Introduction to Basic C programming 02

Programming in C Workshop India Wilson Wingston Sharon wingston.sharon@gmail.com

Hello World • Simple program #include<stdio.h> main() // STRICT ANSI • Traditional { printf("Hello Worldn"); “first” program return 0; for many users } #include<stdio.h> #include<conio> void main() // allowed { printf("Hello Worldn"); getch(); }

User input #include<stdio.h> main() { int number; printf("Enter a numbern"); scanf("%d",&number); printf("Number entered by you is %dn", number); return 0; } • Gets some input from the user. • Scanf function takes a reference to a type. • Exercise: take input of 9 numbers and display them

Command line arguments • Command #include<stdio.h> line main(int argc, char *argv[]) parameters { are passed to int c; the program printf("Number arguments passed: %dn", argc); from cmd command for ( c = 0 ; c < argc ; c++) printf("%d. argument passed is %sn", c+1, argv[c]); prompt return 0; }

If/Else Comparision #include<stdio.h> main() { int x = 1; if ( x == 1 ) printf("x is equal to one.n"); else printf(“Failn"); return 0; } • Comparing variables for conditional operation. • == is comparision; = is assignment operator • Ex: user enters a number. Program informs you if its even, odd or between 67 & 97.

Array I/O Declare an integer array • size of array has to be a constant Input how many elements the user requires in the array as (n) • The user input should not be greater than the array size declared Run a for loop from 0 to n and scanf it into &array[c] Run another for loop and print the complete array

#include<stdio.h> main() { int array[100], n, c; printf("Enter the number of elements in arrayn"); scanf("%d", &n); printf("Enter %d elementsn", n); for ( c = 0 ; c < n ; c++ ) scanf("%d", &array[c]); printf("Array elements entered bu you are:n"); for ( c = 0 ; c < n ; c++ ) printf("array[%d] = %dn", c, array[c]); return 0; }

Function Decide whether you want the function body before or after main • If before main function, write the function • If after, write a function prototype before the main function Decide on the return type and arguments and name them as well as the function appropriately • Appropriate naming is important when sharing functions or using them long after the fact. Write the function task and do a test call before using it in the program.

#include<stdio.h> void my_function(); main() { printf("Main function.n"); my_function(); printf("Back in function main.n"); return 0; } void my_function() { printf("Welcome to my function. Feel at home.n"); }

Structures Structure definitions are placed before the main function • And don’t forget the ; after the closing } Declare an object of the defined structure • struct <structure_name> <obj_name>; Use the . (dot) operator to access structure members • Variable.constant = 3;

#include<stdio.h> struct programming { float constant; char *pointer; }; main() { struct programming variable; char string[] = "Programming in Software Development."; variable.constant = 1.23; variable.pointer = string; printf("%fn", variable.constant); printf("%sn", variable.pointer); return 0; }

Calculator Declare all necessary variables. •Making sure the variable used for result of division operator is float Scanf the numbers Perform all operations and display them •Add •Sub •Mult •div

#include<stdio.h> main() { int first, second, add, subtract, multiply; float divide; printf("Enter two integersn"); scanf("%d%d",&first,&second); add = first + second; subtract = first - second; multiply = first * second; divide = first / (float)second; //typecasting printf("Sum = %dn",add); printf("Difference = %dn",subtract); printf("Multiplication = %dn",multiply); printf("Division = %.2fn",divide); return 0; }

Print current date Include the <dos.h> header file •This file has functions that read the current system clock date and time Declare a struct date d; •This structure is used to store the date information obtained Use getdate(&d); to retrieve the system date Print d.da_day as the current date variable

#include<stdio.h> #include<conio.h> #include<dos.h> main() { struct date d; getdate(&d); printf("Current system date is %d/%d/%d",d.da_day,d.da_mon,d.da_year); getch(); return 0; }

Vowel checking Declare a character (ch) and get an input using scanf() or getch() • Scanf waits for enter • Getch doesn’t wait for enter Check if the inputted character is an upper case or lowercase vowel from aeiou • Else display not a vowel

#include<stdio.h> main() { char ch; printf("Enter a charactern"); scanf("%c",&ch); if ( ch == 'a' || ch == 'A' || ch == 'e' || ch == 'E' || ch == 'i' || ch == 'I' || ch =='o' || ch=='O' || ch == 'u' || ch == 'U') printf("%c is a vowel.n", ch); else printf("%c is not a vowel.n", ch); return 0; }

Vowel checking A better option? •Try switch-case •Convert input character to lowercase and then check only for lowercase •Any other ideas? Make it a function!

