Answer:

It's a little inconvenient to have to declare every variable before it is used, but it's much safer. If the compiler would accept undeclared variables, then it would also accept misspelled names and treat them as valid variables. This can easily lead to incorrect programs. When variables must be declared, the unintentional creation of a variable is simply impossible, and a whole class of possible bugs is avoided.

Answer:

A precondition is a condition that has to hold at a given point in the execution of a program, if the execution of the program is to continue correctly. For example, the statement "x = A[i];" has two preconditions: that A is not null and that 0 <= i < A.length. If either of these preconditions is violated, then the execution of the statement will generate an error.

Also, a precondition of a subroutine is a condition that has to be true when the subroutine is called in order for the subroutine to work correctly.

Answer:

Suppose that a programmer recognizes a precondition at some point in a program. This is a signal to the programmer that an error might occur if the precondition is not met. In order to have a correct and robust program, the programmer must deal with the possible error. There are several approaches that the programmer can take. One approach is to use an if statement to test whether the precondition is satisfied. If not, the programmer can take some other action such as printing an error message and terminating the program. Another approach is to use a try statement to catch and respond to the error. This is really just a cleaner way of accomplishing the same thing as the first approach. The best approach, when it is possible, is to ensure that the precondition is satisfied as a result of what has already been done in the program. For example, if the precondition is that x >= 0, and the preceding statement is "x = Math.abs(y);", then we know that the precondition is satisfied, since the absolute value of any number is greater than or equal to zero.

Answer:

The binarySearch() method searches for an integer N in a sorted array of integers A[]. The loop invariant is implicit in the discussion of the method in Subsection 7.4.1: "If N is in the array, then N is in at one of the indices in the range lowestPossibleLoc and highestPossibleLoc."

This statement is true before the loop is started, since at that point, the range of indices from lowestPossibleLoc to highestPossibleLoc includes the entire array. In the loop, when A[mid] is compared to N, the fact that A is sorted means that half of the range of possible indices can be discarded. After the value of lowestPossibleLoc or highestPossibleLoc is modified to take that into account, the loop invariant remains true.

If at any point N is found, the loop ends and the method returns the correct value. If N is never found, the loop will end when lowestPossibleLoc is larger than highestPossibleLoc. At that point, the truth of the loop invariant implies that N is not in the array.

Answer:

The class Throwable represents all possible objects that can be thrown by a throw statement and caught by a catch clause in a try..catch statement. That is, the thrown object must belong to the class Throwable or to one of its (many) subclasses such as Exception and RuntimeException. The object carries information about an exception from the point where the exception occurs to the point where it is caught and handled.

Answer:

The problem of large values in a 3N+1 sequence was discussed in Section 8.1. In that section, it is pointed out that the test "if (N > 2147483646/3)" can be used to test whether the value of N has become too large. This test is used in the following method.

/** Print the 3N+1 sequence starting from N.  If N
 * is not greater than 0 or if the value of N exceeds
 * the maximum legal value for ints, than an
 * exception will be thrown.
 */
static void printThreeNSequence(int N) {
   if (N < 1) {
      throw new IllegalArgumentException(
                  "Starting value for 3N+1 sequence must be > 0.");
   }
   System.out.println("3N+1 sequence starting from " + N + " is: ");
   System.out.println(N);
   while (N > 1) {
      if (N % 2 == 0) {  // N is even.  Divide by 2.
          N = N / 2;
      }
      else {  // N is odd.  Multiply by 3 and add 1.
          if (N > 2147483646/3) {
             throw new ArithmeticException("Value has exceeded the largest int.");
          }
          N = 3 * N + 1;
      }
      System.out.println(N);
   }
}

(Note that it would be possible to declare that this routine can throw exceptions by adding a "throws" clause to the heading:

static void printThreeNSequence(int N)
           throws IllegalArgumentException, ArithmeticException {

However, this is not required since IllegalArgumentExceptions and ArithmeticExceptions are not checked exceptions.)

Answer:

We can replace the if statements that check for errors with assert statements that give the same error messages:

/** Print the 3N+1 sequence starting from N.
  * Precondition:  N > 0 and the 3N+1 sequence for N does not contain
  * any numbers that are too big to be represented as 32-bit ints.
  */
static void printThreeNSequence(int N) {
   
   assert  N > 0 : "Starting value for 3N+1 sequence must be > 0.";

   System.out.println("3N+1 sequence starting from " + N + " is: ");
   
   System.out.println(N);
   while (N > 1) {
      if (N % 2 == 0) {  // N is even.  Divide by 2.
          N = N / 2;
      }
      else {  // N is odd.  Multiply by 3 and add 1.
          assert  N <= 2147483646/3 : "Value has exceeded the largest int.";
          N = 3 * N + 1;
      }
      System.out.println(N);
   }
   
}

The first version of the method will always check for errors when the program is run. The second version, on the other hand, does not actually do any error checking when the program is run in the ordinary way. In order for assert statements to be executed, the program must be run with assertions enabled. The assert statements are really there only to do error-checking during debugging and testing. (In this particular case, I would say that an exception should definitely be thrown when N exceeds the maximum legal value, but that it's reasonable to use an assert to check whether N > 0.)

Answer:

Subclasses of the class Exception which are not subclasses of RuntimeException are checked exceptions. This has two consequences: First, if a method can throw such an exception, then it must declare this fact by adding a throws clause to the method heading. Second, if a routine includes any code that can generate such an exception, then the routine must deal with the exception. It can do this by including the code in a try statement that has a catch clause to handle the exception. Or it can add a throws clause to the method definition to declare that calling the method might throw the exception.

Answer:

try {
   processData();
}
catch (IOException e) {
   System.out.println("An IOException occurred while processing the data.");
}

Answer:

Terminating the program is too drastic, and this tactic certainly doesn't lead to robust programs! It's likely that the subroutine doesn't know what to do with the error, but that doesn't mean that it should abort the whole program. When the subroutine throws an exception, the subroutine is terminated, but the program that called the subroutine still has a chance to catch the exception and handle it. In effect, the subroutine is saying "Alright, I'm giving up. Let's hope someone else can deal with the problem."

Answer:

In the absence of other information, the second algorithm, with run time Θ(n*log(n)), is the better choice, since n*log(n) is much smaller than n², for most values of n. However, it's not completely certain that the second algorithm is the better choice in a particular case. First of all, although the n*log(n) algorithm is certainly better than the n² algorithm for large enough values of n, that is not necessarily true for n = 100. Second, there is the issue of "average-case" run time. Even if the n*log(n) algorithm is better for most inputs of size 100, it might not be better for all such inputs.

Answer:

The run time of this algorithm is Θ(n). There are several things in the code that are evaluated n times: the test "i < A.length", the increment "i++", and the test in the if statement. The initialization is done once, and nothing is executed more than n times. It follows that both the worst-case and the average case run times are Θ(n).