class A { public A(){ System.out.print ("A "); } } class B extends A{ public B(){ //line n1 System.out.print ("B "); } } class C extends B{ public C(){ //line n2 System.out.print ("C "); } public…

The correct answer is C. A B C. Option C (A B C) is correct because Java implicitly inserts a super() call as the first statement in any constructor that doesn't explicitly call super() or this(). So when new C() is invoked, C's constructor calls B's constructor, which calls A's constructor - execution then…

Working with Inheritance

Question

class A {
public A(){
System.out.print ("A ");
}
}
class B extends A{
public B(){ //line n1
System.out.print ("B ");
}
}
class C extends B{
public C(){ //line n2
System.out.print ("C ");
}
public static void main(String[] args) {
C c = new C();
}
}
What is the result?

Options

AC B A
BC
CA B C
DCompilation fails at line n1 and line n2

How the community answered

(29 responses)

A
14% (4)
B
7% (2)
C
76% (22)
D
3% (1)

Explanation

Option C (A B C) is correct because Java implicitly inserts a super() call as the first statement in any constructor that doesn't explicitly call super() or this(). So when new C() is invoked, C's constructor calls B's constructor, which calls A's constructor - execution then unwinds back down the chain, printing "A", then "B", then "C".

Why the distractors are wrong:

A (C B A) reverses the order - constructors complete from top-down (ancestor first), not bottom-up.
B (C) would only be true if child constructors could skip calling their parent - they cannot; the implicit super() is mandatory.
D (Compilation fails) is wrong because the implicit super() insertion is done by the compiler automatically; neither n1 nor n2 have any issue.

Memory tip: Think of it as "you can't exist before your parents" - Java always constructs the ancestry chain from the oldest ancestor (Object → A → B → C) before the child can finish initializing.

Topics

#constructor chaining#implicit super()#inheritance hierarchy#constructor execution order

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