Given: class ConSuper { protected ConSuper() { this(2); } System.out.print("1"); } protected ConSuper(int a) { System.out.print(a); } } and public class ConSub extends ConSuper{ ConSub(){ this(4)…

The correct answer is B. 2143. Option B is correct assuming the main method calls new ConSub() (the no-arg version) - there appears to be a typo in the transcription. Tracing the call chain: ConSub() immediately delegates via this(4) to ConSub(int a), which triggers an implicit super() call to ConSuper()…

Java Object-Oriented Approach

Question

Given:
class ConSuper {
protected ConSuper() {
this(2);
}
System.out.print("1");
}
protected ConSuper(int a) {
System.out.print(a);
}
}
and
public class ConSub extends ConSuper{
ConSub(){
this(4);
System.out.print("3");
}
ConSub(int a) {
System.out.print(a);
}
public static void main (String[] args){
new ConSub(4);
}
}
What is the result?

Options

A3134
B2143
C214
D244

How the community answered

(35 responses)

A
11% (4)
B
80% (28)
C
3% (1)
D
6% (2)

Explanation

Option B is correct assuming the main method calls new ConSub() (the no-arg version) - there appears to be a typo in the transcription. Tracing the call chain: ConSub() immediately delegates via this(4) to ConSub(int a), which triggers an implicit super() call to ConSuper(), which in turn delegates via this(2) to ConSuper(int a) - printing "2"; unwinding back to ConSuper() prints "1"; unwinding back to ConSub(int a) prints "4"; finally unwinding back to ConSub() prints "3" - giving 2143.

Why distractors are wrong:

A (3134): Would require ConSub()'s print("3") to execute before delegation via this(), but this() must be the first statement and blocks anything after it until the chain unwinds.
C (214): This is actually the output if main calls new ConSub(4) directly - it skips ConSub() entirely, so the "3" is never printed.
D (244): No execution path produces two 4s; this confuses the 4 passed into ConSub(int a) with something printed by ConSuper.

Memory tip: Think of this() and super() constructor chaining as a stack - execution dives all the way down to the innermost constructor before any print statements run on the way back up, so read the output bottom-up through the call chain.

Topics

#constructor chaining#this() and super() calls#inheritance#execution order

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