错误：

G:/PROJECT266/projects/PROJECT266//jni/../jni/SBE/source/Sprite.cpp: In member function'void Sprite::AddTex(TEX::GUItex)':
G:/PROJECT266/projects/PROJECT266//jni/../jni/SBE/source/Sprite.cpp:103:57: error: no match for 'operator+' in '((Sprite*)this)->Sprite::m_uv0 + Vector2((uv0.Vector2::x *((Sprite*)this)->Sprite::m_uvt.Vector2::x), (uv0.Vector2::y * ((Sprite*)this)->Sprite::m_uvt.Vector2::y))'
G:/PROJECT266/projects/PROJECT266//jni/../jni/SBE/source/Sprite.cpp:103:57: note: candidates are:
G:/PROJECT266/projects/PROJECT266//jni/../jni/SBE/source/SBMath.h:38:10: note: Vector2 Vector2::operator+(Vector2&)
G:/PROJECT266/projects/PROJECT266//jni/../jni/SBE/source/SBMath.h:38:10: note:   no known conversion for argument 1 from 'Vector2' to 'Vector2&'

但是,如果我像这样重写上面的代码：

Vector2 uv0 =  textures.back()->m_uv0;
Vector2 uvt =  textures.back()->m_uvt;

Vector2 vec1 = Vector2(uv0.x * m_uvt.x, uv0.y * m_uvt.y);
Vector2 vec2 = Vector2(0.01f,0.01f);

uv0 = m_uv0 + vec1 + vec2;

编译期间不会出现任何错误.
我无法理解,这个愚蠢错误的原因是什么.
如果你解释我如何解决这个问题,我会很高兴的.

解决方法:

无法将r值绑定到非const引用.

这一行：

uv0 = m_uv0 + Vector2(uv0.x * m_uvt.x, uv0.y * m_uvt.y) + Vector2(0.01f,0.01f);

相当于:(我用PARAMS替换了参数,使我的例子更具可读性)：

uv0 = (m_uv0.operator+(Vector2(PARAMS))).operator+(Vector2(PARAMS));

这里Vector2(PARAMS)将创建一个临时对象.那就是你试图将r值引用传递给你的运算符重载,并且编译器找不到匹配项,因为你的运算符被声明为：

Vector2 operator+ (Vector2& vector);

有关为什么temporaries不能绑定到非const引用的更多信息,请参阅：How come a non-const reference cannot bind to a temporary object?

在第二个示例中,首先声明两个Vector2对象,然后将它们作为l值引用传递给运算符,这些引用与运算符重载相匹配.

解决问题并让运算符重载同时接受l值和r值引用的一种方法是将其声明为引用const,因为将r值绑定到const的引用是完全正确的.请参阅krsteeve的答案,了解如何做到这一点.

通常,如果您不打算修改参数,则应始终声明将引用作为引用const的函数.

参考绑定的示例：

Vector2& ref1 = Vector2(); // Error, trying to bind r-value to non-const ref.
Vector2 v;
Vector2& ref2 = v; // OK, v is an l-value reference.

// It is however OK to bind an r-value to a const reference:
const Vector& ref3 = Vector2(); // OK.