CS 1044 Test 2 Form A Key            Spring 2000
 Q  A  Explanation
1.  3  The value of             I   J:
       Prior to the loop:       6   0
       In the first iteration:  5  -5
       In the second iteration: 4  -9
       In the third iteration:  3 -12
       The main thing to remember here is that I counts down and J sums the 
       neagative values of I after it is decremented.
2.  4  (See 1. Explanation above.)
3.  3  The value of              J    varq     newq     iteration
       Prior to the loop:        7    1.0      0.0         0
       In the first iteration:  13    1.0      8.0         1
       In the second iteration: 19    1.0     14.0         2
       In the third iteration:  25    1.0     20.0         3
                                ............................
                                73    1.0     68.0        11
                                79    1.0     73.0        12
       Note that varq never changes, the loop controlled by J counts up by 6 starting at 7,
       and newq equals 1 more than J before it changes in each iteration.
4.  5  (See 3. Explanation above.)
5.  2  The loop executes only 1 time since 0 <= 0 (count <= total). Thus only 2 values from
       the file are read: 4 into mystery and 1 into value. 
6.  7  Before the for loop the value of mystery is 4.  Therefore, the loop is executed 4 
       times.  The loop sums up 1, 7, 5 & 3.  Therefore, the value of sum is 13.
7.  7  Initially the value of mystery read in is 4.  It is then decremented by 1 in the loop.  
       The loop continues until mystery reaches 0 or a negative number is read into value.
       The next 4 numbers after the first number (1, 7, 5, 3) are input until mystery is 0.
       Thus sum = 13 = 1 + 7 + 5 (the 3 is read but not added to sum.)
8.  2  The initial value of j is:  j=28
       After the first iteration:  j=25
       After the second iteration: j=22
       After the third iteration:  j=19
       After the fourth iteration: j=16
       After the fifth iteration:  j=13, therefore the loop executes 5 times before J == 13.
9.  5  (See 8. Explanation above.) Note that when J becomes 13, the loop stops.
10. 6  The outer loop, i, executes for the values 0, 1, 2 the inner loop, j, executes for the
       values 0 & 1 (twice for each outer loop value). Sum is reset to 0 before each execution
       of the inner loop. Thus sum = 1 + 2 * i for each iteration of i. The 1 is the only
       value of j that gets added & sum doesn't add in between loops since it resets to 0. 
11. 8  Similar to the above question, except that sum is NOT reset to 0 before the inner loop.
       Thus sum = sum + 1 + 2 * i for each iteration of i, which means sum = 9.
12. 9  In the function, the first parameter is pass-by-reference.  Therefore, a variable
       must be specified for it.  A value cannot be passed. The second parameter is 
       pass-by-value,  therefore, a numeric expression can be specified for it.
13. 2  Since pass-by-reference parameter communication is in both directions (2-way) and
       pass-by-value parameter communication is only into the called function (1-way),
       the first parameter must be passed by reference and the second by value.
14. 1  Since the variable 'Ben' is passed by value the changes made in the function are
       not reflected in the main program.  Therefore, the value of 'Ben' remains -5.
15. 6  Since the variable 'Jerry' is passed by reference to the function DoThis.  In the 
       function DoThis 'Beta' is set to 999.  Therefore, the value of Jerry becomes 999.
16. 2  Since Tmp is NOT a global variable and it is also defined locally in DoThis the
       changes to Tmp in DoThis has no effect on the Tmp variale in main.
17. 3  The key point to remember here is that rows are first and columns are second in 
       a two dimentional array.  Also the arrays indexes start from 0.
18. 1  The array elements are NOT changed by SmallValue so it should be passed by const int
19. 2  The number of array elements are NOT changed by SmallValue so it should be passed by value.
20. 2  Since the loop starts at 1 the initial value for smallest must be List[0] for all array
       values to be checked.
21. 5  When a smaller value in the array is located then smallest must be reset to that value,
       which is List[Lo].
22. 7  The array locations 0 .. 3 are output by the loop.
23. 3  aRay[4] = bRay[4]; == aRay[4] = 10;
24. 2  cRay has a dimension of 4, but the Size passed to InitArray is 5 which will cause
       the InitArray function to access beyond the bounds of the array.
25. 1  The first parameter to InitArray must be an array and the second must be a simple int.
26. 3  cRay[aRay[1]] = bRay[Size-aRay[0]]; == cRay[2] = bRay[4]; == cRay[2] = 10;
27. 4  Line2 is concatenated on to the end of Line1 and stored in Line1.
28. 2  Line1 contains 7 characters, counting the spaces. 
29. 1  Line2 is copied into (and replaces) Line1. strcpy(Line1, Line2); == Line1 = Line2;
30. 1  The 'i' in "Daniel" comes before the second 'n' in "Danny".