Question

g ++无法编译此程序，因为：＆＃34;

45:20: warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second:
xxx.cpp:27:15: note: candidate 1: const Point2d Point2d::operator*(float) const
 const Point2d Point2d::operator*(float xy) const
xxx.cpp:45:20: note: candidate 2: operator*(double, int) <built-in>
  Point2d x = xxx * 3;

＆＃34;

当我删除＆＃34; operator double（）＆＃34;然后＆＃34;运营商*＆＃34;有效，但是可以选择立即将它们运送给运营商吗？

代码：

class Point2d {
    public :
        Point2d(float x, float y);
        const Point2d operator*(float xy) const;
        operator double(); //when I delete operator double(), then operator* works
    private :
        float x;
        float y;
};

Point2d::operator double()
{
    return sqrt((this->x * this->x) + (this->y * this->y));
}


const Point2d Point2d::operator*(float xy) const
{
    return Point2d(x * xy, y * xy);
}


Point2d::Point2d(float x, float y)
{
    this->x = x;
    this->y = y;
}

int main()
{
    Point2d xxx(3, 5.5);
    Point2d x = xxx * 3;
    return 0;
}

Answer 1

在xxx * 3中，编译器可以使用{{1}将3转换为float并使用operator*(float xy)或将xxx转换为double然后使用内置乘法。

规则说这是不明确的，因此不允许选择。

当您删除转换时，其中一个选项会消失，因此不再含糊不清。

使用运算符double（）重载运算符* ambigious

1 个答案: