Branch coverage and Statement Coverage are form of white box testing techniques.

The main difference between them is, the aim of statement coverage is to traverse all statements at least once, whereas the goal of branch coverage it to traverse all the branches at least once.

Difference between Statement Coverage and Branch Coverage in Tabular form

Statement Coverage

It is also termed as Line coverage.
The goal of this technique is to cover all the statements at least once by executing the program.

It requires test cases that make possible to run all the statement consisting of the program in order to achieve 100% coverage. It only assures that all the statements have executed but not assures whether all the paths have covered or not.

For example:

READ A
IF A == 10
THEN
PRINT I am True
ElSE
PRINT I am False
ENDIF

Test case #1 ( A = 5 )

Statement coverage = (Total Statements covered/Total Statements )* 100

=(5/7)*100

In the above code, 71.5% statement coverage is achieved by test case #1.

Test case #2 ( A = 10 )

Statement coverage = (Total Statements covered/Total Statements )* 100

=(5/7)*100

Again, 71.5% statement coverage is covered, and altogether these two test cases executed all the possible paths with 71.5% statement coverage each.

Executing every statement of a program helps in finding faults, unreachable or useless statements.

Branch Coverage

In a flow graph, an arrow represents edges and counting a statement as a node, where two nodes are connected with an edge.

A diamond symbol represents the decision node where more than one edges are present in an outgoing manner which is termed as Branches.

The main aim of branch coverage is to cover all the branches ( two separate paths) at least once (true and false).
Branch Coverage is a structurally based technique that checks both conditional and unconditional branches.

Branch  coverage = (number of executed branches/ total number of branches) * 100

For example:

READ A 
IF A == 10
THEN 
PRINT I am True 
ElSE 
PRINT I am False 
ENDIF

Branches:

• 2,5,6,7
• 2,3,4,7

To cover all the branches we would require 2 test cases:

Test case #1 ( A = 12 )

Branch coverage = (Total branch covered/Total Branches )* 100

=(1/2)*100

In the above code, 50% branch coverage is achieved by test case #1.

Test case #2 ( A = 10 )

Branch coverage = (Total branch covered/Total Branches )* 100

=(1/2)*100

Again, 50% branch coverage is achieved by test case #2.

Altogether these two test cases executed all the possible branches with 50% branch coverage each.

We will be using Robustness Testing to generate test cases.  The expected output can be  [Invalid input, Invalid date, previous Date].

Program

Here is the “previous date problem” program along with its tested test cases.

Robustness Testing

So, there will be 6N+ 1 test cases that is 6*3 +1 =19 in this case as we are using Robustness Testing.

We will be applying Boundary Value Analysis to generate test cases.  The expected output can be  [Invalid date, previous Date].

Program:

#include<conio.h>
#include<stdio.h>
void main()
{
	int month,date,year, valid=-1,leap_year=-1;
	printf("please enter a date = ");
	scanf("%d",&date);
		printf("please enter a month = ");
	scanf("%d",&month);
		printf("please enter a year = ");
	scanf("%d",&year);
	if((date>0 && date<=31)&&(month>=1 && month<=12)&&(year>=2000 && year<=2050))
	{
	// finding given input is a leap year or not
	if((year%4)==0)
	{
		leap_year=1;
    if((year%100)==0)
    {
	if((year%400)==0)
	{
		leap_year=1;	
	}
	else
	{
	leap_year=-1;	
	}
}
}
if(month==2 && leap_year==1 && date >29)
valid=-1;
else if(month==2 && leap_year==-1 && date >28)
valid=-1;
else 
valid=1;
}

if((month==6 || month==4||month==9||month==11) && date>30)
valid=-1;

// validating & finding output


if(valid==1)
{
	printf("Entered date = %d-%d-%d",date,month,year);

if(date==1)
{
if(month==1)
{
date=31;
month=12;
year--;
}
else if(leap_year=1 && month==3)
{
	date=29;
	month--;
}
else if(leap_year==-1 && month==3)
{
	date=28;
	month--;
}
else if(month==2||month==4||month==6||month==8||month==9||month==11){
	date=31;
	month--;
}else{
	date=30;
	month--;
}
}
else{
	
	date--;
}
printf("\nPrevious date = %d-%d-%d \n",date,month,year);
}
else
printf("\n Not a valid date");

getch();
}

So, there will be 4N+ 1 test cases that is 4*3 +1 =13 in this case as we are using boundary value analysis.

Testing the program:

Invalid Date:

Previous Date: