Java Programming - Java.lang Class
- Java.lang Class - General Questions
- Java.lang Class - Finding the output
- Java.lang Class - Pointing out the correct statements
public class ExamQuestion7
{
static int j;
static void methodA(int i)
{
boolean b;
do
{
b = i<10 | methodB(4); /* Line 9 */
b = i<10 || methodB(8); /* Line 10 */
}while (!b);
}
static boolean methodB(int i)
{
j += i;
return true;
}
public static void main(String[] args)
{
methodA(0);
System.out.println( "j = " + j );
}
}
The lines to watch here are lines 9 & 10. Line 9 features the non-shortcut version of the OR operator so both of its operands will be evaluated and therefore methodB(4) is executed.
However line 10 has the shortcut version of the OR operator and if the 1st of its operands evaluates to true (which in this case is true), then the 2nd operand isn't evaluated, so methodB(8) never gets called.
The loop is only executed once, b is initialized to false and is assigned true on line 9. Thus j = 4.
try
{
Float f1 = new Float("3.0");
int x = f1.intValue();
byte b = f1.byteValue();
double d = f1.doubleValue();
System.out.println(x + b + d);
}
catch (NumberFormatException e) /* Line 9 */
{
System.out.println("bad number"); /* Line 11 */
}
The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.
class Q207
{
public static void main(String[] args)
{
int i1 = 5;
int i2 = 6;
String s1 = "7";
System.out.println(i1 + i2 + s1); /* Line 8 */
}
}
This question is about the + (plus) operator and the overriden + (string cocatanation) operator. The rules that apply when you have a mixed expression of numbers and strings are:
If either operand is a String, the + operator concatenates the operands.
If both operands are numeric, the + operator adds the operands.
The expression on line 6 above can be read as "Add the values i1 and i2 together, then take the sum and convert it to a string and concatenate it with the String from the variable s1". In code, the compiler probably interprets the expression on line 8 above as:
System.out.println( new StringBuffer()
.append(new Integer(i1 + i2).toString())
.append(s1)
.toString() );
public class SqrtExample
{
public static void main(String [] args)
{
double value = -9.0;
System.out.println( Math.sqrt(value));
}
}
The sqrt() method returns NaN (not a number) when it's argument is less than zero.
String s = "ABC";
s.toLowerCase();
s += "def";
System.out.println(s);
String objects are immutable. The object s above is set to "ABC". Now ask yourself if this object is changed and if so where - remember strings are immutable.
Line 2 returns a string object but does not change the originag string object s, so after line 2 s is still "ABC".
So what's happening on line 3? Java will treat line 3 like the following:
s = new StringBuffer().append(s).append("def").toString();
This effectively creates a new String object and stores its reference in the variable s, the old String object containing "ABC" is no longer referenced by a live thread and becomes available for garbage collection.