This document outlines a series of C++ programs designed to evaluate various properties of triangles and date manipulation through boundary and robustness testing. It covers methods to determine triangle type based on given side lengths, calculate the next date considering leap years, and checks the nature of quadratic equations. Each program includes logic for testing both normal and edge cases, illustrating software testing techniques.

1. Name : Ejaz khan
Roll no: 1120725
Section CO-5
Doc file is in office 365 it may behave
differently in older versions of ms office
SOFTWARE TESTING LAB
Program 1.
1. cyclometric complexity of a program(graph).
//cyclomatic complexity
#include<iostream>
using namespace std;
int main()
{
int i,j,v,val,edg, arr[50][50];;
cout<<" number of vertices";
cin>>v;

2. for(i=0;i<v;i++)
{
for(j=0;j<v;j++)
{
cin>>arr[i][j];
}
}
for(i=0;i<v;i++)
{
for(j=0;j<v;j++)
{
if(arr[i][j]==1)
{
edg++;
}
}
}
cout<<"Complexity is:"<<edg-v+2;}

3. program 2.
2. type of triangle given sides of triangle then perform
boundary case analysis :
//type of triangles and boundary value analysis
#include<iostream>
#include<conio.h>
using namespace std;
void triangle (int a,int b,int c)
{
if((a<(b+c))&&(b<(a+c))&&(c<(a+b)))
{
if(a==b&&b==c)
{
cout<<"Equilateral Trianglen";
}
else if ((a==b)||(a==c)||(b==c))
{
cout<<"Isosceles Trianglen";
}
else
{
cout<<"Scalene Trianglen";

4. }
}
else
{
cout<<"Not a Trianglen";
}
}
int main()
{
int x,y,z, no_of_test=0,i,j;
cout<<"Enter the value of sides of triangle in the
range 1-100n";
cin>>x>>y>>z;
triangle (x,y,z);
no_of_test =4*3+1;
int min_range=1,max_range=100;
int arr[4]={min_range,min_range+1,max_range-
1,max_range};
int mid=(max_range-min_range)/2;
cout<<"No. of boundary cases = "<< no_of_test
<<endl;
for(i=0;i<3;i++)
{
for(j=0;j<4;j++)
{
if(i==0)
{

5. cout<<arr[j] <<" "<<(max_range-min_range)/2<<"
"<<(max_range-min_range)/2<<endl;
triangle (arr[j],mid,mid);}
else if(i==1)
{
cout<<(max_range-min_range)/2<<" "<<arr[j] <<"
"<<(max_range-min_range)/2<<endl;
triangle (mid,arr[j],mid);}
else
{
cout<<(max_range-min_range)/2<<" "<<
(max_range-min_range)/2<<" "<<arr[j]<<endl;
triangle (mid,mid,arr[j]);}
}
}
cout<<(max_range-min_range)/2<<" "<<(max_range-min_
range)/2<<" "<<(max_range-min_range)/2<<endl;
triangle (mid,mid,mid);
getch();
}

6. Program 3.
3. type of triangle given sides of triangle and operate
robustness testing :
//type of triangle using robustness testing analysis
#include<iostream>
#include<conio.h>

7. using namespace std;
void triangle (int a,int b,int c)
{
if((a<(b+c))&&(b<(a+c))&&(c<(a+b)))
{
if(a==b&&b==c)
{
cout<<"Equilateral Trianglen";
}
else if ((a==b)||(a==c)||(b==c))
{
cout<<"Isosceles Trianglen";
}
else
{
cout<<"Scalene Trianglen";
}
}
else
{
cout<<"Not a Trianglen";
}
}
int main()
{
int x,y,z, no_of_test =0,i,j;

