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SE 4367 Functional Testing

SE 4367 Functional Testing

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SE 4367 Functional Testing

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  1. SE 4367Functional Testing

  2. Learning Objectives • What is functional testing? • How to perform functional testing? • How to generate test inputs? • What are equivalence partitioning, boundary value testing, domain testing, state testing, and decision table testing?

  3. What is Functional Testing? • When test inputs are generated using functional specifications, we say that we are doing functional testing. • Functional testing tests how well a program meets the functionality requirements.

  4. The Methodology • The derivation of test inputs is based on functional specifications. • Clues are obtained from the specifications. • Clues lead to test requirements. • Test requirements lead to test specifications. • Test specifications are then used to actually execute the program under test.

  5. Clues Expected behavior Test requirements Oracle or Test specifications Program has failed; make a note and proceed with testing or get into the debug mode. Test driver Until specs. Exhausted. Actual behavior Program Test Methodology Specifications Program output is correct

  6. Equivalence Partitioning • The input domain is usually too large for exhaustive testing. • It is therefore partitioned into a finite number of sub-domains for the selection of test inputs. • Each sub-domain is known as an equivalence class and serves as a source of at least one test input.

  7. Input domain partitioned into four sub-domains. Input domain 2 1 3 4 Equivalence Partitioning Four test inputs, one selected from each sub-domain. Too many test inputs.

  8. How to partition? • Inputs to a program provide clues to partitioning. • Example 1: • Suppose that program P takes an integer X as input X. • For X<0 the program is required to perform task T1 and for X>=0 task T2.

  9. How to partition? (contd.) • The input domain is prohibitively large because X can assume a large number of values. • However, we expect P to behave the same way for all X<0. • Similarly, we expect P to perform the same way for all values of X>=0. • We therefore partition the input domain of P into two sub-domains.

  10. One test case: X=-3 Equivalence class X<0 X>=0 Equivalence class Another test case: X=15 Two sub-domains All test inputs in the X<0 sub-domain are considered equivalent. The assumption is that if one test input in this sub-domain reveals an error in the program, so will the others. This is true of the test inputs in the X>=0 sub-domain also.

  11. Non-overlapping Partitions • In the previous example, the two equivalence classes are non-overlapping. In other words the two sub-domains are disjoint. • When the sub-domains are disjoint, it is sufficient to pick one test input from each equivalence class to test the program.

  12. Non-overlapping Partitions • An equivalence class is considered coveredwhen at least one test has been selected from it. • In partition testing our goal is to cover all equivalence classes.

  13. Overlapping Partitions • Example 2: • Suppose that program P takes three integers X, Y and Z. It is known that: • X<Y • Z>Y

  14. X<Y, Z<=Y X=2, Y=3, Z=1 X>=Y, Z<=Y X=15, Y=4, Z=1 X<Y X<Y, Z>Y X=3, Y=4, Z=7 X>=Y Z>Y Z<=Y X>=Y, Z>Y X=15, Y=4, Z=7 Overlapping partitions

  15. Overlapping Partition-Test Selection • In this example, we could select 4 test cases as: • X=4, Y=7, Z=1 satisfies X<Y • X=4, Y=2, Z=1 satisfies X>=Y • X=1, Y=7, Z=9 satisfies Z>Y • X=1, Y=7, Z=2 satisfies Z<=Y • Thus, we have one test case from each equivalence class.

  16. Overlapping Partitions-Test Selection • However, we may also select only 2 test inputs and satisfy all four equivalence classes: • X=4, Y=7, Z=1 satisfies X<Y and Z<=Y • X=4, Y=2, Z=3 satisfies X>=Y and Z>Y • Thus, we have reduced the number of test cases from 4 to 2 while covering each equivalence class.

  17. Partitioning using non-numeric data • In the previous two examples the inputs were integers. One can derive equivalence classes for other types of data also. • Example 3: • Suppose that program P takes one character X and one string Y as inputs. P performs task T1 for all lower case characters and T2 for upper case characters. Also, it performs task T3 for the null string and T4 for all other strings.

  18. X: lc, Y: not null X: UC, Y: not null X: lc X:UC X: lc, Y: null Y: not null Y: null X: UC, Y: null Partitioning using non-numeric data lc: Lower case character UC: Upper case character null: null string.

  19. Non-numeric Data • Once again we have overlapping partitions. • We can select only 2 test inputs to cover all four equivalence classes. These are: • X: lower case, Y: null string • X: upper case, Y: not a null string

  20. Guidelines for equivalence partitioning • Input condition specifies a range: create one for the valid case and two for the invalid cases. • e.g. for a<=X<=b the classes are • a<=X<=b (valid case) • X<a and X>b (the invalid cases)

  21. Guidelines (contd.) • Input condition specifies a member of a set: create one for the valid value and one for the invalid (not in the set) value. • Input condition specifies a value: create one for the valid value and two for incorrect values (below and above the valid value). This may not be possible for certain data types, e.g. for boolean.

  22. Sufficiency of Partitions • In the previous examples we derived equivalence classes based on the conditions satisfied by the input data. • Then we selected just enough tests to cover each partition. • Think of the advantages and disadvantages of this approach!

  23. Boundary Value Analysis (BVA) • Another way to generate test cases is to look for boundary values. • Suppose a program takes an integer X as input. • In the absence of any information, we assume that X=0 is a boundary. Inputs to the program might lie on the boundary or on either side of the boundary.

  24. BVA (contd.) • This leads to 3 test inputs: • X=0, X=-20, and X=14. Note that the values -20 and 14 are on either side of the boundary and are chosen arbitrarily. • Notice that using BVA we get 3 equivalence classes. One of these three classes contains only one value (X=0), the other two are large!

