在C＆＃x2B;＆＃x2B中的布尔上下文中，在布尔上下文中使用户定义的类型为真/错误。

发布于 2025-01-25 02:39:42 字数 338 浏览 2 评论 0原文

我有一个line，代表笛卡尔平面上一条线的相关信息。具有bool的类型，该类型指示是否定义了斜率。我希望能够执行以下操作：

if(my_line){
   double new_slope = my_line.slope * 9;
}

在布尔上下文中，实例my_line本身被隐式转换为true/false值。我正在考虑使用智能指针看到的行为，如果指向nullptr或0，则该实例被视为false as-is。

我将如何模仿这种行为？

原文

I have a Line that represents the relevant information of a line on the cartesian plane. The type that has, among other members, a bool that indicates whether the slope is defined. I would like to be able to do the following:

if(my_line){
   double new_slope = my_line.slope * 9;
}

where the instance my_line itself is implicitly converted to a true/false value in a boolean context. I am thinking of the behavior I see with smart pointers, where if it is pointing to nullptr or 0, the instance is considered false as-is.

How would I go about emulating this behavior?

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说谎友 2025-02-01 02:39:42

在您的line类中，实现bool转换操作员。您还可以选择超载 operator！ 但这在C ++ 11及以后不需要。参见上下文转换。

例如：

class Line {
    bool mSlopeDefined;
    ...

public:
    ...

    explicit operator bool() const noexcept {
        return mSlopeDefined;
    }

    // optional since C++11, but doesn't hurt...
    bool operator!() const {
        return !mSlopeDefined;
    }
};

In your Line class, implement a bool conversion operator. You could also optionally overload the operator!, but that is not required in C++11 and later. See Contextual conversions.

For example:

class Line {
    bool mSlopeDefined;
    ...

public:
    ...

    explicit operator bool() const noexcept {
        return mSlopeDefined;
    }

    // optional since C++11, but doesn't hurt...
    bool operator!() const {
        return !mSlopeDefined;
    }
};

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