8. cout<<"Enter the value of sides of triangle in the
range 1-100n";
cin>>x>>y>>z;
triangle (x,y,z);
no_of_test =6*3+1;
int min_range=1,max_range=100;
int arr[4]={min_range,min_range+1,max_range-
1,max_range};
int mid=(max_range-min_range)/2;
cout<<"No. of boundary cases = "<< no_of_test
<<endl;
for(i=0;i<3;i++)
{
for(j=0;j<6;j++)
{
if(i==0) {cout<<arr[j] <<" "<<(max_range-min_
range)/2<<" "<<(max_range-min_range)/2<<endl;
triangle (arr[j],mid,mid);}
else if(i==1) {cout<<(max_range-min_range)/2<<"
"<<arr[j] <<" "<<(max_range-min_range)/2<<endl;
triangle (mid,arr[j],mid);}
else { cout<<(max_range-min_range)/2<<"
"<< (max_range-min_range)/2<<" "<<arr[j]<<endl;
triangle (mid,mid,arr[j]);}

9. }
}
cout<<(max_range-min_range)/2<<" "<<(max_range-min_
range)/2<<" "<<(max_range-min_range)/2<<endl;
triangle (mid,mid,mid);
getch();
}

10. Program 4
4. find type of triangle and worst case testing :
//type of triangle and Worst case analysis
#include<iostream>
#include<conio.h>
#include<math.h>
using namespace std;
void triangle(int a,int b,int c)
{
if((a<(b+c))&&(b<(a+c))&&(c<(a+b)))
{
if(a==b&&b==c)
{
cout<<"Equilateral Trianglen";
}
else if ((a==b)||(a==c)||(b==c))
{
cout<<"Isosceles Trianglen";
}
else
{
cout<<"Scalene Trianglen";
}
}
else
{
cout<<"Not a Trianglen";

11. }
}
int main()
{
int x,y,z, no_of_test =0,i,j,k;
cout<<"Enter the value of sides of triangle in the
range 1-100n";
cin>>x>>y>>z;
triangle (x,y,z);
no_of_test =pow(5,3);
int min_range=1,max_range=100;
int arr[4]={min_range,min_range+1,max_range-
1,max_range};
int mid=(max_range-min_range)/2;
cout<<"No. of boundary cases = "<< no_of_test endl;
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
for(k=0;k<5;k++)
{
cout<<arr[i]<<” “<<arr[j]<<”
“<<arr[k]<<endl;
triangle (arr[i],arr[j],arr[k]);
}}}
5. find next date and perform boundary case analysis :

12. #include<iostream>
#include<conio.h>
using namespace std;
int leap_year_year(int y)
{
if(y%400==0||(y%4==0 && y%100!=0))
return 1;
else
return 0;
}
void next_date(int d,int m,int y)
{
int
mon1[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int
mon2[12]={31,28,31,30,31,30,31,31,30,31,30,31};
if(leap_year_year)
{if(d==31&&m==12)
{
cout<<"1-01-"<<y+1<<endl;
return;
}
else if(d==mon2[m-1])
{cout<<1<<"-"<<m+1<<"-"<<y<<endl;

13. return;
}
else
cout<<d+1<<"-"<<m<<"-"<<y<<endl;
return;
}
else
{
if(d==31&&m==12)
{
cout<<"1-01-"<<y+1<<endl;
return;
}
else if(d==mon1[m-1])
{cout<<1<<"-"<<m+1<<"-"<<y<<endl;
return;
}
else
cout<<d+1<<"-"<<m<<"-"<<y<<endl;
return;
}

14. }
void next(int d,int m,int y)
{
if(d<1)
{cout<<"enter a valid daten";
return;
}
else if((d==29&&m==2)&& (!leap_year_year (y)))
{
cout<<"enter a valid daten";
return;
}
else if(d>=29&&m==2)
{
cout<<"enter a valid daten";
return;
}
else
if((d==31&&m==4)||(d==31&&m==6)||(d==31&&m==9
)||(d==31&&m==11))
{
cout<<"enter a valid daten";
return;
}
else if(d>31)
{ cout<<"enter a valid daten";
return;

