nov 200291.3913 r mcfadyen1 a traditional software development process unit test integration test...
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Nov 2002 91.3913 R McFadyen 1
A Traditional Software Development Process
Unit test
Integration test
System test
Detailed design
Architectural design
Analysis
Requirements analysis
Business modeling process
Business results
Nov 2002 91.3913 R McFadyen 2
Black Team
Peopleware: Productive Projects and Teams
by Tom Demarco and Timothy Lister
Ch 19: The Black Team
•"legendary" Black Team at IBM in 60’s
•a particularly effective testing team
•the best testers at the company
•testers delighted in finding new ways to break software
•programmers dreaded having their software go through Black Team testing
•promoted themselves by dressing all in black
•the spirit of the Black Team has survived
Nov 2002 91.3913 R McFadyen 4
White-box Testing
•Tests are derived from knowledge of the software’s construction
•aka structural, glass-box, or clear-box testing
•Tester analyzes the code, the algorithm, to derive tests
Nov 2002 91.3913 R McFadyen 5
Whitebox testing
Tests are derived from knowledge of the software’s construction
Path testing: A whitebox testing technique based on flow of control
•Construct a flow graph for Cyclomatic Complexity
•gives us the maximum number of tests necessary to cover all edges in the flow graph
•Design test cases such that each transition is traversed at least once - examine each condition and select an input for the true branch and another for the false branch
•Each test should contain an edge that is not contained in any other test
Nov 2002 91.3913 R McFadyen 6
White-box Testing Example
FindMean(float Mean, FILE ScoreFile) …see next slide
What is the flow graph for FindMean?
What is the CC?
What tests can we generate from the flowgraph/CC … to know all paths have been executed?
Nov 2002 91.3913 R McFadyen 7
/*Read in and sum the scores*/
White-box Testing Example
FindMean(float Mean, FILE ScoreFile)
{ SumOfScores = 0.0; NumberOfScores = 0; Mean = 0;
Read(ScoreFile, Score);
while (! EOF(ScoreFile) {
if ( Score > 0.0 ) {
SumOfScores = SumOfScores + Score;
NumberOfScores++;
} Read(ScoreFile, Score);
}
/* Compute the mean and print the result */
if (NumberOfScores > 0 ) {
Mean = SumOfScores/NumberOfScores;
printf("The mean score is %f \n", Mean);
} else
printf("No scores found in file\n");
}
Nov 2002 91.3913 R McFadyen 8
Prepare for Flow Graph
FindMean (FILE ScoreFile){ float SumOfScores = 0.0;
int NumberOfScores = 0; float Mean=0.0; float Score;Read(ScoreFile, Score);while (! EOF(ScoreFile) {
if (Score > 0.0 ) {SumOfScores = SumOfScores + Score;NumberOfScores++;}
Read(ScoreFile, Score);}/* Compute the mean and print the result */if (NumberOfScores > 0) {
Mean = SumOfScores / NumberOfScores;printf(“ The mean score is %f\n”, Mean);
} elseprintf (“No scores found in file\n”);
}
1
2
3
4
5
7
6
89
Nov 2002 91.3913 R McFadyen 9
Constructing the Logic Flow Diagram
4
3
2
1
5
6
7 8
9
CC = 11-9+2 = 4
1, 2, 6, 8, 9
1, 2, 6, 7, 9 is not possible
1, 2, 3, 4, 5, 2, 6, 7, 9
1, 2, 3, 4, 5, 2, 6, 8, 9 is not possible
1, 2, 3, 5, 2, 6, 8, 9
1, 2, 3, 5, 2, 6, 7, 9 is not possible
ok
ok
ok
Note: If node 4 is included then node 7 must be included
These 3 tests cover all paths! We didn’t need 4!
Every line of code is tested!
