在开关中声明类对象，然后在开关外使用该变量 [英] declaring class objects in a switch and later using that variable outside the switch

问题描述

有没有一种方法可以解决这个问题?我在switch语句中声明类对象，然后在switch外使用该变量，只有在每种情况下我都将其余代码都放在那儿时才有效，这才有效.代码

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

推荐答案

具有这样的IPrintable接口

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

然后，从IPrintable派生您的类型Quad，Rectangle等，即实现该接口.然后您的代码如下:

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();

当然，如果常用功能不只是打印功能，还可以将这些功能添加到界面中.还要看看访客模式.根据您问题的具体情况，它可能对您有帮助也可能没有帮助.

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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