在开关中声明类对象，然后在开关外部使用该变量

发布于 2024-12-17 13:32:29 字数 432 浏览 1 评论 0原文

有没有办法解决这个问题？我在 switch 语句中声明类对象，然后在 switch 之外使用该变量，只有当我将其余代码放在每种情况下时才有效，这不是很有效。这是我的代码

switch (shape)
{
case 'q':
{
    Quad geo(a,b,c,d);
}
break;
case 'r':
{
    Rectangle geo(a,b,c,d);
}
break;
case 't':
{
    Trapezoid geo(a,b,c,d);
}
break;
case 'p':
{
    Parrelogram geo(a,b,c,d);
}
break;
case 's':
{
    Square geo(a,b,c,d);

}
break;
default:
    break;
}

 geo.print();//obviously wont work

原文

Is there a way to work around this?Im declaring class objects in a switch statement and later using that variable outside the switch, it only works if i put the rest of my code in each case which isnt very efficient.Heres my code

switch (shape)
{
case 'q':
{
    Quad geo(a,b,c,d);
}
break;
case 'r':
{
    Rectangle geo(a,b,c,d);
}
break;
case 't':
{
    Trapezoid geo(a,b,c,d);
}
break;
case 'p':
{
    Parrelogram geo(a,b,c,d);
}
break;
case 's':
{
    Square geo(a,b,c,d);

}
break;
default:
    break;
}

 geo.print();//obviously wont work

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千秋岁 2024-12-24 13:32:29

有一个 IPrintable 接口，如下所示

struct IPrintable
{
    virtual ~IPrintable() {}
    virtual void Print() = 0;
};

然后，从 IPrintable 派生您的类型 Quad、Rectangle 等，即实现那个界面。那么你的代码看起来像这样：

std::unique_ptr<IPrintable> pShape;
switch(shape)
{
    case quad:
       pShape.reset(new Quad(...));
    case rect
       pShape.reset(new Rect(...));
}
if(pShape)
    pShape->Print();

当然，如果常用功能不仅仅是打印，你也可以将这些功能添加到界面中。还要看看访客模式。它可能对您有帮助，也可能没有帮助，具体取决于您问题的具体情况。

Have an IPrintable interface, like this

struct IPrintable
{
    virtual ~IPrintable() {}
    virtual void Print() = 0;
};

Then, derive your types Quad, Rectangle, etc from IPrintable, i.e. implement that interface. Then your code looks like this:

std::unique_ptr<IPrintable> pShape;
switch(shape)
{
    case quad:
       pShape.reset(new Quad(...));
    case rect
       pShape.reset(new Rect(...));
}
if(pShape)
    pShape->Print();

Of course, if the common functionality is more than print, you can add those functions to the interface as well. Also take a look at the visitor pattern. It may or may not be of help to you depending on the specifics of your problem.

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∞梦里开花 2024-12-24 13:32:29

不，这是不可能的。 geo 在编译时只能有一种类型，并且在运行时不能改变。

您可以使用动态分配和多态性做类似的事情，但这可能不是解决您的问题的最佳解决方案。

知道 Quad 是其他类的基类后，以下可能是一个可用的解决方案：

Quad* geo = 0;
switch (shape) {
case 'q':
    geo = new Quad(a,b,c,d);
    break;
case 'r':
    geo = new Rectangle(a,b,c,d);
...
default:
    break;
}
if (geo) geo->print();
delete geo; // Ok if geo is 0.

该解决方案不是特别漂亮，主要是因为它使用原始指针以及 new 和 delete 直接。更精致的版本将使用工厂模式，返回智能指针。

No, it is not possible. geo can only have one type at compile time, and it cannot change at runtime.

You could do something similar with dynamic allocation and polymorphism, but it might not be the best solution to your problem.

With the knowledge that Quad is the base class of the others, the following might be a usable solution:

Quad* geo = 0;
switch (shape) {
case 'q':
    geo = new Quad(a,b,c,d);
    break;
case 'r':
    geo = new Rectangle(a,b,c,d);
...
default:
    break;
}
if (geo) geo->print();
delete geo; // Ok if geo is 0.

This solution is not particularly pretty, mainly because it uses raw pointers and new and delete directly. A more polished version would use the Factorypattern, returning a smart pointer.

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