## CS 1401, Quiz 8

**Date:** Monday, October 30, 2006

**Name** (please type legibly, ideally in block letters):
______________________________________________________________________
1. Write a piece of code that, given an array *h* of real
numbers, computes the average value of elements in this array.

Solution:
double sum = 0.0;
for (int i = 0; i < h.length; i++)
{sum = sum + h[i];}
double average = sum/h.length;

2. To check the correctness of the code you wrote in Problem
1, write a piece of code that defines a new array *h*
consisting of 2 elements of type double and assigns, to elements of
this array, values 1.0 and 2.0. Trace step-by-step how the piece of
code you wrote in Problem 1 will compute the average value of this
array.

Solution:
double[] h = {1.0, 2.0};
Tracing:
The fact that we have defined an array h with 2 elements means, in
effect, that we have defined two variables h[0] and h[1]:
----
| \|
---\
h \
\|
- -------
| 1.0 | h[0]
-------
| 2.0 | h[1]
-------
Then, we define a new variable sum and assigning to it a value 0.0:
-------
| 0.0 |
-------
sum
Next, we get to the for-loop. First, we define a new integer variable
i and assign to it the value 0:
------- -------
| 0.0 | | 0 |
------- -------
sum i
We check the condition: the length of our array h is 2, so i is smaller
than this length. Thus, the condition i < h.length is satisfied, and
we implement the operation in the body of the loop. Since i = 0,
we thus perform the operation
sum = sum + h[0];
In other words, we add the values of sum (0.0) and h[0] (1.0) and place
the result 0.0 + 1.0 = 1.0 into the variable sum:
------- -------
| 1.0 | | 0 |
------- -------
sum i
Then, we go to the update statement i++; this statement adds 1 to i, so
we get:
------- -------
| 1.0 | | 1 |
------- -------
sum i
The condition i < h.length is still satisfied, so we again go inside
the loop. This time, i is 1, so the statement in the body of the loop
becomes
sum = sum + h[1];
In other words, we add the values of sum (1.0) and h[1] (2.0) and place
the result 1.0 + 2.0 = 3.0 into the variable sum:
------- -------
| 3.0 | | 1 |
------- -------
sum i
Then, we go to the update statement i++; this statement adds 1 to i, so
we get:
------- -------
| 3.0 | | 2 |
------- -------
sum i
The condition i < h.length is no longer satisfied, since i = 2 and the
length of h is also 2. Thus, we get out of the loop.
Finally, we declare a new variable average and assign to it the value
3.0 / 2 = 1.5:
------- ------- -------------
| 3.0 | | 2 | | 1.5 |
------- ------- -------------
sum i average