Exercise: Java Lang Class

Questions for: Java Lang Class

What will be the output of the program? 
class Tree { } 
class Pine extends Tree { } 
class Oak extends Tree { } 
public class Forest1 
{ 
    public static void main (String [] args)
    { 
        Tree tree = new Pine(); 
        if( tree instanceof Pine ) 
            System.out.println ("Pine"); 
        else if( tree instanceof Tree ) 
            System.out.println ("Tree"); 
        else if( tree instanceof Oak ) 
            System.out.println ( "Oak" ); 
        else 
            System.out.println ("Oops "); 
    } 
}
A:
Pine
B:
Tree
C:
Forest
D:
Oops
Answer: A

The program prints "Pine".

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What will be the output of the program? 
public class Test138 
{ 
    public static void stringReplace (String text) 
    {
        text = text.replace ('j' , 'c'); /* Line 5 */
    } 
    public static void bufferReplace (StringBuffer text) 
    { 
        text = text.append ("c");  /* Line 9 */
    } 
    public static void main (String args[]) 
    { 
        String textString = new String ("java"); 
        StringBuffer textBuffer = new StringBuffer ("java"); /* Line 14 */
        stringReplace(textString); 
        bufferReplace(textBuffer); 
        System.out.println (textString + textBuffer); 
    } 
}
A:
java
B:
javac
C:
javajavac
D:
Compile error
Answer: C

A string is immutable, it cannot be changed, that's the reason for the StringBuffer class. The stringReplace method does not change the string declared on line 14, so this remains set to "java".

Method parameters are always passed by value - a copy is passed into the method - if the copy changes, the original remains intact, line 5 changes the reference i.e. text points to a new String object, however this is lost when the method completes. The textBuffer is a StringBuffer so it can be changed.

This change is carried out on line 9, so "java" becomes "javac", the text reference on line 9 remains unchanged. This gives us the output of "javajavac"

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What will be the output of the program? 
public class StringRef 
{
    public static void main(String [] args) 
    {
        String s1 = "abc";
        String s2 = "def";
        String s3 = s2;   /* Line 7 */
        s2 = "ghi";
        System.out.println(s1 + s2 + s3);
    }
}
A:
abcdefghi
B:
abcdefdef
C:
abcghidef
D:
abcghighi
Answer: C

After line 7 executes, both s2 and s3 refer to a String object that contains the value "def". When line 8 executes, a new String object is created with the value "ghi", to which s2 refers. The reference variable s3 still refers to the (immutable) String object with the value "def".

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What will be the output of the program? 
String a = "newspaper";
a = a.substring(5,7);
char b = a.charAt(1);
a = a + b;
System.out.println(a);
A:
apa
B:
app
C:
apea
D:
apep
Answer: B

Both substring() and charAt() methods are indexed with a zero-base, and substring() returns a String of length arg2 - arg1.

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What will be the output of the program? 
interface Foo141 
{ 
    int k = 0; /* Line 3 */
} 
public class Test141 implements Foo141 
{
    public static void main(String args[]) 
    {
        int i; 
        Test141 test141 = new Test141(); 
        i = test141.k; /* Line 11 */
        i = Test141.k; 
        i = Foo141.k; 
    } 
}
A:
Compilation fails.
B:
Compiles and runs ok.
C:
Compiles but throws an Exception at runtime.
D:
Compiles but throws a RuntimeException at runtime.
Answer: B

The variable k on line 3 is an interface constant, it is implicitly public, static, and final. Static variables can be referenced in two ways:

Via a reference to any instance of the class (line 11)

Via the class name (line 12).

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