switch(ch) { case 'a': case 'A': case 'e': case 'E': case 'i': case 'I': case 'o': case 'O': case 'u': case 'U': printf("%c is a vowel.n", ch); break; default: printf("%c is not a vowel.n", ch); }

int check_vowel(char a) { if ( a >= 'A' && a <= 'Z' ) a = a + 'a' - 'A'; /* Converting to lower case */ if ( a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u') return 1; return 0; }

Leap year What is a leap year? • Divisible by 4, 100 AND 400 Input a year into an integer type Use an if – else – if ladder to check for the given trye conditions • Algorithm for checking follows • http://en.wikipedia.org/wiki/Leap_year

Algorithm for leap year Is year % 4 = 0? •Yes – check next condition •No – not leap year Is year%100 = 0 •Yes – check next condition •No – Leap year Is year %400 = 0 •Yes – leap year •No – not leap year

int year; printf("Enter a year to check if it is a leap yearn"); scanf("%d", &year); if ( year%4 == 0) { if(year%100 == 0) { if(year%400==0) printf(“Leap year!”); else printf(“not Leap!”); } else printf(“Leap”); } else printf(“not leap year”);

Fibonacci Declare int variables : first second and next • Also get number of terms to generate (n) Loop from o to n incrementing by 1 if c<=1 don’t add, just put next = c Else: generate the next fibonacci term • Next = first + second • First = second • Second = next Print next and continue loop

printf("Enter the number of terms "); scanf("%d",&n); printf("First %d terms of fibonacci series are :-n",n); for ( c = 0 ; c < n ; c++ ) { if ( c <= 1 ) next = c; else { next = first + second; first = second; second = next; } printf("%dn",next); }

Sum of n numbers Declare sum = 0 and n, var • Scan n Loop from 0 to n Scan a number into var • Var is a temporary variable used for the scanf function Add sum and var and store it in var Repeat loop till n Print sum

#include<stdio.h> #include<conio.h> main() { int n, sum = 0, c, var; printf("Enter the number of integers you want to addn"); scanf("%d",&n); printf("Enter %d numbersn",n); for ( c = 1 ; c <= n ; c++ ) { scanf("%d",&var); sum = sum + var; } printf("Sum of entered numbers = %dn",sum); getch(); return 0; }

Reversing a number Declare n and rev as int • Scan n and initialise rev = 0 run a loop while n is not 0 rev = rev * 10 • Multiplication by 10 shifts the number left and puts a 0 in the ones place rev = rev + n%10 • n%10 extracts the last digit of n and then we add it to rev n = n/10 • Division by 10 and storing result in an integer types removes the ones digit of the number and shifts the number right Loop ends and then you can print rev

#include<stdio.h> main() { int n, reverse = 0; printf("Enter a number to reversen"); scanf("%d",&n); while( n != 0 ) { reverse = reverse * 10; reverse = reverse + n%10; n = n/10; } printf("Reverse of entered number is = %dn", reverse); return 0; }

factorial Input a number n and give it to the factorial function • Make sure it is not 0 or negative Make a factorial function • Long return type and one int n argument If n == 0 • Yes - return 1 • No – return n*factorial(n-1)

How this works Say you call fact(5) • n is 5 so if condition fails Function should return 5*fact(4) • Function is frozen in stack and fact(4) is called. Its result will be appended and then only fact(5) can return. Fact(4) is called and should return 4*fact(3) • Fact(5) is still waiting because fact(4) needs to call fact(3) before it can return Fact(3) must return 3*fact(2) and Fact(2) must return 2*fact(1) Fact(1) will return 1 • No more function calls so function can start returning

Fact(1) returns 1 to fact(2) Fact(2) return 2*1 to fact(3) Fact(3) returns 3*2*1 to fact(4) Fact(4) returns 4*3*2*1 to fact(5) Fact(5) returns 5*4*3*2*1 to user

Recursive factorial long factorial(int n) { if(n==0) return(1); else return(n*factorial(n-1)); }

#include<stdio.h> #include<conio.h> main() { int c, n, fact = 1; printf("Enter a number to calculate it's factorialn"); scanf("%d",&n); for( c = 1 ; c <= n ; c++ ) fact = fact*c; printf("Factorial of %d = %dn",n,fact); getch(); return 0; }

Pattern making a triangle * *** ***** ******* *********

Lets say we want a 5 row triangle Line 1 : 4 spaces and 1 star Line 2: 3 spaces and 3 stars Line 3: 2 spaces and 5 stars Line 4: 1 spaces and 7 stars Line 5: 0 space and 9 stars

