c programs

Name : Ejaz khan

Roll no: 1120725

Section CO-5

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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;

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;}

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 Triangle\n";

} else if ((a==b)||(a==c)||(b==c))

{

cout<<"Isosceles Triangle\n";

}

else {

cout<<"Scalene Triangle\n";

}

} else

{

cout<<"Not a Triangle\n"; }

} int main()

{

int x,y,z, no_of_test=0,i,j; cout<<"Enter the value of sides of triangle in the

range 1-100\n";

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)

{

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(); }

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>


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 Triangle\n"; }

else if ((a==b)||(a==c)||(b==c))

{


} else

{

cout<<"Scalene Triangle\n"; }

}

else {

cout<<"Not a Triangle\n";

}

}

int main() {

int x,y,z, no_of_test =0,i,j;

cout<<"Enter the value of sides of triangle in the

range 1-100\n"; 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]);}

}

}

cout<<(max_range-min_range)/2<<" "<<(max_range-

min_range)/2<<" "<<(max_range-min_range)/2<<endl;

triangle (mid,mid,mid); getch();

}

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>


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 Triangle\n"; }

else if ((a==b)||(a==c)||(b==c))

{


} else

{

cout<<"Scalene Triangle\n"; }

}

else {

cout<<"Not a Triangle\n";

}

}

int main()

{ int x,y,z, no_of_test =0,i,j,k;

cout<<"Enter the value of sides of triangle in the

range 1-100\n"; 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 :

#include<iostream> #include<conio.h>


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;

return;

}

else

cout<<d+1<<"-"<<m<<"-"<<y<<endl; return;

}

else

{

if(d==31&&m==12)

{


}


{cout<<1<<"-"<<m+1<<"-"<<y<<endl;

return;

}

else cout<<d+1<<"-"<<m<<"-"<<y<<endl;

return;

}

}

void next(int d,int m,int y) {

if(d<1) {cout<<"enter a valid date\n";

return;

} else if((d==29&&m==2)&& (!leap_year_year (y)))

{

cout<<"enter a valid date\n"; return;

}

else if(d>=29&&m==2) {

cout<<"enter a valid date\n";

return; }

else

if((d==31&&m==4)||(d==31&&m==6)||(d==31&&m==9)||(d==31&&m==11))

{


}

else if(d>31) { cout<<"enter a valid date\n";

return;

} else if(m<1||m>12)

{ cout<<"enter a valid date\n";

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++)

{

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();

}

6. WAP to find next date and perform robustness

testing :

#include<iostream>


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 (y))

{if(d==31&&m==12)

{ cout<<"1-01-"<<y+1<<endl;

return;

}


{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;

}


{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 date\n"; }

if(d<1)

{cout<<"enter a valid date\n"; return;

}

else if((d==29&&m==2)&& (!leap_year_year (y))) {


return; }

else if(d>=29&&m==2)


return;

} else

if((d==31&&m==4)||(d==31&&m==6)||(d==31&&m==9

)||(d==31&&m==11)) {


return; }

else if(d>31)

{ cout<<"enter a valid date\n"; return;



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++)

{

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);

}

7. WAP to find next date and perform worst case

analysis : #include<iostream>

#include<conio.h>


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};

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


}

else

{

if(d==31&&m==12)

{


}


{cout<<1<<"-"<<m+1<<"-"<<y<<endl;

return;

}

else


}

}

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 date\n";

return;

} else if((d==29&&m==2)&& (!leap_year(y)))

{


}

else if(d>=29&&m==2)


return;

} else

if((d==31&&m==4)||(d==31&&m==6)||(d==31&&m==9

)||(d==31&&m==11)) {


return; }

else if(d>31)

{ cout<<"enter a valid date\n"; return;



return;

}

find_next(d,m,y);

}

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]); }

}

8. Check whether the given equation is quadratic and

also check the nature of its roots and preform worst case analysis: #include<iostream>


void quad(int a, int b,int c) {

int d=b*b-4*a*c;

if(a==0)

{

cout<<"Given equation is not quadratic\n"; }

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);

cout<<r1<<" "<<r2<<endl; }

else

cout<<”roots are imaginary\n”; }

}

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;

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]);

}

}

}

}

c programs

Environment