Path Testing

Presentation onPath Testing

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AGENDA

Path Testing

Path testing is a White box testing technique which ensures that every path in a program has

been executed at least once.

• It requires complete knowledge of program structure.

• It is most often used by programmers to test their own code.

Why do we need Path Testing?

Goal of the path testing is to pick enough paths to ensure that every source statement is

executed at least once and to eliminate the redundant code.


• Converting the Programming Language to directed graph• Converting Directed graph to DD-Path (Decision to Decision)• Determine the Cyclomatic Complexity• Determine the Basis set of Independent paths


Path Testing Steps


• 1 setDays = 0

• 2 availDays = 4

• 3 input = InputBox ("enter the selection 1: Paid Leave 2: LOP")

• 4 if (input=1) then

• 5 setDays = InputBox ("Enter the required number of paid leave")

• 6 if (setDays<=4 and setDays>0) Then

• 7 availDays = availDays - setDays

• 8 MsgBox("Paid Leave approved." &chr(13) &"Remaining Paid Leaves : "&availDays)

• 9 elseIf ((setDays > availDays) AND (availDays<>0)) Then

• 10 MsgBox ("Enter Valid days for leave")

• 11 elseif (setDays<=0) then

• 12 MsgBox("Enter Valid days for leave")

• 13 End If

• 14 elseif (input=2) then

• 15 setDays = InputBox ("Enter the required number of LOP ")

• 16 if (setDays>0) then

• 17 MsgBox ("LOP approved.")

• 18 else MsgBox ("Enter valid days for leave")

• 19 end if

• 20 else msgbox ("no selection")

• 21 end if

HRM Application with Directed Graph


DD Path

Directed Graph DD Path

Case1: It consist of a single node with indeg=0Case2: It consist of a single node with outdeg=0Case3: It consist of a single node with indeg>=2 or Outdeg>=2Case4: It consist of a single node with indeg=1 and Outdeg=1Case5: It is maximal chain of length>=1


DD path case:

Cyclomatic complexity is a software metric (measurement), used to indicate the complexity of a program.

V(G)=E-N+2P

E-Number of EdgesN-Number of NodesP-Number of connected Region


Determine Cyclomatic Complexity:

No of Edges=20No of Nodes=16Connected Region=1

V(G)=E-N+2PV(G)=20-16+2(1) = 4+2

V(G) = 6

Hence the number of paths for the given program will be


Determine Cyclomatic Complexity:

i. P1: First, A,B,C,D,E,H,Last ii. P2: First, A,B,C,D,F,H,Last iii. P3: First, A,B,C,D,G,H,Last iv. P4: First,A,B,I,J,K,M,Lastv. P5: First,A,B,I,J,L,M,Lastvi. P6: First,A,B,N,Last


Paths covered in the program

THANK YOU


Software

Path Testing