Triangle pattern First we take in as input number of rows we want in n Make temp variable = n Loop 1 : iterates over every line of the triangle • Starts from row = 1 till row<=n Inside, loop 1 first we print spaces for every line • Print temp - 1 number of spaces “ “ • Decrement temp Then we print 2*row – 1 stars Then a new line is printed • row is incremented and loop1 starts again

#include<stdio.h> main() { int row, c, n, temp; printf("Enter the number of rows in pyramid of stars you wish to see "); scanf("%d",&n); temp = n; for ( row = 1 ; row <= n ; row++ ) { for ( c = 1 ; c < temp ; c++ ) printf(" "); temp--; for ( c = 1 ; c <= 2*row - 1 ; c++ ) printf("*"); printf("n"); } return 0; }

Diamond pattern Diamond pattern is like two triangles upside down Same code for triangle is used except when drawing the second triangle we go backwards •Spaces start from 1 and get incremented •No of stars is 2*(n-rows)-1 Otherwise code is the same! Take a look!

#include<stdio.h> main() { int n, c, k, space = 1; printf("Enter number of rowsn"); scanf("%d",&n); space = n - 1; for ( k = 1 ; k <= n ; k++ ) { for ( c = 1 ; c <= space ; c++ ) printf(" "); space--; for ( c = 1 ; c <= 2*k-1 ; c++) printf("*"); printf("n"); }

space = 1; for ( k = 1 ; k <= n - 1 ; k++ ) { for ( c = 1 ; c <= space; c++) printf(" "); space++; for ( c = 1 ; c <= 2*(n-k)-1 ; c++ ) printf("*"); printf("n"); } return 0; }

Primality checking • Problem statement: determine if a number is prime or not. • Input the number required to be checked for primality What is a prime number? • Give it to the checkprime() function Loop from 2 to num-1 • Loop variable is c If num %c =0 • Num is not prime return 0 After the loop if loop variable c =a • Then the num is prime return 1

int check_prime(int a) { int c; for ( c = 2 ; c <= a - 1 ; c++ ) { if ( a%c == 0 ) return 0; } if ( c == a ) return 1; }

Armstrong number • A number is armstrong if the sum of cubes of individual digits of a number is equal to the number itself. • For example 371 is an armstrong number as 33 + 73 + 13 = 371. • Some other armstrong numbers are: 0, 1, 153, 370, 407

armstrong number Initialize num ,temp, remainder and sum=0 take a num from user Store the num in temp Take a loop till temp its not equal to 0 • Temp%10 store it in remainder • Now add sum and (remainder to the power 3 ) and store it in sum • Now temp/10 store it in temp After the loop Check wether the num= sum • If yes then its armstrong num • Or else its not armstrong num

#include<stdio.h> #include<conio.h> main() { int number, sum = 0, temp, remainder; printf("Enter a numbern"); scanf("%d",&number); temp = number; while( temp != 0 ) { remainder = temp%10; sum = sum + remainder*remainder*remainder; temp = temp/10; } if ( number == sum ) printf("Entered number is an armstrong number."); else printf("Entered number is not an armstrong number."); getch(); }

Arrays – max and min Declare the array,max,size,c and location=1 Take the num of elements from the user First a loop must input all aray elements from the user • The loop goes from 0 till array size Initalize a variable called maximum to array[0] Loop from 1 till array size to find maximum element • Check with a condition each time whether the array is greater than maximum • Store the array variable in maximum and increment the value of the size of the array each time After the loop ends, maximum will hold the maximum element

#include<stdio.h> main() { int array[100], maximum, size, c, location = 1; printf("Enter the number of elements in arrayn"); scanf("%d",&size); printf("Enter %d integersn", size); for ( c = 0 ; c < size ; c++ ) scanf("%d", &array[c]); maximum = array[0]; for ( c = 1 ; c < size ; c++ ) { if ( array[c] > maximum ) { maximum = array[c]; location = c+1; } } printf("Maximum element is present at location number %d and it's value is %d.n", location, maximum); return 0; }

Binary search Declare as variables first,last,middle ,n,search and array • Scan array from user and sort it! Scan variable to be searched and store it in search variable Initialize the index variables first,last and middle • Middle = (first + last)/2 start a while loop until first <= last Check if search is on right side or left side of middle (x > a[mid] or x<a[mid]) • If x > a[mid] then min = mid+1 • If x<a[mid] then max = mid -1 • If x = a[mid] then we have found element and we stop loop Then update mid itself and continue loop If loop completes and first>last, then we have not found element