  25. 14 5 2 1 y2 10 11 6 9 8 13 12 y1 3 4 7 x1 x2 BVA (contd.) • Now suppose that a program takes two integers X and Y and that x1<=X<=x2 and y1<=Y<=y2.

  26. BVA (contd.) • In this case the four sides of the rectangle represent the boundary. • The heuristic for test selection in this case is: • Select one test at each corner (1, 2, 3, 4). • Select one test just outside of each of the four sides of the boundary (5, 6, 7, 8)

  27. BVA (contd.) • Select one test just inside of each of the four sides of the boundary (10, 11, 12, 13). • Select one test case inside of the bounded region (9). • Select one test case outside of the bounded region (14). • How many equivalence classes do we get?

  28. BVA (contd.) • In the previous examples we considered only numeric data. • BVA can be done on any type of data. • For example, suppose that a program takes a string S and an integer X as inputs. The constraints on inputs are: • length(S)<=100 and a<=X<=b • Can you derive the test cases using BVA?

  29. BVA Applied to Output Variables • Just as we applied BVA to input data, we can apply it to output data. • Doing so gives us equivalence classes for the output domain. • We then try to find test inputs that will cover each output equivalence class.

  30. Finite State Machines (FSMs) • A state machine is an abstract representation of actions taken by a program or anything else that functions! • It is specified as a quintuple: • A: a finite input alphabet • Q: a finite set of states • q0: initial state which is a member of Q.

  31. FSMs (contd.) • T: state transitions which is a mapping Q x A--> Q • F: A finite set of final states, F is a subset of Q. • Example: Here is a finite state machine that recognizes integers ending with a carriage return character. • A={0,1,2,3,4,5,6,7,8,9, CR} • Q={q0,q1,q2} • q0: initial state

  32. FSMs (contd.) • T: {((q0,d),q1),(q1,d),q1), (q1,CR),q2)} • F: {q2} • A state diagram is an easier to understand specification of a state machine. For the above machine, the state diagram appears on the next slide.

  33. State transitions indicated by labeled arrows from one state the another (which could be the same). Each label must be from the alphabet. It is also known as an event. d CR d Final state indicated by concentric circles. q2 q0 q1 States indicated by circles. State diagram d: denotes a digit

  34. q2 q0 q1 State Diagram-Actions x/y: x is an element of the alphabet and y is an action. d/i=i+d*10j ; j=j+1 d/ i= d j = 1 CR/output i i is initialized to d and j to 1 when the machine moves from state q0 to q1. i is incremented by 10j*d and j by 1 when the machine moves from q1 to q1. The current value of i is output when a CR is encountered. Can you describe what this machine computes?

  35. State Machine: Languages • Each state machine recognizes a language. • The language recognized by a state machine is the set S of all strings such that: • when any string s in S is input to the state machine the machine goes through a sequence of transitions and ends up in the final state after having scanned all elements of s.

  36. q2 q0 q1 State diagram-Errors d/i=i+d*10j ; j=j+1 d/ i= d j = 1 CR/output I CR/output error reset q4 q4 has been added to the set of states. It represents an error state. Notice that reset is a new member added to the alphabet.

  37. State Diagram-Code • A state diagram can be transformed into a program using case analysis. Let us look at a C program fragment that embodies logic represented by the previous state diagram. • There is one function for each action. • digit is assumed to be provided by the lexical analyzer.

  38. Program for “integer” state machine /* state is global, with values q0, q1, q2. i is also global.*/ void event_digit() case q0: i=digit; j = 1 ; /* perform action. */ state=q1; /* set next state. */ break; /* event digit is done. */ case q1: i=i+10^j*digit; j++; /* Add the next digit. */ state=q1; break; /*…complete the program. */ { switch (state)

  39. Checking State Diagrams • Unreachable state: One that cannot be reached from q0 using any sequence of transitions. • Dead state: One that cannot be left once it is reached.

  40. Test Requirements • Every state must be reached at least once, Obtain 100% state coverage. • Every transition must be exercised at least once. Obtain 100% transition coverage.

  41. Example: Test Requirements • For the “integer” state machine: • state machine transitions: • event digit in state q0 • event CR in state q0 • event digit in state q1 • event CR in state q1 • event reset in state q4

  42. More testing of state machines? • Yes, it is possible! • When we learn about path coverage we will discuss how more test requirements can be derived from a state diagram.

  43. Test Specifications • As before, test specifications are derived from test requirements. • In the absence of dead states, all states and transitions might be reachable by one test. • It is advisable not to test the entire machine with one test case.

  44. Decision Tables • Requirements of certain programs are specified by decision tables. • A decision table is useful when specifying complex decision logic

  45. Decision Tables • A decision table has two parts: • condition part • action part • The two together specify under what condition will an action be performed.

  46. Decision Table-Nomenclature • C: denotes a condition • A: denotes an action • Y: denotes true • N:denotes false • X: denotes action to be taken. • Blank in condition: denotes “don’t care” • Blank in action: denotes “do not take the action”

  47. Bank Example • Consider a bank software responsible for debiting from an account. The relevant conditions and actions are: • C1: The account number is correct • C2: The signature matches • C3: There is enough money in the account • A1: Give money • A2: Give statement indicating insufficient funds • A3: Call vigilance to check for fraud!

  48. Decision tables

  49. Example (contd.) • A2 is to be performed when C1 and C2 are true and C3 is false. • A1 is to be performed when C1, C2, and C3 are true. • A3 is to be performed when C1 is true and C2 is false.

  50. Default Rules • Are all possible combinations of conditions covered? • No! Which ones are not covered? • We need a default action for the uncovered combinations. A default action could be an error report or a reset.