blocks|key|4853951|text|IShape+s+=+(Square)shape;+
s.PrintMore();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4853952|有了这段代码，s仍然是一个IShape，而不是Square。变量保留了您在其声明中定义的类型，无论您尝试填充它。|unstyled|offset|length|style|CODE|4853953|要工作，代码应该是：|4853954|Square+s+=+(Square)shape;+
s.PrintMore();|4853955|entityMap^0|0|7|1|D|6|N|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@$G|S|H|T|I|J]|$G|U|H|V|I|J]|$G|W|H|X|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$B|C]]|$1|O|3|-4|5|F|7|10|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>IShape s = (Square)shape; 
s.PrintMore();
</code></pre>

With this code <code>s</code> is still an <code>IShape</code>, not a <code>Square</code>. A variable keeps the type you have defined at his declaration, whatever you try to fill in it.

To work the code should be :

<pre><code>Square s = (Square)shape; 
s.PrintMore();
</code></pre>

blocks|key|3239388|text|在IShape+s+=+(Square)shape;行中，您告诉编译器s是一个IShape，因此只有IShape上的方法可用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3239389|如果您使用：|3239390|Square+s+=+(Square)shape;
s.PrintMore()|code-block|syntax|javascript|3239391|那么这应该是你想要的。|3239392|这里的关键点是，当你声明一个变量时，你告诉编译器它是什么，然后你把什么对象放进那个变量中并不重要。|3239393|entityMap^0|1|P|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|Y|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

In the line <code>IShape s = (Square)shape;</code> you are telling the compiler that s is an IShape so only methods on IShape are available.

If you user:

<pre><code>Square s = (Square)shape;
s.PrintMore()
</code></pre>

Then that should do what you want.

The key point here is that when you declare a variable you tell the compiler what it is and it doesn't then matter what object you put into that.

blocks|key|4854001|text|问题是您试图通过IShape引用访问PrintMore()，IShape引用只能看到它们声明的Print()方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4854002|因此，您拥有的(Square)+shape类型转换没有做任何事情，因为它会立即将其存储到IShape引用中。您需要将引用本身转换为Square，方法是将其转换并存储在Square引用中，或者在调用之前进行转换：|4854003|Square+s+=+(Square)+shape;
s.PrintMore();|code-block|syntax|javascript|4854004|或|4854005|IShape+s+=+shape;
((Square)s).PrintMore();|4854006|或者，如果您不确定它是否是Square，可以使用as强制转换：|4854007|Square+s+=+shape+as+Square;

//+will+cast+if+it+is+a+Square,+otherwise,+returns+null+
//+this+doesn't+work+for+value+types+(int,+struct,+etc)+
if+(s+!=+null)
{
++++s.PrintMore();
}|4854008|entityMap^0|8|6|I|B|U|6|1B|7|0|7|E|18|6|1T|6|2B|6|0|0|0|0|D|6|O|2|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1F|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|S|3|T|5|J|7|1M|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|V|$]]

The problem is you are trying to access <code>PrintMore()</code> off of an <code>IShape</code> reference, <code>IShape</code> references only see the <code>Print()</code> method they declare.

So the cast you have <code>(Square) shape</code> is doing nothing because it is immediately storing it into an <code>IShape</code> reference. You need the reference itself to be <code>Square</code> by either casting and storing it in a <code>Square</code> reference, or by casting before the invoke:

<pre><code>Square s = (Square) shape;
s.PrintMore();
</code></pre>

Or

<pre><code>IShape s = shape;
((Square)s).PrintMore();
</code></pre>

Or, if you aren't sure if it's a <code>Square</code> or not, use an <code>as</code> cast:

<pre><code>Square s = shape as Square;

// will cast if it is a Square, otherwise, returns null 
// this doesn't work for value types (int, struct, etc) 
if (s != null)
{
 s.PrintMore();
}
</code></pre>

blocks|key|5137383|text|无法访问PrintMore方法的原因是，您正在将类型转换的形状重新赋值给定义为IShape的变量。为了能够在Square类上使用方法，您需要将其存储在Square类型的变量中，例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5137384|Square+s+=+(Square)shape;
s.PrintMore();+|code-block|syntax|javascript|5137385|或者：|5137386|((Square)+shape).PrintMore();|5137387|尽管可能值得仔细检查一下您的代码，但像这样的类型转换通常是一个很好的警告信号，表明它可能并不理想。也许IShape应该有PrintMore方法，或者您现在应该只接受Square对象？至少，我建议在进行这种类型转换之前，确保shape的类型实际上是Square。|5137388|例如：|5137389|Square+s+=+shape+as+Square;

if+(s+!=+null)
++++s.PrintMore();+|5137390|entityMap^0|4|9|13|6|1I|6|23|6|0|0|0|0|1F|6|1O|9|2A|6|33|5|3F|6|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1J|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|V|$]]

The reason you are unable to access the <code>PrintMore</code> method is because you're reassigning the type-casted shape, back to a variable defined as <code>IShape</code>. To be able to use methods on the <code>Square</code> class you need to store it in a variable of type <code>Square</code> for example:

<pre><code>Square s = (Square)shape;
s.PrintMore(); 
</code></pre>

or alternatively:

<pre><code>((Square) shape).PrintMore();
</code></pre>

Although it might be worth taking a good look at your code, type-casts such as these are usually a good warning sign that perhaps it is not ideal. Perhaps <code>IShape</code> should have the <code>PrintMore</code> method, or perhaps you should only be accepting <code>Square</code> objects at this point? At the very least I'd suggest making sure that the <code>shape</code> is actually of type <code>Square</code> before doing this type cast.