#include<stdio.h> main() { int c, first, last, middle, n, search, array[100]; printf("Enter number of elementsn"); scanf("%d",&n); printf("Enter %d integersn", n); for ( c = 0 ; c < n ; c++ ) scanf("%d",&array[c]); printf("Enter value to findn"); scanf("%d",&search); first = 0; last = n - 1; middle = (first+last)/2;

while( first <= last ) { if ( array[middle] < search ) first = middle + 1; else if ( array[middle] == search ) { printf("%d found at location %d.n", search, middle+1); break; } else last = middle - 1; middle = (first + last)/2; } if ( first > last ) printf("Not found! %d is not present in the list.n", search); return 0; }

Reverse an array d Declare the variables n,c,d,temp and 2 arrays with same size • Take the number of elements from the user (n) • Take the array elements from the user We start a single loop that goes forward through one array and backward through the other • C = n+1 to 0 decrementing by 1 used for array a • D = 0 to n-1 incrementing by 1used for array b b[d] = a[c] copies a in reverse to b We can copy b to a again so our original array a is reversed

#include<stdio.h> main() { int n, c, d, temp, a[100], b[100]; printf("Enter the number of elements in arrayn"); scanf("%d",&n); printf("Enter the array elementsn"); for ( c = 0 ; c < n ; c++ ) scanf("%d",&a[c]); for ( c = n - 1, d = 0 ; c >= 0 ; c--, d++ ) b[d] = a[c]; for ( c = 0 ; c < n ; c++ ) a[c] = b[c]; printf("Reverse array isn"); for( c = 0 ; c < n ; c++ ) printf("%dn", a[c]); return 0; }

Insert elements in array Declare the variables array,position,c,n,value • Take the number of elements from the user • Take the values of the array Take the position from user where the element to be added and store it in position • Enter the value of the element to be added Take a loop starting from n-1 to (position -1) and decrement each the value • Increase the size of the array by 1 as we add a new element • Now place the value to be added in that position. The resultant array can be displayed

#include<stdio.h> main() { int array[100], position, c, n, value; printf("Enter number of elements in arrayn"); scanf("%d", &n); printf("Enter %d elementsn", n); for ( c = 0 ; c < n ; c++ ) scanf("%d", &array[c]); printf("Enter the location where you wish to insert an elementn"); scanf("%d", &position); printf("Enter the value to insertn"); scanf("%d", &value);

for ( c = n - 1 ; c >= position - 1 ; c-- ) array[c+1] = array[c]; array[position-1] = value; printf("Resultant array isn"); for( c = 0 ; c <= n ; c++ ) printf("%dn", array[c]); return 0; }

#include<stdio.h> main() { int array[100], position, c, n; printf("Enter number of elements in arrayn"); scanf("%d", &n); printf("Enter %d elementsn", n); for ( c = 0 ; c < n ; c++ ) scanf("%d", &array[c]); printf("Enter the location where you wish to delete elementn"); scanf("%d", &position);

if ( position >= n+1 ) printf("Deletion not possible.n"); else { for ( c = position - 1 ; c < n - 1 ; c++ ) array[c] = array[c+1]; printf("Resultant array isn"); for( c = 0 ; c < n - 1 ; c++ ) printf("%dn", array[c]); } return 0; }

Matrix Addition Declare all matrixes and matrix sizes, first, second and sum • Matrix are 2D arrays and are declared as first[10][10] A nested for loop from 0 to m and 0 to n is used to input matrix 1 and 2 • m and n are the sizes of the matrix A nested for loop is used for matrix addition • Loop 1 variable is c, goes from 0 to m • Loop 2 variable is d, goes from 0 to n sum[c][d] = first[c][d] + second[c][d] • Print the sum matrix

#include<stdio.h> #include<conio.h> main() { int m, n, c, d, first[10][10], second[10][10], sum[10][10]; printf("Enter the number of rows and columns of matrix "); scanf("%d%d",&m,&n); printf("Enter the elements of first matrixn"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) scanf("%d",&first[c][d]); printf("Enter the elements of second matrixn");

for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) sum[c][d] = first[c][d]+ second[c][d]; printf("Sum of entered matrices:-n"); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < n ; d++ ) printf("%dt",sum[c][d]); printf("n"); } getch(); return 0; }

Matrix Transpose Declare 2 matrixes and sizes • A nested for loop from 0 to m and 0 to n is used to input matrix • m and n are the sizes of the matrix A nested for loop is used to transpose • Loop 1 variable is c, goes from 0 to m • Loop 2 variable is d, goes from 0 to n Transpose[c][d] = matrix[d][c] Print the transposed matrix

