C ++与运营商不匹配*(C++ no match for operator*)

C ++与运营商不匹配*(C++ no match for operator*)

我想重载运算符：

static Vector operator*(float s, Vector right){ Vector result(right.getX()*s, right.getY()*s, right.getZ()*s); return result; }

当我想使用它时：

Vector a(0,1,5) Vector v(4*a);

我有：

error: no match for 'operator*' (operand types are 'int' and 'Vector')

哪里不对？

I want to overload operator:

static Vector operator*(float s, Vector right){ Vector result(right.getX()*s, right.getY()*s, right.getZ()*s); return result; }

When I want to use it:

Vector a(0,1,5) Vector v(4*a);

I got:

error: no match for 'operator*' (operand types are 'int' and 'Vector')

What is wrong?

最满意答案

你需要声明（在类声明中）operator *作为非静态非成员和朋友，如：

friend Vector operator*(float s, Vector right);

并定义为（外部类声明）：

Vector operator*(float s, Vector right){ return Vector(right.getX()*s, right.getY()*s, right.getZ()*s); }

Operator *在此处用作二元运算符。 如果你作为成员，那么第一个参数被隐式地当作当前对象（*适用于*，例如，如果你使用x * y，那么operator *适用于成员的x）。 但是，对于非成员，两个参数都可以是非Vector类型，并且可以根据需要转换为Vector。 如果您作为非成员非朋友和静态（在外部类声明中定义），这甚至可以工作

You need to declare (within class declaration) operator* as non-static non-member and friend like:

friend Vector operator*(float s, Vector right);

and define as (outside class declaration):

Vector operator*(float s, Vector right){ return Vector(right.getX()*s, right.getY()*s, right.getZ()*s); }

Operator * is used as binary operator here. If you make as member, then first argument is implicitly taken as the current object (on which * applies, for example, if you use x * y, then operator * applies on x for member). However, for non-member both the arguments can be non-Vector type and can be converted to Vector if needed. This will even work if you make as non-member non-friend and as static (tobe defined outside class declaration)