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Type: Posts; User: dlorde

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    Re: problem with interface

    That's pretty much it, although to be more precise:

    AnagramFinder finder = declare a reference variable called 'finder', of type AnagramFinder
    finder = new areAnagramsx(); = create a new object...
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    Re: problem with interface

    Looks like you've got very confused...

    To use the interface, you need to create an instance of a class that implements it (e.g. areAnagramsx) and then call a method on that object, for example:...
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