#include<stdio.h> main() { int m, n, c, d, matrix[10][10], transpose[10][10]; printf("Enter the number of rows and columns of matrix "); scanf("%d%d",&m,&n); printf("Enter the elements of matrix n"); for( c = 0 ; c < m ; c++ ) { for( d = 0 ; d < n ; d++ ) { scanf("%d",&matrix[c][d]); } }

for( c = 0 ; c < m ; c++ ) { for( d = 0 ; d < n ; d++ ) { transpose[d][c] = matrix[c][d]; } } printf("Transpose of entered matrix :-n"); for( c = 0 ; c < n ; c++ ) { for( d = 0 ; d < m ; d++ ) { printf("%dt",transpose[c][d]); } printf("n"); } return 0; }

Matrix multiplication Declare all matrixes and matrix sizes, first, second and mul •Input sizes and the two input matrixes Check sizes of matrix for conformity •m,n are sizes of matrix 1 •p,q are sizes of matrix 2 •Then if n == p only multiplication can take place 3 nested for loops are used •Loop1 c from 0 to m •Loop2 d from o to n •Loop3 k from 0 to p In loop3 sum = sum+first[c][k]*second[k][d] After loop3 ends, in loop2 •mul[c][d] = sum •Sum = 0 After all loops terminate, mul can be printed as resultant matrix

#include<stdio.h> #include<conio.h> main() { int m, n, p, q, c, d, k, sum = 0; int first[10][10], second[10][10], mul[10][10]; printf("Enter the number of rows and columns of first matrixn"); scanf("%d%d",&m,&n); printf("Enter the elements of first matrixn"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) scanf("%d",&first[c][d]); printf("Enter the number of rows and columns of second matrixn"); scanf("%d%d",&p,&q);

if ( n != p ) printf("Matrices with entered orders can't be multiplied with each other.n"); else { printf("Enter the elements of second matrixn"); for ( c = 0 ; c < p ; c++ ) for ( d = 0 ; d < q ; d++ ) scanf("%d",&second[c][d]); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < n ; d++ ) { for ( k = 0 ; k < p ; k++ ) { sum = sum + first[c][k]*second[k][d]; } mul[c][d] = sum; sum = 0; } }

String I/O Declare a character array to store the string •Take the string from the user and scan it by %s Now enter the value of the string as %s with the array name

#include<stdio.h> main() { char array[100]; printf("Enter a stringn"); scanf("%s",&array); printf("You entered the string %sn",array); return 0; }

Length of strings Initialize the variables length and string name Take the string from the user • Use the function strlen for the string and store it in length Now print the length of the string

#include<stdio.h> #include<string.h> main() { char a[100]; int length; printf("Enter a string to calculate it's lengthn"); gets(a); length = strlen(a); printf("Length of entered string is = %dn",length); return 0; }

Concatenation strings Function takes two arguments as pointers •These are character pointers that will point to the strings that are passed to the function First we have to move the original pointer to the end of the first string •When original is pointing to the end of the first string the while loop will fail. (last character is NULL) Then we start another while loop that parses through the next string •The while loop copies it character by character to the first string •Both pointers are incremented While loop will fail when the end of the add string is reached •The string end NULL must be appended to the new string After the function returns, the original string that was passed to the function would be changed

void concatenate_string(char *original, char *add) { while(*original) original++; while(*add) { *original = *add; add++; original++; } *original = '0'; }

Palindrome check Initialize 2 strings a and b • Get the string to be checked whether its palindrome or not Use strcpy to copy the string to b Now use strrev for b to reverse the string a Now check with a loop using strcmp bt sting a and b if the resultant is 0 • Then it’s a palindrome • Or else its not a palindrome

#include<stdio.h> #include<string.h> main() { char a[100], b[100]; printf("Enter the string to check if it is a palindromen"); gets(a); strcpy(b,a); strrev(b); if( strcmp(a,b) == 0 ) printf("Entered string is a palindrome.n"); else printf("Entered string is not a pailndrome.n"); return 0; }

Read & write files Create a file pointer • FILE *fp Open the file object • Fp = fopen(“filename”,’mode’); Fgetc() will return a character from the file EOF will signify end of file.

include<stdio.h> #include<conio.h> #include<stdlib.h> main() { char ch, fname[25]; FILE *fp; printf("Enter the name of file you wish to see "); gets(fname); fp = fopen(fname,"r"); printf("The contents of %s file are :- nn",fname); while( ( ch = fgetc(fp) ) != EOF) printf("%c",ch); getch(); fclose(fp); return 0; }

Introduction to Basic C programming 02

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Introduction to Basic C programming 02