15. }
else if(m<1||m>12)
{ cout<<"enter a valid daten";
return;
}
next_date(d,m,y);
}
int main()
{
int d,m,y,dd,mm,yy,i,j,k;
int d1[4]={1,2,30,31};
int mid1=15,mid2=6,mid3=2500;
int m1[4]={1,2,11,12};
int y1[4]={1,2,4999,5000};
cout<<"enter a date:";
cin>>d>>m>>y;
next(d,m,y);
int total_cases=4*3+1;
cout<<"No. of boundary cases = "<<total_cases<<endl;
for(i=0;i<3;i++)

16. {
for(j=0;j<4;j++)
{
if(i==0) {cout<<d1[j] <<" "<<mid2<<"
"<<mid3<<" ";
next(d1[j],mid2,mid3);}
else if(i==1) {cout<<mid1<<" "<<m1[j]<<"
"<<mid3<<" ";
next(mid1,m1[j],mid3);}
else { cout<<mid1<<" "<< mid2<<"
"<<y1[j]<<" ";
next(mid1,mid2,y1[j]);}
}
}
cout<<mid1<<" "<<mid2<<" "<<mid3<<" ";
next(mid1,mid2,mid3);
getch();
}

17. 6. WAP to find next date and perform robustness
testing :
#include<iostream>
#include<conio.h>
using namespace std;
int leap_year_year (int y)
{
if(y%400==0||(y%4==0 && y%100!=0))
return 1;

18. else
return 0;
}
void next_date(int d,int m,int y)
{
int mon1[12]={31,28,31,30,31,30,31,31,30,31,30,3
1};
int
mon2[12]={31,28,31,30,31,30,31,31,30,31,30,31};
if(leap_year_year (y))
{if(d==31&&m==12)
{
cout<<"1-01-"<<y+1<<endl;
return;
}
else if(d==mon2[m-1])
{cout<<1<<"-"<<m+1<<"-"<<y<<endl;
return;
}
else
cout<<d+1<<"-"<<m<<"-"<<y<<endl;
return;
}

19. else
{
if(d==31&&m==12)
{
cout<<"1-01-"<<y+1<<endl;
return;
}
else if(d==mon1[m-1])
{cout<<1<<"-"<<m+1<<"-"<<y<<endl;
return;
}
else
cout<<d+1<<"-"<<m<<"-"<<y<<endl;
return;
}
}
void next(int d,int m,int y)
{
if(y<0||y>5000)
{
cout<<"enter a valid daten";
}

20. if(d<1)
{cout<<"enter a valid daten";
return;
}
else if((d==29&&m==2)&& (!leap_year_year (y)))
{
cout<<"enter a valid daten";
return;
}
else if(d>=29&&m==2)
{
cout<<"enter a valid daten";
return;
}
else
if((d==31&&m==4)||(d==31&&m==6)||(d==31&&m==9
)||(d==31&&m==11))
{
cout<<"enter a valid daten";
return;
}
else if(d>31)
{ cout<<"enter a valid daten";
return;
}
else if(m<1||m>12)
{ cout<<"enter a valid daten";

21. return;
}
next_date(d,m,y);
}
int main()
{
int d,m,y,dd,mm,yy,i,j,k;
int d1[6]={0,1,2,30,31,32};
int mid1=15,mid2=6,mid3=2500;
int m1[6]={0,1,2,11,12,13};
int y1[6]={0,1,2,4999,5000,5001};
cout<<"enter a date:";
cin>>d>>m>>y;
next(d,m,y);
int total_cases=6*3+1;
cout<<"No. of boundary cases = "<<total_cases<<endl;
for(i=0;i<3;i++)
{
for(j=0;j<6;j++)
{

22. if(i==0) {cout<<d1[j] <<" "<<mid2<<"
"<<mid3<<" ";
next(d1[j],mid2,mid3);}
else if(i==1) {cout<<mid1<<" "<<m1[j]<<"
"<<mid3<<" ";
next(mid1,m1[j],mid3);}
else { cout<<mid1<<" "<< mid2<<"
"<<y1[j]<<" ";
next(mid1,mid2,y1[j]);}
}
}
cout<<mid1<<" "<<mid2<<" "<<mid3<<" ";
next(mid1,mid2,mid3);
}

