PESIT Bangalore South Campus SOLUTION

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1 USN 1 P E PESIT Bangalore South Campus Hosur road, 1km before Electronic City, Bengaluru -100 Department of Information Science & Engineering INTERNAL ASSESSMENT TEST 2 Date : 02/04/2018 Max Marks: 40 Subject & Code: Software Testing / 15IS63 Section: 6 th A & B Name of faculty: Prof. Animesh Giri Time: 11:30 AM 1:00 PM SOLUTION Note: ANSWER FIVE QUESTIONS, CHOOSING ONE QUESTION FROM EACH PART Problem definitions have been discussed during Theory Classes & Lab Sessions. PART - I 1. a) Explain the steps involved to determine Basis Paths by applying McCabe s Algorithm, then do the following i) Find the Cyclomatic Complexity for the program graph given below. ii) Identify a set of Basis Paths for the program graph (5 Marks) Program Graph Answer: Mc Cabes s Algorithm: The method begins with the selection of a baseline path, which should correspond to a normal execution of a program (from start node to end node, and has as many decisions as possible).

2 The algorithm proceeds by retracing the paths visited and flipping the conditions one at a time. The process repeats up to the point all flips have been considered. i) Cyclomatic Complexity: V(G)=e-n+2p From the given program graph e=9,n=7 & p=1 Therefore,V(G)=9-7+2(1) ii) Basis Paths P1: P2: P3: P4:1-7 Answer: =2+2 = 4 b) Illustrate with a neat program graphs the Violations of Structured Programming Constructs such as i) Branching into a loop ii) Branching out of a decision (3 Marks)

3 Answer: 2. Explain Condensation with respect to the Structured Programming Constructs until the final try and illustrate Essential Complexity of Structured Program is 1 for the DD-Path Graph drawn by considering the Pseudo-Code for triangle program. (8 Marks) The basic idea is to look for a structured programming construct and to collapse it into a single node, and repeat it until no more programming construct can be found. This process is followed in the graphs above of a triangle program to final arrive at the graph with "e" as the final node.

4 The if-then-else construct involving nodes A,B,C,D is condensed into node a, and then the three if-then constructs are condensed onto nodes b, c and d. The remaining if-then-else construct is condensed into node e, resulting in a condensed graph with cyclomatic complexity V(G)=1. 3. PART II Define the following test coverage metrics: a. Statement (C 0 ) Coverage b. Decision (DD Path C 1 ) Coverage c. Condition Testing (C 1p ) Coverage d. Multiple Compound Condition (C MCC ) Coverage Apply these coverage techniques on the program graph shown and mention the path traversed for each test case input identified. Program Graph i) Statement Coverage (C0) Let T be a test suite for a Program P, then T satisfies the statement criterion for P, iff, for each statement S of P, there exists at least one test case in T that causes the execution of S. In terms of Flow graph model of program P, it is the same as visiting each single node on some execution path exercised by a test case in T C0: path: ace T1: A=2, B=0, X=3 ii) Decision (DD Path C 1 ) Coverage

5 Test suite T for a program P satisfies the decision adequacy criterion for P, iff, for each branch B (or predicate P), there exists at least test case in T that causes execution of B. This is the same as stating that every edge in the flow graph model of program P belongs to some execution path exercised by a test case in T C1: path1: acd and path2: abe iii) Condition Testing (C 1p ) Coverage T1: A=3, B=0, X=1 T2: A=2, B=1, X=1 A test suite T for a program P covers all C1P iff for each atomic condition in P, it has at least two test cases in T : one forcing P to have true out come and the other one forcing P to have a false outcome Four conditions: A>1, B=0, and A=2, X>1 1) A>1, and A<=1 2) B=0, and B<> 0 3) A=2, and A<> 2 4) X>1, and X<=1 Test cases: T1: A=2, B=0, X=4 ( True Path) T2: A=1, B=1, X=1 ( False Path) iv) Multiple Compound Condition (C MCC ) Coverage CMCC, A more complete extension that includes both the criteria basic condition and branch adequacy CMCC requires a test case T for each possible evaluation of compound conditions

6 1)A>1,B=0 2)A>1,B< >0 3)A<=1,B=0 4)A<=1,B< >0 5)A=2,X>1 6)A=2,X<=1 7)A< >2,X>1 8)A< >2,X<=1 Test Cases with input values: A=2, B=0,X=4 covers 1, 5 A=2, B=1, X=1 Covers 2,6 A=1, B=0,X=2 Covers 3,7 A=1,B=1,X=1 Covers 4,8 4. Illustrate the significance of DU- Paths and DC- Paths with the help of code fragments & the program graph of Commission problem. (8 Marks) i) Considering three variables, show few selected DU- Paths and ii) Mention whether those DU- Paths are Definition-Clear or not Answer: 1 int main() 2 { 3 int locks, stocks, barrels, tlocks, tstocks, tbarrels; 4 float lprice,sprice,bprice,sales; 5 lprice=45.0; 6 sprice=30.0; 7 bprice=25.0; 8 tlocks=0;

