stqa_exp1.doc

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EXPERIMENT NO 1

Write a ‘c’ program to demonstrate the working of the fallowing constructs:

i) do…while ii) while… iii) if …else iv) Switch v) for Loops in C language

AIM: To demonstrate the working of do..while construct

Objective: To understand the working of do while with different range of values and test cases

Program:

#include <stdio.h>

int main( )

{

int i,j=0,n=5;

printf("enter a number");

scanf("%d",&i);

{

if(i%2==0)

{

printf("%d",i);

printf("is a even number\n");

i++;

j++;

}

else

{

printf("%d",i);

printf("is a odd nuumber\n");

i++;

j++;

}

}while(i>0 && j<n);

return 0;

}

Output:

Input = 56	Actual output
	56 is even number
	57 is odd number
	58 is even number
	59 is odd number
	60 is even number

Test cases:

Test case no: 1

Test case name: Positive values within range

Input	Expected output	Actual output	Remarks
Input = 56	56 is even number	56 is even number	Success
	57 is odd number	57 is odd number
	58 is even number	58 is even number
	59 is odd number	59 is odd number
	60 is even number	60 is even number

Test case no: 2

Test case name: Negative values within a range

Input	Expected output	Actual output	Remarks
Input = -56	-56 is even number	-56 is even number	Fail
	-57 is odd number
	-58 is even number
	-59 is odd number
	-60 is even number

Test case no: 3

Test case name: Out of range values testing

Input	Expected output	Actual output	Remarks
Input = 54687123416345151	54687123416345151 is a odd number	2147483647 is a odd number	Fail

Aim: To demonstrate the working of while construct

Objective: To understand the working of while with different range of values and test cases

Program:

#include <stdio.h>

int main( )

{

int i,j=0,n=5;

printf("enter a number\n");

scanf("%d",&i);

while(i>0 && j<n)

{

if(i%2==0)

{

printf("%d",i);

printf("is a even number\n");

i++;

j++;

}

else

{

printf("%d",i);

printf("is a odd number\n");

i++;

j++;

}

return 0;

}

Output:

Input	Actual output
Input = 7	7 is odd number
	8 is even number
	9 is odd number
	10 is even number
	11 is odd number

Test cases:

Test case no: 1

Test case name: Positive values within range

Input	Expected output	Actual output	Remarks
Input = 7	7 is odd number	7 is odd number	Success
	8 is even number	8 is even number
	9 is odd number	9 is odd number
	10 is even number	10 is even number
	11 is odd number	11 is odd number

$C:\Windows\TEMP\Rar$DRa0.190\p1wl1.png$

Test case no: 2

Test case name: Negative values within a range

Input	Expected output	Actual output	Remarks
Input = -85	-85 is odd number	//No output While condition fails//	Fail
	-86 is even number
	-87 is odd number
	-88 is even number
	-89 is odd number

$C:\Windows\TEMP\Rar$DRa0.689\p1wl2.png$

Test case no: 3

Test case name: Out of range values testing

Input	Expected output	Actual output	Remarks
Input =123456987456321023321144563200000111223333333636666666666	123456987456321023321144563200000111223333333636666666666 is a even number	21474836472147 is a odd number	Fail

$C:\Windows\TEMP\Rar$DIa0.091\p1wl3.png$

Aim: To demonstrate the working of if else construct

Objective: To understand the working of if else with different range of values and test cases

Program:

#include <stdio.h>

int main()

{

int a;

printf("Enter a: ");

scanf("%d",&a);

if (a%2 == 0)

{

printf("The given number is EVEN");

}

else

{

printf("The given number is ODD");

}

return 0;

}

Output:

Input = 124	Actual output
	The given number is EVEN

Test cases:

Test case no: 1

Test case name: Positive values within range

Input	Expected output	Actual output	Remarks
Input = 124	The given number is EVEN	The given number is EVEN	Success

$C:\Windows\TEMP\Rar$DRa0.643\p1if1.png$

Test case no: 2

Test case name: Negative values within a range

Input	Expected output	Actual output	Remarks
Input = -84123	The given number is ODD	The given number is ODD	Fails

$C:\Windows\TEMP\Rar$DRa0.994\p1if2.png$

Test case no: 3

Test case name: Out of range values testing

Input

Expected output

Actual output

Remarks

Input = 22369978452435235365614652

3335664561461413031303

The given number is ODD

Fails

$C:\Windows\TEMP\Rar$DRa0.126\p1if3.png$

Aim: To demonstrate the working of switch constructs

Objective: To understand the working of switch with different range of values and test cases

Program:

#include<stdio.h>

int main()

