Blinded by the lights

Example

Can you predict the output of the following code?

class SubTest extends Test {
    public int aNumber;

    public SubTest() {
        aNumber = 17;
    }

    public void doubleANumber() {
        System.out.println("Inside SubTest.doubleANumber()");
        aNumber *= 2;
    }
}

public class Test {
    public int aNumber;

    public Test() {
        aNumber = 6;
    }

    public void doubleANumber() {
        System.out.println("Inside Test.doubleANumber()");
        aNumber *= 2;
    }

    public static void main(String[] args) {
        Test t = new SubTest();
        t.doubleANumber();
        System.out.println(”The value of aNumber is ” + t.aNumber);
    }
}

Result

The output is:
Inside SubTest.doubleANumber()
The value of aNumber is 6

As you see, it was not the reference type but the real object type of t variable that decided which method was invoked at runtime. This way, aNumber of SubTest class was modified. This variable shadowed aNumber of Test class. Therefore, Test.aNumber variable was untouched (didn’t change at all).

A few weeks ago I described how to format output using System.out.printl() and System.out.println() methods…

Try to run the code below:

double avgAge = 245 / 34;
System.out.format("Average age is %d.", avgAge);

Without doubts you’ll get an exception thrown at runtime:

Average age is Exception in thread "main"
java.util.IllegalFormatConversionException:
d != java.lang.Double
    at java.util.Formatter$FormatSpecifier.failConversion(Unknown Source)
    at java.util.Formatter$FormatSpecifier.printInteger(Unknown Source)
    at java.util.Formatter$FormatSpecifier.print(Unknown Source)
    at java.util.Formatter.format(Unknown Source)
    at java.io.PrintStream.format(Unknown Source)

The reason is becase wrong conversion label was used. To recall, conversion label is ‘%’ sign followed by a letter b, c, d, f, or s inside System.out.format().

In the above example, %d meant that an integral type (byte, Byte, short, Short, int, Integer, long, Long, BigInteger) would be used. However, finally there was a floating point passed (avgAge). This example would work if this conversion label were used: %f.

So, remember to be careful with conversion label!

Can you predict the output/result of the following code?

String s = "A String ";
s += 12345;
System.out.print(s);

The answer is: “A String 12345“.

The reason for that is described in the desciption of String class in Java 2 Platform SE 5.0 API:

The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings. String concatenation is implemented through the StringBuilder(or StringBuffer) class and its append method. String conversions are implemented through the method toString, defined by Object and inherited by all classes in Java.