7 9 tstocks=0; 10 tbarrels=0; 11 printf("\nenter the number of locks and to exit the loop enter -1 for locks\n");scanf("%d", &locks); 12 while(locks!=-1) { 13printf("enter the number of stocks and barrels\n");scanf("%d%d",&stocks,&barrels); 14 tlocks=tlocks+locks; 15 tstocks=tstocks+stocks; 16 tbarrels=btarrels+barrels; 17 printf("\nenter the number of locks and to exit the loop enter -1 for locks\n");scanf("%d",&locks); } 18 printf("\ntotal locks = %d\,tlocks); 19 printf( total stocks =%d\n,tstocks); 20 printf( total barrels =%d\n",tbarrels); 21sales= lprice*tlocks + sprice*tstocks + bprice*tbarrels; 22 printf("\nthe total sales=%f\n",sales); }

8 Variable name Description Variable path(begining,end nodes) Du-paths Definiti on clear? lprice DEF(lprice,5) and USE(lprice 21) (5,21) < > sprice DEF(sprice,6) and USE(sprice,21) (6,21) < > Yes Yes bprice DEF(bprice,7) and USE(bprice,21) (7,21) < > Yes tlocks DEF((tlocks,8) and DEF(tlocks,14)) and USE(tlocks,14),(tlocks,18) and USE(tlocks,21) (8,14) < > Yes (8,18) < > (8,21) < > No No (14,14) <14-14> Yes (14,18) < > No (14,21) < > No tstocks DEF((tstocks,9) and DEF(tstocks,15)) and USE(tstocks,15),USE(tstocks,19), USE(tstocks,21), (9,15) < > Yes (9,19) < > No (9,21) < > No locks DEF((locks,11) and DEF(locks,17)) and USE(locks,12),USE(locks,17) (15,15) <15-15> Yes (15,19) < > No (15,21) < > No (11,12) <11-12> Yes (11,14) < > Yes (17,12) < > Yes

9 stocks DEF(stocks,13) and USE(stocks,15) (17,17) <17-17> Yes (13,15) < > Yes sales DEF(sales,21) and USE(sales,22) (21,22) <21-22> Yes Answer: PART - III 5. Explain with purpose of Scaffolding in test execution by defining the role of Test Drivers, Test Stubs & Test Harness. (8 Marks) Code produced to support development activities (especially testing) Not part of the product as seen by the end user May be temporary (like scaffolding in construction of buildings Includes Test harnesses, drivers, and stubs In complex systems science, "scaffolding" are those structures necessary to move from an initial state to an emerged form. For example, when recapturing an area of land and bringing it back to its natural state (e.g. prairie), you will find that certain birds and plants have to be present during the transition time in order for the old form to re-emerge. Once the new state has emerged, the scaffolding comes down... BUT, without the scaffolding, no change would have occurred. Test driver A main program for running a test May be produced before a real main program Provides more control than the real main program To driver program under test through test cases Test stubs Substitute for called functions/methods/objects Test harness Substitutes for other parts of the deployed environment Ex: Software simulation of hardware devices Example: Interactive Systems developed by MVC Pattern Controllability & Observability

10 6. What are Test Oracles? Explain with neat diagrams, Comparison-Based Oracle and Self- Checking Code Oracle (8 Marks) Answer:

11 PART - IV

12 7. Why Loop Testing is important? Mention the strategy to the Nested Loops, Concatenated Loops and the Knotted Loops (Unstructured Loop). (8 Marks) Answer: Loop testing has been studied extensively, and with good reason loops are a highly fault-prone portion of source code. To start an amusing taxonomy of loops occurs: concatenated, nested, and horrible, shown in Figure- 3.6 Concatenated loops are simply a sequence of disjoint loops. Nested loops are such that one loop is contained inside another. Knotted loops cannot occur when the structured programming percepts are followed, but they can occur in languages like Java with try/catch. When it is possible to branch into (or out from) the middle of a loop, and these branches are internal to other loops, the result is Beizer s knotted loop. The simple view of loop testing is that every loop involves a decision, and we need to test both outcomes of the decision: one is to traverse the loop, and the other is to exit (or not enter) the loop. We can also take a modified boundary value approach, where the loop index is given its minimum, nominal, and maximum values. We can push this further to full boundary value testing and even robustness testing. Once a loop has been tested, the rest condenses it into a single node.

