Learning to Program

A positive integer is entered through the keyboard. Write a function to obtain the prime factors of this number. For example, prime factors of 24 are 2, 2, 2 and 3, whereas prime factors of 35 are 5 and 7. #include <stdio.h> main ( ) { int number ; int prime ( int number ) ; int primefactor ( int number ) ; printf ( "Enter the number whose prime factors are to be calculated:" ) ; scanf ( "%d" , & number ) ; primefactor ( number ) ; } // The following function detects a Prime number . prime ( int num ) { int i , ifprime ; for ( i = 2 ; i < = num - 1 ; i + + ) { if ( num % i = = 0 ) { ifprime = 0 ; } else ifprime = 1 ; } return ( ifprime ) ; } // The following function prints the prime factors of a number . primefactor ( int num ) { int factor , ifprime ; for ( factor = 2 ; factor < = num ; ) { prime ( factor ) ; // so that the factors are only prime and nothing else . if ( ifprim

Write a function power(a,b), to calculate the value of a raised to b. #include <stdio.h> main ( ) { int power ( a , b ) ; int a , b , result ; printf ( "Enter the value of a and b:" ) ; scanf ( "%d %d" , & a , & b ) ; result = power ( a , b ) ; printf ( "%d raised to %d is %d" , a , b , result ) ; } power ( int a , int b ) { int calculation = 1 , calc ; for ( calc = 1 ; calc < = b ; calc + + ) { calculation = calculation * a ; continue ; } return ( calculation ) ; } The File can be found at: Download File

Write a general-purpose function to convert any given year into its roman equivalent. The following table shows the roman equivalents of decimal numbers: Decimal:........Roman 1.....................i 5....................v 10..................x 50..................l 100................c 500...............d 1000.............m Example: Roman equivalent of 1988 is mdcccclxxxviii Roman equivalent of 1525 is mdxxv This program is a big lengthy owing to the use of Case Statements. This program can also be rewritten using Arrays, which will reduce the length considerably. #include <stdio.h> main ( ) { int year ; int convert ( int year ) ; { printf ( "Note:Enter a four year digit year. \n \n " ) ; printf ( "Enter the year that you wanna convert to Roman: " ) ; scanf ( "%d" , & year ) ; if ( year > 1999 ) { printf ( "Invalid Year.Please enter again. \n \n " ) ; } } convert ( year ) ; } convert ( int year ) {

Any year is entered through the keyboard. Write a function to determine whether the year is a leap year or not. #include <stdio.h> main ( ) { int leap_year ( year ) ; int year , lp ; printf ( "Enter the year:" ) ; scanf ( "%d" , & year ) ; lp = leap_year ( year ) ; if ( lp ) { printf ( " \n The entered year is a leap year." ) ; } else { printf ( " \n The entered year is not a leap year." ) ; } } leap_year ( int y ) { int lp ; if ( y % 4 = = 0 ) { lp = 1 ; } else lp = 0 ; return ( lp ) ; } The file can be found at: Download File

Predict the output of the following Code Snippets. This is an exercise based on Functions. The answers are provided at the bottom of the exercise. Be honest with yourself. :P // 1 main ( ) { printf ( " \n Only stupids use c?" ) ; display ( ) ; } display ( ) { printf ( " \n Fools too use C!" ) ; main ( ) ; } // 2 main ( ) { printf ( " \n C to it that C survives." ) ; main ( ) ; } // 3 main ( ) { int i = 45 , c ; c = check ( i ) ; printf ( " \n %d" , c ) ; } check ( int ch ) { if ( ch > = 45 ) return ( 100 ) ; else return ( 10 * 10 ) ; } // 4 main ( ) { int i = 45 , c ; c = check ( i * 1000 ) ; printf ( " \n %d" , c ) ; } check ( int ch ) { if ( ch > = 40000 ) return ( ch / 10 ) ; else return ( 10 ) ; } // 1 ) infinite loop of both the statements // 2 ) The message " C to it that C survives" is iterated infinitesimally ! // 3 ) 100 // 4 ) 4500 The file can be downloaded at:

