C PROGRAMMING EXERCISE 3
Exercise 3: Decision making using If and If-else
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CODE IS AS FOLLOWS:
// LETS START WITH EXERCISE 3 WHICH IS BASED ON IF AND ELSE STATEMENTS.
//LINK FOR THE CODE OF EXERCISE 1,2 AND 3 WILL BE GIVEN IN THE DESCRIPTION
#include
#include
void main()
{
int opt,n,tax,x,y,pl,cp,sp;
char c;
clrscr();
while(1)
{
printf("\nEnter your choice");
printf("\n1.\t2.\t3.\n4.\t5.\t6.\n7.\t8.\t9.\t(Enter 9 for Exit\n)");
scanf("%d",&opt);
switch (opt)
{
case 1: printf("Q.1:-WRITE A PROGRAM TO ACCEPT INTEGER FROM THE USER AND CHECK WHETHER THE INTEGER IS EVEN OR ODD");
printf("\nSolution\n");
printf("Enter the integer");
scanf("%d",&n);
if(n%2==0)
{
printf("even");
}
else{
printf("odd");
}
break;
case 2: printf("Q.2:-ACCEPT A CHARACTER FROM THE USER AND CHECK WHETHER THE CHARACTER IS A DIGIT");
printf("\nSolution\n");
printf("Enter the character");
scanf("%c",&c);
if(c>=0 && c<=9)
{
printf("Digit is entered");
}
else{
printf("Non digit");
}
break;
case 3: printf("Q.3:-WRITE A C PROGRAM WHICH ACCEPTS ANNUAL BASIC SALARY OF AN EMPLOYEE AND CALCULATES AND DISPLAYS THE INCOME TAX");
printf("\nSolution\n");
/*TAX RULES:
<150000 TAX=0
150000 TO 300000 TAX=20%
>300000 TAX=30%*/
printf("Enter your annual income/salary");
scanf("%d",&n);
//will use nested if i.e else if
if(n<150000)
{
printf("there is 0 tax");
}
else if(n>=150000 && n<=300000)
{
tax=(n*20)/100;
printf("tax=%d",tax);
}
else
{
tax=(n*30)/100;
printf("tax=%d",tax);
}
break;
case 4: printf("Q.4:-ACCEPT CHARACTER FROM THE USER AND CHECK WHETHER THE CHARACTER IS A VOWEL OR CONSTANT");
printf("\nSolution\n");
printf("Enter the character");
scanf("%c",&c);
if (c=='a' || c=='e' || c=='i' ||c=='o' || c=='u' || c=='A' || c=='E' || c=='I' || c=='O' || c=='U')
{
printf("character is a vowel");
}
else{
printf("character is not a vowel");
}
break;
case 5: printf("Q.5:-ACCEPT YEARS IN INPUT THROUGH A KEYBOARD.WRITE A PROGRAM TO CHECK WHETHER THE YEAR IS A LEAP YEAR OR NOT");
printf("\nSolution\n");
printf("Enter the year");
scanf("%d",&n);
// note that leap year is divisible by 4 and not divisible by 100 or 400
if(n%4==0 && n%100!=0)
{
printf("The year is a leap year");
}
else
{
printf("The year is not a leap year");
}
break;
case 6: printf("Q.6:-WRITE A PROGRAM TO CHECK WHETHER GIVEN CHARACTER IS A DIGIT OR A CHARACTER IN LOWER CASE OR UPPER CASE ALPHABET");
printf("\nSolution\n");
printf("Enter the character");
scanf("%d",&c);
// note: ASCII value of digit is between 48 to 58 and lower case character is 97 to112 and uppercase is between 65 to 90
if(c>=48 || c<=58)
{
printf("Value entered is a digit");
}
else if(c>=97 || c<=112)
{
printf("Value entered is in Lower case");
}
else
{
printf("Value entered is in uppercase");
}
break;
case 7: printf("Q.7:-ACCEPT THE X AND Y CORDINATE OF A POINT AND FIND WHICH QUADRANT DOES THE POINT LIES");
printf("\nSolution\n");
/* QUADRAND 1=X>0 Y>0
QUADRANT 2=X<0 Y>0
3=X<0 Y0 Y<0*/
printf("Enter x and y points");
scanf("%d%d",&x,&y);
if(x>0 && y>0)
{
printf("Points lies in quadrant 1");
}
else if(x<0 && y>0)
{
printf("Points lies in quadrant 2");
}
else if(x<0 && y<0)
{
printf("Points lies in quadrant 3");
}
else
{
printf("Points lies in quadrant 4");
}
break;
case 8: printf("Q.8 ACCEPT THE COST PRICE AND SELLING PRICE FROM THE KEYBOARD AND FINDOUT WHETHER THE SELLER HAS MADE PROFIT OR NOT");
printf("\nSolution\n");
printf("Enter cost price and selinf price");
scanf("%d%d",&cp,&sp);
pl=sp-cp;
if(pl>0)
{
printf("it is profit");
}
else
{
printf("loss!");
}
break;
case 9: exit(1);
}
}
//time to check errors..lets swith to another app
getch();
}
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