For example:

<pre><code>Square s = shape as Square;

if (s != null)
 s.PrintMore(); 
</code></pre>

blocks|key|5137437|text|您的s变量的类型仍然是IShape。仅仅因为您在赋值时恰好使用了强制转换，并不会改变编译器所关心的类型。您需要：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5137438|Square+s+=+(Square)+shape;
s.PrintMore();|code-block|syntax|javascript|5137439|当然，只有当shape真的是一个Square时，这才能起作用(假设没有进行自定义转换)。|5137440|我建议你在走这条路之前仔细考虑一下。通常，这样的强制转换表明您在某种程度上破坏了抽象-如果您只知道将shape作为一个IShape，那么您应该(通常)能够仅使用IShape的成员来做您需要的事情。如果不是这样，您可以：|5137441|5137442|使IShape变得更强大(如果绝对必要，让它更多地接受您的代码以接受Square而不是|unordered-list-item|5137443|5137444|entityMap^0|2|1|B|6|0|0|6|5|G|6|0|1E|5|1N|6|28|6|0|0|1|6|Y|6|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|U|$]]

Your <code>s</code> variable is still of type <code>IShape</code>. Just because you happen to have used a cast when assigning to it doesn't change the type as far as the compiler's concerned. You'd need:

<pre><code>Square s = (Square) shape;
s.PrintMore();
</code></pre>

Of course, this will only work if <code>shape</code> really is a <code>Square</code> (assuming there aren't custom conversions going on).

I would advise you to think carefully before going down this route though. Usually a cast like this indicates that you're breaking abstraction somewhat - if you only know about <code>shape</code> as an <code>IShape</code>, you should (usually) be able to do what you need just with the members of <code>IShape</code>. If that's not the case, you can:

<ul>
<li>Make <code>IShape</code> more powerful (give it more members)</li>
<li>Change your code to accept a <code>Square</code> instead of an <code>IShape</code></li>
<li>Cast if absolutely necessary</li>
</ul>

blocks|key|4854152|text|因为当您尝试使用时，s.PrintMore()+"s“的类型是IShape，所以它只知道接口函数。你需要做像这样的事情|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4854153|Square+s+=+(Square)shape;++
s.PrintMore();|code-block|syntax|javascript|4854154|或|4854155|((Square)shape).PrintMore();+//+assuming+you're+positive+its+a+Square+type|4854156|可以把接口想象成对象上的包装器，它只公开接口中定义的函数。它们仍然在那里，如果没有转换到适当的对象，您就无法访问它们。|4854157|entityMap^0|A|D|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Because when you try <code>s.PrintMore()</code> "s" is of type IShape, so it only knows about the interface functions. You'd need to do something like 

<pre><code>Square s = (Square)shape; 
s.PrintMore();
</code></pre>

or

<pre><code>((Square)shape).PrintMore(); // assuming you're positive its a Square type
</code></pre>

Think of the interface like a wrapper over your object that only exposes only the functions defined in the interface. They're still there, you just can't access them without a cast to the appropriate object.

I'm having a bit of an issue casting my objects in C# to be able to use the additional object methods besides the ones declared in the interface. Below is a simple example of what I am speaking about.

<pre><code>public interface IShape
{
 void Print();
}

public class Square : IShape
{
 #region IShape Members

 public void Print()
 {
 HttpContext.Current.Response.Write("Square Print Called"); 
 } 

 #endregion

 public void PrintMore()
 {
 HttpContext.Current.Response.Write("Square Print More Called"); 
 } 
}
</code></pre>

Why when this code below is called am I not able to access PrintMore()?

<pre><code>IShape s = (Square)shape;
s.PrintMore() // This is not available. only Print() is. 
</code></pre>

Any help and explanation would be helpful?

Casting in C# with interfaces to be able to use all object functionality

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在C#中转换我的对象时遇到了一点问题，这样除了在接口中声明的方法之外，我还可以使用其他的对象方法。下面是我所说的一个简单的例子。public interface IShape{    void Print();}public class Square : IShape{    #region IShape Membe...

问在带有接口的C#中进行强制转换，以便能够使用所有对象功能
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问在带有接口的C#中进行强制转换，以便能够使用所有对象功能EN