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meyer q.61

 
Ken Truitt
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Here's another one I didn't get from the Meyer mock:

class sup {
void method() throws Exception { }
}

class test extends sup {
public static void main(String[] args) { //1
sup s = new test();
s.method();
}

void method() {} // 2
}

What can be done to avoid compiler error?

---------------

a. Add a "throws Exception" clause at line 1 <---- first correct answ.
b. Add a "throws Exception" clause at line 2
c. Use a try catch block when calling s.method(). <--- second correct answ.
d. There is no compile error. <--- my answer
e. None of the above.

----------------

As I recall, override rules say that you can use the same, fewer, or
sub-types of exceptions from the overridden method in the overriding
method. The overriding version states no exceptions, which is valid.
But the compiler still requires that an exception be handled or declared.
Why?
 
Bob Ruth
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Because the reference declared is a sup reference (superclass) and it is handed a reference to a test (subclass) object, polymorphism is in play. When the is the case, the compiler is going to require that you construct the code to handle each possible type of object whether YOUR code does it or not. The compiler isn't going to assume that the super reference will ONLY hold the subclass type....rather, it will have you code the try catch block so that either object could be put into the reference.
 
Ken Truitt
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Thanks--excellent explanation
 
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