{

int a,b,c,i;

printf("1.add\n2.sub\n3.mul\n4.div\n5.enter your choice\n");

scanf("%d",&i);

printf("Enter value of a and b");

scanf("%d%d",&a,&b);

switch(i)

{

case 1:

c=a+b;

printf("Sum of no is:%d",c);

break;

case 2:

c=a-b;

printf("Substraction of no is:%d",c);

break;

case 3:

c=a*b;

printf("multiplication of no is:%d",c);

break;

case 4:

c=a/b;

printf("division of no is:%d",c);

break;

default:

printf("Enter your choice");

break;

}

return 0;

}

Output:

Input

Output

Enter Ur choice: 1

Enter a, b Values: 5,6

Sum of no. is: 11

Enter Ur choice: 2

Enter a, b Values: 9,5

Subtraction of no. is: 4

Enter Ur choice: 3

Enter a, b Values: 5,6

Multiplication of no. is: 30

Enter Ur choice: 4

Enter a, b Values: 20,5

Division of no. is: 4

Test cases:

Test case no: 1

Test case name: Positive values within range

Input	Expected output	Actual output	Remarks
Enter Ur choice: 1 Enter a, b Values: 5,6	Sum of no. is: 11	Sum of no. is: 11	Success
Enter Ur choice: 2 Enter a, b Values: 9,5	Subtraction of no. is: 4	Subtraction of no. is: 4
Enter Ur choice: 3 Enter a, b Values: 5,6	Multiplication of no. is: 30	Multiplication of no. is: 30
Enter Ur choice: 4 Enter a, b Values: 20,5	Division of no. is: 4	Division of no. is: 4

$C:\Windows\TEMP\Rar$DIa0.885\Screenshot from 2016-01-19 17_06_09.png$ $C:\Windows\TEMP\Rar$DRa0.117\Screenshot from 2016-01-19 17_08_10.png$

$C:\Windows\TEMP\Rar$DRa0.781\Screenshot from 2016-01-19 17_08_37.png$

$C:\Windows\TEMP\Rar$DRa0.323\Screenshot from 2016-01-19 17_09_51.png$

Test case no:2

Test case name: Out of range values testing

Input

Expected output

Actual output

Remarks

Enter Ur choice: 1

Enter a, b Values:

4865121355651131256456231122315,

8452312355465612321355461213545

Sum of no. is: 24569875351868756480120

Sum of no. is: -2

Fails

$C:\Windows\TEMP\Rar$DRa0.653\Screenshot from 2016-01-19 17_11_52.png$

Test case no: 3

Test case name: Divide by zero

Input

Expected output

Actual output

Remarks

Enter Ur choice: 4

Enter a, b Values: 25, 0

Division of no. is: 0

Floating point exception(core dumped)

Fails

$C:\Windows\TEMP\Rar$DRa0.081\Screenshot from 2016-01-19 17_13_19.png$

Aim: To demonstrate working of for construct

Objective: To understand the working of for with different range of values and test cases

Program:

#include <stdio.h>

int main()

{

int i,j;

printf("enter a number\n");

scanf("%d",&j);

for(i=1;i<=5;i++)

{

if(i%2==0)

{

printf("%d",j);

printf("is a even no\n");

j++;

}

else

{

printf("%d",j);

printf("is a odd no\n");

j++;

}

return 0;

}

Output:

Input	Actual output
Input = 63	63 is odd no
	64 is even no
	65 is odd no
	66 is even no
	67 is odd no

Test cases:

Test case no: 1

Test case name: Positive values within range

Input	Expected output	Actual output	Remarks
Input = 63	63 is odd no	63 is odd no	Success
	64 is even no	64 is even no
	65 is odd no	65 is odd no
	66 is even no	66 is even no
	67 is odd no	67 is odd no

$C:\Windows\TEMP\Rar$DRa0.393\p1for1.png$

Test case no:2

Test case name: Negative values within a range

Input	Expected output	Actual output	Remarks
Input = -36	-36 is even no	-36 is odd no	Fails
	-37 is odd no	-35 is even no
	-38 is even no	-34 is odd no
	-39 is odd no	-33 is even no
	-40 is even no	-32 is odd no

$C:\Windows\TEMP\Rar$DRa0.971\p1for2.png$

Test case no: 3

Test case name: Out of range values testing

Input	Expected output	Actual output	Remarks
Input = 54878845445544545454 545444445454455445545 454554554444545455454 454544	2147483647 is odd no	2147483647 is odd no	Fails
	2147483648 is even no	-2147483648 is even no
	2147483647 is odd no	-2147483647 is odd no
	2147483646 is even no	-2147483646 is even no
	2147483645 is odd no	-2147483645 is odd no

$C:\Windows\TEMP\Rar$DRa0.887\p1for3.png$