Irem In-Line: Fill & Download for Free

While standing in line at Walmart, A.K.A. “Ghetto Mart, “ there was a young girl in front of me with $20 bill and 1 item when some P.O.S. woman just walked in front of her, me and 4 other people behind me. I told her, “ There's a young lady standing in front of me ( about 10,) with a $20 bill in her hand and 1 irem, don't you see her? “ She then apologized to me. I told her to “ apologize to the young lady you walked in front of, to the 4 people behind me and then me,” which that P.O.S. then did ! Mission acomplished !

Entering a try block and executing the code within it is not expensive as long there is no Exception. Creating and throwing Exception in Java is relatively more expensive due to the requirement of filling stack trace and possible stack unwinding.Byte code generated for a method contains an exception table. The table contain an entry for each exception that the method catches and the program counter offset to the code to run when the exception occurs.The exception table is not generated at run time and there is no extra penalty for entering a try block.You can see this for yourself by looking at the byte code a compiler generates, with and without a try block.For example -Below are two methods to calculate remainder given dividend and divisor.First method called doStuffNoE does not handle ArithmeticException (runtime exception) that occurs when a Java program divides by zero.The second called doStuff handles ArithmeticException, prints the stack trace and throws the exception again -public int doStuffNoE(int dividend, int divisor) {  return dividend % divisor; } public int doStuff(int dividend, int divisor) {  try {  return dividend % divisor;  } catch (ArithmeticException e) {  e.printStackTrace();  throw e;  } } Following bytecode is generated for the above -For doStuffNoE (does not catch exception)// Method descriptor #15 (II)I  // Stack: 2, Locals: 3  public int doStuffNoE(int dividend, int divisor);  0 iload_1 [dividend]  1 iload_2 [divisor]  2 irem  3 ireturn  Line numbers:  [pc: 0, line: 5]  Local variable table:  [pc: 0, pc: 4] local: this index: 0 type: ExceptionTest  [pc: 0, pc: 4] local: dividend index: 1 type: int  [pc: 0, pc: 4] local: divisor index: 2 type: int The actual remainder calculation is done from program counter (PC) offset 0 to 3 and fairly straight forward (consult Chapter 6. The Java Virtual Machine Instruction Set for reference)For doStuff (catches exception)// Method descriptor #15 (II)I  // Stack: 2, Locals: 4  public int doStuff(int dividend, int divisor);  0 iload_1 [dividend]  1 iload_2 [divisor]  2 irem  3 ireturn  4 astore_3 [e]  5 aload_3 [e]  6 invokevirtual java.lang.ArithmeticException.printStackTrace() : void [20]  9 aload_3 [e]  10 athrow  Exception Table:  [pc: 0, pc: 3] -> 4 when : java.lang.ArithmeticException  Line numbers:  [pc: 0, line: 10]  [pc: 4, line: 11]  [pc: 5, line: 12]  [pc: 9, line: 13]  Local variable table:  [pc: 0, pc: 11] local: this index: 0 type: ExceptionTest  [pc: 0, pc: 11] local: dividend index: 1 type: int  [pc: 0, pc: 11] local: divisor index: 2 type: int  [pc: 5, pc: 11] local: e index: 3 type: java.lang.ArithmeticException  Stack map table: number of frames 1  [pc: 4, same_locals_1_stack_item, stack: {java.lang.ArithmeticException}] Just like the previous version, here too, the actual remainder calculation is done from PC offset 0 to 3. There is no extra code executed at the start (No try block penalty)PC offset 4 to 10 handles the exception and is executed only if an exception occurs.Note the presence of an exception table in this version in line 13.The exception table tells the run time environment that -The code block from PC 0 to PC 3 throws ArithmeticExceptionJump to PC 4 if ArithmeticException occursPC 4 to PC 10 contains instructions for handling the exceptionIf there is no error, both methods will execute code from PC offset 0 to 3 and there will be no difference in performance.There is good detail here with multiple catch block example - Chapter 3. Compiling for the Java Virtual Machine

The reason is that because you never change the value of the boolean attempt. You just change value2 on line 8, but you never re-evaluate attempt, therefore attempt always remains true and for that reason the while loop never ends.