我有一个vector3类,我需要实现不同的乘法选项(所以我重载了运算符*),这取决于im乘法的类型。重载运算符*
的问题是,在最后一节我得到的错误:
Description Resource Path Location Type
ambiguating new declaration of 'Pang::vector3 Pang::operator*(const Pang::vector3&, const Pang::vector3&)' vector3.h /PangGame/src line 130
C/C++问题
但我只有一个operator
超载返回vector
和muyltiplies 2 vector
秒。
希望你能帮助(我只想澄清类载体3具有threee双号),例如:vector3(double x, double y, double z);
)
friend vector3 operator* (const double& number, const vector3& vector)
{
vector3 result;
result.x = number*vector.x;
result.y = number*vector.y;
result.z = number*vector.z;
return result;
}
friend vector3 operator* (const vector3& vector, const double& number)
{
vector3 result;
result.x = number*vector.x;
result.y = number*vector.y;
result.z = number*vector.z;
return result;
}
//Scalar product: If a = a1i + a2j + a3k and b = b1i + b2j + b3k then
// a · b = a1*b1 + a2*b2 + a3*b3
friend double operator* (const vector3& vector1, const vector3& vector2)
{
double result;
result= (vector1.x)*(vector2.x)+(vector1.y)*(vector2.y) + (vector1.z)*(vector2.z);
return result;
}
/* Product: Vector x Vector
* Example: The cross product of a = (2,3,4) and b = (5,6,7)
cx = aybz - azby = 3×7 - 4×6 = -3
cy = azbx - axbz = 4×5 - 2×7 = 6
cz = axby - aybx = 2×6 - 3×5 = -3
Answer: a × b = (-3,6,-3)*/
friend vector3 operator* (const vector3& vector,const vector3& vector2)
{
vector3 result;
result.x = (vector.y)*(vector2.z) - (vector.z)*(vector2.y);
result.y = (vector.z)*(vector2.x) - (vector.x)*(vector2.z);
result.z = (vector.x)*(vector2.y) - (vector.y)*(vector2.x);
return result;
}
如果我没有弄错,你的两个'operator *'函数具有相同的签名(参数)。上次我尝试过,C++不支持单独返回类型的重载。 – domsson