C Program to print ASCII values of all characters.

C Program to print ASCII values of all characters.
ASCII value is the American Standard Code for Information Interchange.
ASCII value is the numerical value, or order, of an character. There are 128 standard ASCII characters, numbered from 0 to 127. Extended ASCII adds another 128 values and goes to 255. The numbers are typically represented in decimal or in hexadecimal. Read more about C Programming Language .

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*
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***********************************************************/

#include<stdio.h>
#include<conio.h>
int main(){

int i;
clrscr();
for(i=0;i<=255;i++)
printf("ASCII value of
character %c: %dn",i,i);
getch();
return 0;

}
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Number System

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C Program to find the strong number.

Write a C Program to find the given number is strong or not?
A number is called strong number if sum of the factorial of its digit is equal to number itself.
Example: 145 since 1! + 4! + 5! = 1 + 24 + 120 = 145.
Read more about C Programming Language .

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*
* Happy Coding
***********************************************************/
#include<stdio.h>
#include<conio.h>

int main()
{
int num,i,f,r,sum=0,num1;
clrscr();
printf("Enter a number: ");
scanf("%d",&num);

num1=num;
while(num){
i=1,f=1;
r=num%10;

while(i<=r){
f=f*i;
i++;
}
sum=sum+f;
num=num/10;

}
if(sum==num1)
printf("%d is a strong number",num1);
else
printf("%d is not a strong number",num1);
getch();
return 0;

}
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C Program to find the two sets Intersection and Union

Write a C Program to find the two sets Intersection and Union
A Set is a collection of well defined and distinct objects.
Intersection of two sets A and B is defined as, all the elements of set A, which are also elements of set B.Union of two sets A and B is defined as, all the elements of A and B, but not belonged to both.
Read more about C Programming Language .

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***********************************************************/

#include<stdio.h>
#include<conio.h>
void Union(int set1[10],int set2[10],int m,int n);
void Intersection(int set1[10],int set2[10],int m,int n);
void main()
{
int a[10],b[10],m,n,i,j;
int ch;
clrscr();
printf("nEnter the number of elements in first set:n");
scanf("%d",&m);
printf("nEnter the elements:n");
for(i=0;i<m;i++)
{
scanf("%d",&a[i]);
}
printf("nElement of First set:nn");
for(i=0;i<m;i++)
{
printf("%dt",a[i]);
}
printf("nEnter the number of elements in second set:n");
scanf("%d",&n);
printf("nEnter the elements:n");
for(i=0;i<n;i++)
{
scanf("%d",&b[i]);
}
printf("nElement of second setn");
for(i=0;i<n;i++)
{
printf("%dt",b[i]);
}
for(;;)
{
printf("nnMenunn1.Unionn2.Intersection");
printf("n3.exit");
printf("nEnter your choice:n");
scanf("%d",&ch);
switch(ch) {
case 1:
Union(a,b,m,n);
break;
case 2:
Intersection(a,b,m,n);
break;
case 3:
exit(0);
}
getch();
}
}

void Union(int a[10],int b[10],int m,int n)
{
int c[20],i,j,k=0,flag=0;
for(i=0;i<m;i++)
{
c[k]=a[i];
k++;
}
for(i=0;i<n;i++)
{
flag=0;
for(j=0;j<m;j++)
{
if(b[i]==c[j])
{
flag=1;
break;
}
}
if(flag==0)
{
c[k]=b[i];
k++;
}
}
printf("nElement of resultant setnn");
for(i=0;i<k;i++)
{
printf("t%d",c[i]);
}
}
void Intersection(int a[10],int b[10],int m,int n)
{
int c[20],i,j,k=0,flag=0;
for(i=0;i<m;i++)
{
flag=0;
for(j=0;j<n;j++)
{
if(a[i]==b[j])
{
flag=1;
break;
}
}
if(flag==1)
{
c[k]=a[i];
k++;
}
}
if(k==0)
{
printf("nnResultant set is null set!n");
}else{
printf("nElement of resultant setn");
for(i=0;i<k;i++)
{
printf("t%d",c[i]);
}
}
}


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K & R C Programs Exercise 4-6.

