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Hi friends,
public class Child extends Base {

Child(int i) { test (); }

Child(float f) { this ((int)f); }

void test() {
System.out.println("Child.test()");
}

static public void main(String[] a) {
new Child(10.8f).test();
}
}
Select most appropriate answer.

a) Child.test()
Child.test()
b) Compilation Error: No default constructor ( constructor matching Base())
found in class Base.
c) Runtime Error: No default constructor ( constructor matching Base())
found in class Base.
d) Compilation Error: Cannot call this() from a constructor.
The Answer to this question in given as ...a)
Why does Child.test() print twice ?
-Thanks
 
Ranch Hand
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Hey, where's the base class man?
 
Greenhorn
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Originally posted by Makarand Akdar:
Hi friends,
public class Child extends Base {

Child(int i) { test (); }

Child(float f) { this ((int)f); }

void test() {
System.out.println("Child.test()");
}

static public void main(String[] a) {
new Child(10.8f).test();
}
}
Select most appropriate answer.

a) Child.test()
Child.test()
b) Compilation Error: No default constructor ( constructor matching Base())
found in class Base.
c) Runtime Error: No default constructor ( constructor matching Base())
found in class Base.
d) Compilation Error: Cannot call this() from a constructor.
The Answer to this question in given as ...a)
Why does Child.test() print twice ?
-Thanks


test() is called twice.
1. Child(int i) { test (); }
2. new Child(10.8f).test();
 
Ranch Hand
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priya - in this code there is no ref to the superclass(base)
so can we assume that the int constructor is making a call (super) to the superclass and that the superclass has to have a int constructor
OR
that the superclass has a no args constructor , hence there is no problem of super
 
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