Answered Essay: What will the program display when the ID is the number 11?

What will the program display when the ID is the number 11?

2. How can you write the multiple-alternative selection structure using the longer form of the if statement?

3. How can you write the multiple-alternative selection structure using the shorter form of the if statement?

4. How can you write the multiple-alternative selection structure using the switch statement?

5. What changes would you need to make to the code from Question 4 so that each case clause displays the appropriate price and the default clause displays the invalid id “Invalid ID” messagstart enter ID ID 2,9 5, 7, 11 Other price = 50.55 | | price = 12.35 | | price = 11.46 | | price- l ! =-1 price isT display price display Invalid stop

start enter ID ID 2,9 5, 7, 11 Other price = 50.55 | | price = 12.35 | | price = 11.46 | | price- l ! =-1 price isT display price display Invalid stop

Expert Answer

 

Dear Student,

here i have written the C program to demonstrate you the if structure in long form, short.And using switch keyword.

Note: Please note that the below program has been tested on ubuntu 16.04 system and compiled under gcc compiler. This code will also work on other IDE’s and C++ microsoft visual studio express.

——————————————————————————————————————————————

Program: 2: How can you write the multiple-alternative selection structure using the longer form of the if statement?

——————————————————————————————————————————————

//Hedaer file declration

#include<stdio.h>

//start of main function

int main()

{

float price;

int ID;

printf(“Enter Id: “);

scanf(“%d”, &ID);

//if ID is 1

if(ID == 1)

{

price = 50.55;

}

//if ID is either 2 or 9

if(ID == 2 )

{

price = 12.35;

}

if(ID == 9)

{

price = 12.35;

}

//if ID is either 5, 7 or 11

if(ID == 5 )

{

price = 11.46;

}

if(ID == 7 )

{

price = 11.46;

}

if(ID == 11 )

{

price = 11.46;

}

//else

else

{

price = -1;

}

//if price == -1

if(price)

{

printf(“Price is %lfn”, price);

}

else

{

printf(“Invalid IDn”);

}

return 0;

}

——————————————————————————————————————————————-

Output:

//Hedaer file declration

#include<stdio.h>

//start of main function

int main()

{

float price;

int ID;

printf(“Enter Id: “);

scanf(“%d”, &ID);

//if ID is 1

if(ID == 1)

{

price = 50.55;

}

//if ID is either 2 or 9

if(ID == 2 )

{

price = 12.35;

}

if(ID == 9)

{

price = 12.35;

}

//if ID is either 5, 7 or 11

if(ID == 5 )

{

price = 11.46;

}

if(ID == 7 )

{

price = 11.46;

}

if(ID == 11 )

{

price = 11.46;

}

//if price == -1

if(price)

{

printf(“Price is %lfn”, price);

}

else

{

printf(“Invalid IDn”);

}

return 0;

}

——————————————————————————————————————————————

——————————————————————————————————————————————

Program: 3: How can you write the multiple-alternative selection structure using the shorter form of the if statement?

————————————————————————————————————————————-

//Hedaer file declration

#include<stdio.h>

//start of main function

int main()

{

float price;

int ID;

printf(“Enter Id: “);

scanf(“%d”, &ID);

//if ID is 1

if(ID == 1)

{

price = 50.55;

}

//if ID is either 2 or 9

if(ID == 2 | ID == 9)

{

price = 12.35;

}

//if ID is either 5, 7 or 11

if(ID == 5 | ID == 7 | ID == 11)

{

price = 11.46;

}

//else

else

{

price = -1;

}

//if price == -1

if(price)

{

printf(“Price is %lfn”, price);

}

else

{

printf(“Invalid IDn”);

}

return 0;

}

—————————————————————————————————————————————–

here is the sample run of the program:

Output: 2

——————————————————————————————————————————————-

Program: 3: How can you write the multiple-alternative selection structure using the switch statement?

——————————————————————————————————————————————

//Hedaer file declration

#include<stdio.h>

//start of main function

int main()

{

float price;

int ID;

printf(“Enter Id: “);

scanf(“%d”, &ID);

//switch to case as per the input

switch(ID)

{

case 1:

price = 50.55;

printf(“Price is: %lfn”, price);

break;

//if ID is either 2 or 9

case 2:
case 9:

price = 12.35;

printf(“Price is: %lfn”, price);

break;

//if ID is either 5, 7 or 11

case 5:
case 7:
case 11:

price = 11.46;

printf(“Price is: %lfn”, price);

break;

case -1:

printf(“Invalid ID.n”);

break;

}

return 0;

}

——————————————————————————————————————————————

here is the sample run of the program:

Output: 3

——————————————————————————————————————————————–

Answer : 5

What changes would you need to make to the code from Question 4 so that each case clause displays the appropriate price and the default clause displays the invalid id “Invalid ID” message

Just add this statement in the last default case:

change these lines

case -1:

printf(“Invalid ID.n”);

break;

to

default:

printf(“Invalid ID.n”);

break;

—————————————————————————————-

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