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POCreating an instance of derived class through the base class without hardcoding
 

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    <p>My problem is as follows:</p> <p>I have a base class that needs to be abstract. It has several derived classes, each with their own special properties that are contained in the Properties member.</p> <p>I need to be able to create a new instance of one of these derived classes, so that all the members are equivalent but modifying the new instance doesn't modify the original.</p> <p>Finally, I want to do it without having to hardcode in every derived type of the base class. (Which would, admittedly, be the easiest solution, however that isn't the point)</p> <p>All the derived classes satisfy an "is-a" relationship to the base class.</p> <p>Here is the code:</p> <pre><code>public abstract class BaseClass { //Default properties here int x, y, z, ...; //Custom made class to hold custom properties protected Attributes Properties; public BaseClass createNewInstance() { return createNewInstanceStep1(); } //Each derived class implements their own version of this, //to handle copying any custom members contained in Properties. protected abstract BaseClass createNewInstanceStep2(); protected BaseClass createNewInstanceStep1() { BaseClass newInstance = new BaseClass(); // &lt;- Doesn't work because class is abstract //Copy default properties newInstance.x = x; newInstance.y = y; newInstance.z = z; //Call the new instance's step 2 method, and return the result. return newInstance.createNewInstanceStep2(); } } </code></pre> <p>The issue with this code is the BaseClass newKeyFrame = new BaseClass(); line. As the class is abstract, you cannot create an instance of it.</p> <p>The problem is that I need to be able to call the constructor of whatever type the derived class is, as they all have different code in their constructors that cannot be shared.</p> <p>I've heard that using Reflection might be a viable solution, however I have no idea how.</p> <p>How can I solve this without having to hardcode in a case for every derived type?</p>
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