c programs
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DESCRIPTION
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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;
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>
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 =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>
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,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>
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;
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;
}
}
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))
{
cout<<"enter a valid date\n"; return;
}
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>
#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 (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; 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))) {
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)) {
cout<<"enter a valid date\n";
return; }
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[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>
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};
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;
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 date\n";
return;
} else if((d==29&&m==2)&& (!leap_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)) {
cout<<"enter a valid date\n";
return; }
else if(d>31)
{ cout<<"enter a valid date\n"; return;
} else if(m<1||m>12)
{ cout<<"enter a valid date\n";
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>
#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 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]);
}
}
}
}
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