Write a function to calculate the factorial value of any integer entered through the keyboard. #include <stdio.h> main ( ) { int i , f ; int factorial ( ) ; printf ( "Enter the number to evaluate its factorial:" ) ; scanf ( "%d" , & i ) ; f = factorial ( i ) ; printf ( "%d! = %d \n " , i , f ) ; } factorial ( int num ) { int temp , fact ; for ( temp = 1 , fact = 1 ; temp < = num ; temp + + ) { fact = fact * temp ; continue ; } return ( fact ) ; } The file can be found at: Download File

These code snippets will help you understand the requirement of debugging a program. Without a program being debugged, they generate errors. And the heart of debugging lies in the process to find out what causes these errors. // 1 . main ( ) { int suite = 1 ; switch ( suite ) ; { case 0 : printf ( " \n Club" ) ; case 1 : printf ( " \n Diamond" ) ; } } // 2 . main ( ) { int temp ; scanf ( "%d" , & temp ) ; switch ( temp ) { case ( temp < = 20 ) : printf ( " \n Oooooooohhhhh! Damn Cool!" ) ; case ( temp > 20 & & temp < = 30 ) : printf ( " \n Rain rain here again! " ) ; case ( temp > 30 & & temp < = 40 ) : printf ( " \n Wish I am on Everest" ) ; default : printf ( " \n Good old nagpur weather" ) ; } } // 3 . main ( ) { float a = 3.5 ; switch ( a ) { case 0.5 : printf ( " \n The art of C" ) ; break ; case 1.5 : printf ( " \n The spirit of C&

Write a menu driven program which has the following options: 1. Factorial of a Number 2. Prime or Not 3. Odd or Even 4. Exit Make use of a switch Statement. Outline of the program: //a menu driven program main() { int choice; while (1) { printf("\n1. Factorial"); printf("\n2. Prime"); printf("\n3. Odd/ Even"); printf("\n4. Exit"); printf("\n Your Choice? : "); scanf ("%d", &choice); switch (choice) { case 1: //logic for factorial of a number break; case 2: //logic for deciding a prime number case 3: //logic for odd/even break; case 4: exit(); } } } NOTES: 1) The statement while(1) puts the entire logic in an infinite loop. This is necessary since the menu must keep reappearing on the screen once an item is selected and an appropriate action is taken. 2) ex

The goto statement makes a C Programmer's life only worse. Because of goto, it becomes hard to debug programs and find out and correct errors. With goto, the sequence of program execution changes abruptly, and so it also becomes very difficult to understand the programs. Of course, we can do without goto. But the only thing is that we need to combine the other instructions. One such example is provided here. The first program makes use of the goto statement, which appears to be fascinating. But then, as said earlier, the effect of goto can be realized by using other statements as well. That is being demonstrated in the next program. Program with goto: #include <stdio.h> main ( ) { int i , j , k ; for ( i = 1 ; i < = 3 ; i + + ) { for ( j = 1 ; j < = 3 ; j + + ) { for ( k = 1 ; k < = 3 ; k + + ) { if ( i = = 3 & & j = = 3 & & k = = 3 ) goto out ; else printf ( "%d %d %d \n " , i , j , k ) ; } } } out : printf ( "

Following are the problems, whose outputs you should be able to guess without running these code snippets on the computer. Good knowledge of "switch and case" statements is necessary for solving these. These problems help you know where you stand... // 1 main ( ) { char suite = 3 ; switch ( suite ) { case 1 : printf ( " \n Diamond" ) ; case 2 : printf ( " \n Spade" ) ; default : printf ( " \n Heart" ) ; } printf ( " \n I thought one wears a suite." ) ; } // 2 main ( ) { int k , j = 2 ; switch ( k = j + 1 ) { case 0 : printf ( " \n Tailor" ) ; case 1 : printf ( " \n Tutor" ) ; case 2 : printf ( " \n Tramp" ) ; default : printf ( "Pure simple Egghead!" ) ; } } // 3 main ( ) { int i = 0 ; switch ( i ) { case 0 : printf ( " \n Temple is a non-issue" ) ; case 1 : printf ( " \n Aandhi is never stable" ) ; case 2 : printf ( " \n Mandal will