23. 7. WAP to find next date and perform worst case
analysis :
#include<iostream>
#include<conio.h>
using namespace std;
int leap_year(int y)
{
if(y%400==0||(y%4==0 && y%100!=0))
return 1;
else
return 0;
}
void find_next(int d,int m,int y)
{
int mon1[12]={31,28,31,30,31,30,31,31,30,31,30,3
1};
int
mon2[12]={31,28,31,30,31,30,31,31,30,31,30,31};

24. if(leap_year(y))
{if(d==31&&m==12)
{
cout<<"1-01-"<<y+1<<endl;
return;
}
else if(d==mon2[m-1])
{cout<<1<<"-"<<m+1<<"-"<<y<<endl;
return;
}
else
cout<<d+1<<"-"<<m<<"-"<<y<<endl;
return;
}
else
{
if(d==31&&m==12)
{
cout<<"1-01-"<<y+1<<endl;
return;
}
else if(d==mon1[m-1])
{cout<<1<<"-"<<m+1<<"-"<<y<<endl;

25. return;
}
else
cout<<d+1<<"-"<<m<<"-"<<y<<endl;
return;
}
}
void next(int d,int m,int y)
{
if(y<0&&y>5000)
{
cout<<"enter a valid date";
return;
}
if(d<1)
{cout<<"enter a valid daten";
return;
}
else if((d==29&&m==2)&& (!leap_year(y)))
{
cout<<"enter a valid daten";
return;
}

26. else if(d>=29&&m==2)
{
cout<<"enter a valid daten";
return;
}
else
if((d==31&&m==4)||(d==31&&m==6)||(d==31&&m==9
)||(d==31&&m==11))
{
cout<<"enter a valid daten";
return;
}
else if(d>31)
{ cout<<"enter a valid daten";
return;
}
else if(m<1||m>12)
{ cout<<"enter a valid daten";
return;
}
find_next(d,m,y);
}

27. int main()
{
int d,m,y,dd,mm,yy,i,j,k;
int d1[5]={1,2,15,30,31};
int mid1=15,mid2=6,mid3=2500;
int m1[5]={1,2,6,11,12};
int y1[5]={1,2,2500,4999,5000};
//cout<<"enter a date:";
//cin>>d>>m>>y;
//next(d,m,y);
int total_cases=5*5*5;
cout<<"No. of boundary cases = "<<total_cases<<endl;
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
for(k=0;k<5;k++)
{
cout<<d1[i]<<"-"<<"-"<<m1[j]<<"-"<<y1[k]<<"
";
next(d1[i],m1[j],y1[k]);
}
}

28. }

29. 8. Check whether the given equation is quadratic and
also check the nature of its roots and preform worst
case analysis:
#include<iostream>
#include<conio.h>
using namespace std;
void quad(int a, int b,int c)
{
int d=b*b-4*a*c;
if(a==0)
{
cout<<"Given equation is not quadraticn";
}
else
{
if(d>=0)
{
if(d==0)
cout<<"Roots are real and equal ";
else
cout<<"Root are real and distinct ";
float r1,r2;
r1=(-b+sqrt(d))/(2*a);
r2=(-b-sqrt(d))/(2*a);

30. cout<<r1<<" "<<r2<<endl;
}
else
cout<<”roots are imaginaryn”;
}
}
int main()
{
int a,b,c,total_cases=0,i,j,k;
int min_range=1,max_range=100;
int mid=(max_range-min_range)/2;
int arr[5]={min_range,min_range+1,mid,max_range-
1,max_range};
cout<<"enter coefficeint a,b,c of the quadratic
equation"<<endl;
cin>>a>>b>>c;
check(a,b,c);
total_cases=5*5*5;
cout<<"No. of boundary cases = "<<total_cases<<endl;

31. for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
for(k=0;k<5;k++)
{
cout<<arr[i]<<" "<<arr[j]<<" "<<arr[k]<<" ";
check(arr[i],arr[j],arr[k]);
}
}
}
}