The scores on the quiz were as follows.

Even though many of the scores were not very good, only three of the questions were missed by more than 4 people. Of those 3, two were almost straight from the videos. There is no excuse for getting them wrong!

1. What is wrong with the following code?

    

   int count = 2000 * 3000 * 4000;

   • Wrong data type 3
   • Variable is undefined 1
   • Integer overflow 15 This is the right answer. This question was almost straight from the videos.
   • Illegal expression 2
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8. Assume you have a String, bookText, representing the text contents of the book Alice In the Wonderland. Which of the following will find the number of occurrences of "Mad Hat" in bookText?

    

   a)

   public int occurrencesOfMadHat() {

   int occurrences = bookText.length();

   return occurrences;

   }

   b)

   public int occurrencesOfMadHat() {

   String word = "Mad Hat";

   int bookLength = bookText.length( );

   int lengthAfterReplace = bookText.replace(word , "").length( );

   int occurrences = (bookLength - lengthAfterReplace) / word.length( );        

   return occurrences;

   }

   c)

   public int occurrencesOfMadHat() {

   int bookLength = bookText.length();

   int afterReplace = bookText.indexOf("Mad Hat");

   int occurrences = (int) ( (bookLength - afterReplace) / word.length() );        

   return occurrences;

   }

   d)

   public int occurrencesOfMadHat() {

   return bookText.indexOf("Mad Hat");

   }
   • a 1
   • b 11 This is the right answer. This question was almost straight from the videos.
   • c 4
   • d 5
9.  

10. Does casting distribute over arithmetic operations? If it did, the following would be mathematical identities, i.e., they would be true for all values of x and y. 

    In these examples, imagine that x and y are both double variables. Do these equalities hold?

    • (int) (x + y) = ((int) x) + ((int) y)
    • (int) (x - y) = ((int) x) - ((int) y)
    • (int) (x * y) = ((int) x) * ((int) y)
    • (int) (x / y) = ((int) x) / ((int) y)

    
    In these examples, imagine that x and y are both int variables. Do these equalities hold?

    • (double) (x + y) = ((double) x) + ((double) y)
    • (double) (x - y) = ((double) x) - ((double) y)
    • (double) (x * y) = ((double) x) * ((double) y)
    • (double) (x / y) = ((double) x) / ((double) y)
    • Casting to int distributes over arithmetic operations; casting to double distributes over arithmetic operations 4
    • Casting to int distributes over arithmetic operations; casting to double does not distribute over arithmetic operations 7
    • Casting to int does not distribute over arithmetic operations; casting to double distributes over arithmetic operations 8
    • Casting to int does not distribute over arithmetic operations; casting to double does not distribute over arithmetic operations 2  This is the right answer.  
      
                               Only 2 people got this right. Congratulations to the two of you!                              
      
                  (int) (x / y) != ((int) x) / ((int) y)
      If double x = 3.0 and double y = 1.5  then (int) (x/y) = 2 but ((int) x) / ((int) y) = 3.
      
                 (double) (x / y) != ((double) x) / ((double) y)
      If int x = 3 and int y = 2 then (double) (x/y) = 1.0 but ((double) x) / ((double) y) = 1.5.