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White-Box Testing Techniques II

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White-Box Testing Techniques II

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White-Box Testing Techniques II

Dataflow Testing

Originals prepared by

Stephen M. Thebaut, Ph.D.

University of Florida

- Logic coverage
- Dataflow coverage
- Path conditions and symbolic execution (lecture III)
- Other white-box testing strategies
- e.g. fault-based testing

- Basic idea:
- Program paths along which variables are defined and then used should be covered

- A family of path selection criteria has been defined, each providing a different degree of coverage
- CASE tool support is very desirable

- A program variable is DEFINED when it appears:
- on the left hand side of an assignment statement egy = 17
- in an input statementeg read(y)
- as an call-by-reference parameter in a subroutine call egupdate(x, &y);

- A program variable is USED when it appears:
- on the right hand side of an assignment statement eg y = x+17
- as an call-by-value parameter in a subroutine or function call eg y = sqrt(x)
- in the predicate of a branch statement eg if ( x > 0 ) { … }

- Use in the predicate of a branch statement is a predicate-use or “p-use”
- Any other use is a computation-use or “c-use”
- For example, in the program fragment:
if ( x > 0 ) {

print(y); }

there is a p-use ofxand a c-use ofy

- A variable can also be used and then re-defined in a single statement when it appears:
- on both sides of an assignment statement eg y = y + x
- as an call-by-reference parameter in a subroutine call eg increment( &y )

- A path is definition clear(“def-clear”) with respect to a variable v if it has no variable re-definition of v on the path
- A complete pathis a path whose initial node is a start node and whose final node is an exit node

- A definition-use pair(“du-pair”) with respect to a variable v is a double (d,u) such that
- d is a node in the program’s flow graph at which v is defined,
- u is a node or edge at which v is used and
- there is a def-clear path with respect to vfrom d to u

- Note that the definition of a du-pair does not require the existence of a feasibledef-clear path from d to u

1. input(A,B)

if (B>1) {

2. A = A+7

}

3. if (A>10) {

4. B = A+B

}

5. output(A,B)

input(A,B)

1

B>1

B1

2

A = A+7

3

A>10

A10

4

B = A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

- All-Defs for every program variable v, at least onedef-clear path from every definition of vto at least one c-use or one p-use of v must be covered

- Consider a test case executing path:
1. <1,2,3,4,5>

- Identify all def-clear paths covered (ie subsumed) by this path for each variable
- Are all definitions for each variable associated with at least one of the subsumed def-clear paths?

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

- Since<1,2,3,4,5>covers at least one def-clear path from every definition of A or B to at least one c-use or p-use of A or B, All-Defs coverage is achieved

- All-Uses: for every program variable v,at least onedef-clear pathfrom everydefinition of v to everyc-use and every p-use of v must be covered
- Consider additional test cases executing paths:
2. <1,3,4,5>

3. <1,2,3,5>

- Do all three test cases provide All-Uses coverage?

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

input(A,B)

1

B>1

B1

2

A := A+7

3

A>10

A10

4

B := A+B

5

output(A,B)

- None of the three test cases covers the du-pair (1,<3,5>) for variable A,
- All-Uses Coverage is not achieved

1. input(X,Y)

2. while (Y>0) {

3. if (X>0)

4. Y := Y-X

else

5. input(X)

6. }

7. output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y = Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

1

input(X,Y))

2

Y>0

Y0

3

X0

X>0

input(X)

5

4

Y := Y-X

6

Y0

Y>0

7

output(X,Y)

- A path (either partial or complete) is simpleif all edges within the path are distinct ie different
- A path is loop-freeif all nodes within the path are distinct ie different

A path <n1,n2,...,nj,nk> is a du-pathwith respect to a variable v if v is defined at node n1 and either:

- there is a c-use of v at node nk and <n1,n2,...,nj,nk> is a def-clear simplepath, or
- there is a p-use of v at edge <nj,nk> and <n1,n2,...nj> is a def-clear loop-freepath.

A path <n1,n2,...,nj,nk> is a du-pathwith respect to a variable v if v is defined at node n1 and either:

- there is a c-use of v at node nk and <n1,n2,...,nj,nk> is a def-clear simplepath, or
- there is a p-use of v at edge <nj,nk> and <n1,n2,...nj> is a def-clear loop-freepath.

NOTE!

- All-DU-Paths:for every program variable v,every du-path from every definition of v to every c-use and every p-use of v must be covered

- Identify all c-uses and p-uses for variable Y in Example 2
- For each c-use or p-use, identify (using the “ * ” notation) all def-clear paths
- Identify whether or not each def-clear path is feasible, and whether or not it is a du-path

Path

Compound Condition

All du-paths

Branch / Condition

Basis Paths

All-Uses

Loop

Condition

Branch

All-Defs

Statement