K and R C, Solution to Exercise 4-6:
K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition). You can learn and solve K&R C Programs Exercise.
C Program to Add commands for handling variables.(It’s easy to provide twenty-six variables with single-letter names.). Add a variable for the most recently printed value.

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* and browse!
*
* Happy Coding
***********************************************************/

#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <string.h>

#define MAXOP 100
#define NUMBER 0
#define IDENTIFIER 1
#define ENDSTRING 2
#define TRUE 1
#define FALSE 0
#define MAX_ID_LEN 32
#define MAXVARS 30


struct varType{
char name[MAX_ID_LEN];
double val;
};


int Getop(char s[]);
void push(double val);
double pop(void);
void showTop(void);
void duplicate(void);
void swapItems(void);
void clearStacks(struct varType var[]);
void dealWithName(char s[], struct varType var[]);
void dealWithVar(char s[], struct varType var[]);

int pos = 0;
struct varType last;

int main(void)
{
int type;
double op2;
char s[MAXOP];
struct varType var[MAXVARS];

clearStacks(var);

while((type = Getop(s)) != EOF)
{
switch(type)
{
case NUMBER:
push(atof(s));
break;
case IDENTIFIER:
dealWithName(s, var);
break;
case '+':
push(pop() + pop());
break;
case '*':
push(pop() * pop());
break;
case '-':
op2 = pop();
push(pop()- op2);
break;
case '/':
op2 = pop();
if(op2)
push(pop() / op2);
else
printf("nError: division by zero!");
break;
case '%':
op2 = pop();
if(op2)
push(fmod(pop(), op2));
else
printf("nError: division by zero!");
break;
case '?':
showTop();
break;
case '#':
duplicate();
break;
case '~':
swapItems();
break;
case '!':
clearStacks(var);
break;
case 'n':
printf("nt%.8gn", pop());
break;
case ENDSTRING:
break;
case '=':
pop();
var[pos].val = pop();
last.val = var[pos].val;
push(last.val);
break;
case '<':
printf("The last variable used was: %s (value == %g)n",
last.name, last.val);
break;

default:
printf("nError: unknown command %s.n", s);
break;
}
}
return EXIT_SUCCESS;
}

#define MAXVAL 100

int sp = 0;
double val[MAXVAL];

/* push: push f onto stack. */
void push(double f)
{
if(sp < MAXVAL)
val[sp++] = f;
else
printf("nError: stack full can't push %gn", f);
}

/*pop: pop and return top value from stack.*/
double pop(void)
{
if(sp > 0)
{
return val[--sp];
}
else
{
printf("nError: stack emptyn");
return 0.0;
}
}

void showTop(void)
{
if(sp > 0)
printf("Top of stack contains: %8gn", val[sp-1]);
else
printf("The stack is empty!n");
}


void duplicate(void)
{
double temp = pop();

push(temp);
push(temp);
}

void swapItems(void)
{
double item1 = pop();
double item2 = pop();

push(item1);
push(item2);
}


void clearStacks(struct varType var[])
{
int i;
sp = 0;

for( i = 0; i < MAXVARS; ++i)
{
var[i].name[0] = '';
var[i].val = 0.0;
}
}

void dealWithName(char s[], struct varType var[])
{
double op2;

if(!strcmp(s, "sin"))
push(sin(pop()));
else if(!strcmp(s, "cos"))
push(cos(pop()));
else if (!strcmp(s, "exp"))
push(exp(pop()));
else if(!strcmp(s, "pow"))
{
op2 = pop();
push(pow(pop(), op2));
}

else
{
dealWithVar(s, var);
}
}


void dealWithVar(char s[], struct varType var[])
{
int i = 0;

while(var[i].name[0] != '' && i < MAXVARS-1)
{
if(!strcmp(s, var[i].name))
{
strcpy(last.name, s);
last.val = var[i].val;
push(var[i].val);
pos = i;
return;
}
i++;
}