Nov 2002 91.3913 R McFadyen 10
Constructing the Logic Flow Diagram
4
3
2
1
5
6
7 8
9
CC = 11-9+2 = 4
1, 2, 6, 8, 9
Test with no records in the file
Nov 2002 91.3913 R McFadyen 11
Constructing the Logic Flow Diagram
4
3
2
1
5
6
7 8
9
CC = 11-9+2 = 4
1, 2, 3, 4, 5, 2, 6, 7, 9
A test with one positive score
Nov 2002 91.3913 R McFadyen 12
Constructing the Logic Flow Diagram
4
3
2
1
5
6
7 8
9
CC = 11-9+2 = 4
1, 2, 3, 5, 2, 6, 8, 9
A test with one negative score
Nov 2002 91.3913 R McFadyen 13
Black-box Testing
• A blackbox test is one that focuses on the input/output behaviour of a component without considering its implementation
• The system being tested is a black box, whose behaviour is determined by studying its inputs and outputs
• aka functional testing since it is concerned with functionality and not implementation
• Can use equivalence classes and boundary conditions
Nov 2002 91.3913 R McFadyen 14
Equivalence class testing
• A technique to minimize the number of test cases• An equivalence class is a set of tests with common characteristics• A test case is selected for each class• We use our domain knowledge to generate equivalence classes• Example: suppose a method returns the number of days in a
month, given the month and year.– When we consider months, we could arrive at 3 equivalence
classes:• months with 31 days• months with 30 days• months with 28 or 29 days
– When we consider years, we have two equivalence classes: leap years and non-leap years
– Combining, we have 6 equivalence classes
Nov 2002 91.3913 R McFadyen 15
Six equivalence classes
Equivalence classInput values
month year
31 day month, non-leap year 7 1901
31 day month, leap year 7 1904
30 day month, non-leap year 6 1901
30 day month, leap year 6 1904
28/29 day month, non-leap year 2 1901
28/29 day month, leap year 2 1904
Nov 2002 91.3913 R McFadyen 16
Boundary class testing
•We focus on the boundary conditions of equivalence classes
•Rather than selecting any element in an equivalence class, boundary testing requires that elements be selected from the “edges”
Example
•In general, years that are multiples of 4 are leap years; years that are multiples of 100 are not leap years, unless they are multiples of 400.
•2000 is a leap year, but 1900 is not
•For Year, 2000 and 1900 are good boundary cases
•0 and 13 would be good boundary cases for month
Nov 2002 91.3913 R McFadyen 17
Ten testing classes
Equivalence classInput values
month year31 day month, non-leap year 7 1901
31 day month, leap year 7 1904
30 day month, non-leap year 6 1901
30 day month, leap year 6 1904
28/29 day month, non-leap year 2 1901
28/29 day month, leap year 2 1904
Leap year divisible by 400 2 2000
Non-leap year divisible by 100 2 1900
Non-positive invalid month 0 1904
positive invalid month 13 1904
Nov 2002 91.3913 R McFadyen 18
Example
How would knowledge of the algorithm (the code) affect your choices of equivalence classes and boundary classes?
• First, suppose you are testing a search method. What equivalence and boundary test classes would you use? What are your black-box test classes?
• Next, suppose you learn the search is a binary search … how might this knowledge affect your set of test classes …
Nov 2002 91.3913 R McFadyen 19
Example (continued):
Suppose we have a search module to test. We know:
– the list, elt, must have at least one element
– if the element, key, is found, found will be true and elt[L] = key
– if the key is not found, found will be false
How do we structure, or organize, our test cases?
Nov 2002 91.3913 R McFadyen 20
• How do we structure, or organize, our test cases?
• We have two types of searches: successful and unsuccessful. So, we can partition our search test cases into two classifications: found and not found
• When it comes to lists of elements, we know that lists of length 1, are special cases or boundary points. So, we should have 2 partitions of lists: length 1, and length of more than one.
• When an element is searched for, it can be found in boundary positions. So, we can should have 3 partitions: found at the start, found at the end, and then we should include where it is found in the middle.
Example (continued):
Nov 2002 91.3913 R McFadyen 21
Found Not Found
List of 1 element List of 1 element
•combining the partitions
List of >1 element
In 1st position
In last position
In middle position
List of >1 element
1 2
3
4
5
6
Example (continued):
Nov 2002 91.3913 R McFadyen 22
55
44
•sample tests for the 6 partitions
55 33 88 44
22 33 6 77 88 44
33 66 77 88 22
22 55 88 66 77
1
2
6
4
5
3
55 true, 1
55 false, ?
55 true, 1
44 true, 6
77 true, 3
99 false, ?
Outputs Found, L
Inputs List Key
Example (continued):
Nov 2002 91.3913 R McFadyen 23
•Now, suppose we are examining the algorithm that searches an ordered list for a specific value in Key.
•On examination we see it’s a binary search and we see that it really treats the List as having 3 sections:
elements < midmid-point
elements > mid
Example (continued):
Nov 2002 91.3913 R McFadyen 24
•:We can use this knowledge to further refine our partitioning - we need to test for where the Key we are looking for is at the boundary points for these partitions.
elements < midmid-point
elements > mid
We’ll add two more tests relating to the boundary points – finding a key just before, and just after the midpoint.
(Could there be more that we should add related to not finding a key? What is the control flowgraph?)
Example (continued):
Nov 2002 91.3913 R McFadyen 25
Example (continued)
55
44
55 66 77 88 99
22 33 44 50 56 61 76
22 33 44 50 56 61 76
22 33 44 50 56 61 76
1
2
6
4
5
7
55 true, 1
55 false, ?
55 true, 1
50 true, 4
76 true, 7
44 true, 3
Outputs Found, L
Inputs List Key
22 33 44 50 56 61 768 56 true, 522 55 88 66 773 99 false, ?