13 If loops are nested, this process is repeated starting with the innermost loop and working around outward. These results in the same multiplicity of test cases we found with boundary value analysis, which makes sense, because each loop index variable acts like an input variable. If loops are knotted, it will be necessary to carefully analyze them in terms of the dataflow methods. Answer: 8. Explain Mutation Analysis & Testing; illustrate with the help of code-snippet some of the mutation operator to produce a mutant program. (8 Marks) Mutation testing is done by selecting a set of mutation operators and then applying them to the source program one at a time for each applicable piece of the source code. The result of applying one mutation operator to the program is called a mutant. If the test suite is able to detect the change (i.e. one of the tests fails), then the mutant is said to be killed. For example, consider the following C++ code fragment: if (a && b) { c = 1; } else { c = 0; } The condition mutation operator would replace&&with and produce the following mutant: if (a b) { c = 1; } else { c = 0; } Now, for the test to kill this mutant, the following condition should be met: Test input data should cause different program states for the mutant and the original program. For example, a test with a = 1and b = 0would do this. The value of 'c' should be propagated to the program's output and checked by the test. Mutation testing A mutant is a copy of a program with a mutation A mutation is a syntactic change (a seeded bug) Example: change (i < 0) to (i <= 0) Run test suite on all the mutant programs A mutant is killed if it fails on at least one test case If many mutants are killed, infer that the test suite is also effective at finding real bugs

14 Mutation Operators Syntactic change from legal program to legal program So: Specific to each programming language. C++ mutations don t work for Java, Java mutations don t work for Python Examples: crp: constant for constant replacement for instance: from (x < 5) to (x < 12) select from constants found somewhere in program text ror: relational operator replacement for instance: from (x <= 5) to (x < 5) vie: variable initialization elimination change int x =5; to int x; A variety of mutation operators were explored by researchers. Here are some examples of mutation operators for imperative languages: Statement deletion. Replace each Boolean sub expression with true and false. Replace each arithmetic operation with another one, e.g. + with *, - and /. Replace each Boolean relation with another one, e.g.>with>=,==and<=. Replace each variable with another variable declared in the same scope (variable types should be the same). PART - V 9. Using the Mc-Cabe s Strongly Connected Graph, Write all the Path traversal. (8 Marks) Answer: It is a directed graph that we might take to be the program graph (or the DD-Path graph) of some program. McCabe based his view of testing on a major result from graph theory, which states that the cyclomatic number of a strongly connected graph is the number of linearly independent circuits in the graph. (We can always create a strongly connected graph by adding an edge from the (every) sink node to the (every) source node.) Some sources give the formula for cyclomatic complexity as V(G) = e n + p, while others use the formula V(G) = e n + 2p, where e is the number of edges. n is the number of nodes, and p is the number of connected regions.

15 The cyclomatic complexity of the strongly connected graph if Figure is %; thus, there are five linearly independent circuits. If we now delete the added edge from node G to node A, these five circuits become five linearly independent paths from node A to node G. p1: A, B, C, G p2: A, B, C, B, C, GG p3: A, B, E, F, G p4: A, D, E, F, G p5: A, D, F, G The algorithm is straightforward: The method begins with the selection of a baseline path, which should correspond to some normal case program execution. This can be somewhat arbitrary; McCabe advises choosing a path with as many decisions nodes as possible. Next the baseline path is retraced, and in turn each decision is flipped ; that is, when a node of outdegree 2 is reached, a different edge must be taken. The process repeats up to the point all flips have been considered.

16 Here we follow McCabe s example, in which he first postulates the path through nodes A, B, C, B, E, F, G as the baseline. The first decision node (outdegree 2) in this path is node A; so for the next basis path we traverse edge 2 instead of edge 1. We get the path A, D, E, F, G, where we retrace nodes E, F, G in path 1 to be as minimally different as possible. For the next path, we can follow the second path, and take the other decision outcome of node D, which gives us the path A, D, F, G. Now only decision nodes B and C have not been flipped; doing so yields the last two basis paths, A, B, E, F, G and A, B, C, G. 10. What is Fault based Testing? Distinguish between: i) Competent Programmer hypothesis and Coupling effect hypothesis ii) Distinguished mutant and Equivalent mutant (8 Marks) Answer: If the program under test has an actual fault, we hypothesize that it differs from the another, corrected program by only a small textual change. If so, then we merely distinguish the program from all such small variants to ensure detection of all such faults. This is known as competent programmer hypothesis, an assumption that the program under test is "close to" a correct program. Sometimes an error of logic will result in much more complex difference in program text. This may not invalidate fault-based testing with a simpler fault model, provided test cases sufficient for detecting the simpler faults are sufficient also detecting the more complex faults. This is known as coupling effect. Coupling effect hypothesis: A complex change is equivalent to several smaller changes in program text. If the effect of one of these small changes is not masked by the effect of others, then a test case that differentiates a variant based on a single change may also serve to detect the more complex error. The mutant can be distinguished from the original program,but the test suite under consideration is not adequate with respect to the mutant. If the mutant could have been killed, but was not, it indicates a weakness in the test suite. A mutant may be equivalent to the original program. Maybe changing (x < 0) to (x <= 0) didn t change the output at all! The seeded fault is not really a fault. Determining whether a mutant is equivalent maybe easy or hard; in the worst case it is undecideable. For eg, int index = 0; while ( ) { ; index++; if (index == 10)

17 { break; } } Boolean relation mutation operator will replace == with >= and produce the following mutant: int index = 0; while ( ) { ; index++; if (index >= 10) { break; } }