posted 16 years ago
You're calling PrintStream's println() method and passing in the return value of the return_Number() method. However, the return type of the return_Number() method is void, which is why it's not valid to pass the result of this method as an argument to the println() method - the return_Number() method doesn't yield a return value.
[ October 07, 2008: Message edited by: Jelle Klap ]
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