/* variable name not found so add it */
strcpy(var[i].name, s);
/* And save it to the last variable */
strcpy(last.name, s);
push(var[i].val);
pos = i;
}

int getch(void);
void unGetch(int);

/* Getop: get next operator or numeric operand. */
int Getop(char s[])
{
int i = 0;
int c;
int next;

/* Skip whitespace */
while((s[0] = c = getch()) == ' ' || c == 't')
{
;
}
s[1] = '';

if(isalpha(c))
{
i = 0;
while(isalpha(s[i++] = c ))
{
c = getch();
}
s[i - 1] = '';
if(c != EOF)
unGetch(c);
return IDENTIFIER;
}

/* Not a number but may contain a unary minus. */
if(!isdigit(c) && c != '.' && c != '-')
{
/* 4-6 Deal with assigning a variable. */
if('=' == c && 'n' == (next = getch()))
{
unGetch('');
return c;
}
if('' == c)
return ENDSTRING;

return c;
}

if(c == '-')
{
next = getch();
if(!isdigit(next) && next != '.')
{
return c;
}
c = next;
}
else
{
c = getch();
}

while(isdigit(s[++i] = c))
{
c = getch();
}
if(c == '.') /* Collect fraction part. */
{
while(isdigit(s[++i] = c = getch()))
;
}
s[i] = '';
if(c != EOF)
unGetch(c);
return NUMBER;
}

#define BUFSIZE 100

int buf[BUFSIZE];
int bufp = 0;

/* Getch: get a ( possibly pushed back) character. */
int getch(void)
{
return (bufp > 0) ? buf[--bufp]: getchar();
}

/* unGetch: push character back on input. */
void unGetch(int c)
{
if(bufp >= BUFSIZE)
printf("nUnGetch: too many charactersn");
else
buf[bufp++] = c;
}



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K & R C Programs Exercise 4-5.

K and R C, Solution to Exercise 4-5:
K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition). You can learn and solve K&R C Programs Exercise.
Write a C program to access to library functions like sin, exp, and pow. See “math.h” for more details.Read more about C Programming Language .

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
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* and browse!
*
* Happy Coding
***********************************************************/

#include<stdlib.h>
#include<stdio.h>
#include<ctype.h>
#include<math.h>
#include <string.h>

#define MAXOP 100
#define NUMBER 0
#define IDENTIFIER 1
#define TRUE 1
#define FALSE 0



int Getop(char s[]);
void push(double val);
double pop(void);
void showTop(void);
void duplicate(void);
void swapItems(void);
void clearStack();
void mathfnc(char s[]);

int main(void)
{
int type;
double op2;
char s[MAXOP];
int flag = TRUE;

while((type = Getop(s)) != EOF)
{
switch(type)
{
case NUMBER:
push(atof(s));
break;
case IDENTIFIER:
mathfnc(s);
break;
case '+':
push(pop() + pop());
break;
case '*':
push(pop() * pop());
break;
case '-':
op2 = pop();
push(pop()- op2);
break;
case '/':
op2 = pop();
if(op2)
push(pop() / op2);
else
printf("nError: division by zero!");
break;
case '%':
op2 = pop();
if(op2)
push(fmod(pop(), op2));
else
printf("nError: division by zero!");
break;
case '?':
showTop();
break;
case '#':
duplicate();
break;
case '~':
swapItems();
break;
case '!':
clearStack();
case 'n':
printf("nt%.8gn", pop());
break;
default:
printf("nError: unknown command %s.n", s);
break;
}
}
return EXIT_SUCCESS;
}

#define MAXVAL 100

int sp = 0;
double val[MAXVAL];

/* push: push f onto stack. */
void push(double f)
{
if(sp < MAXVAL)
val[sp++] = f;
else
printf("nError: stack full can't push %gn", f);
}

/*pop: pop and return top value from stack.*/
double pop(void)
{
if(sp > 0)
return val[--sp];
else
{
printf("nError: stack emptyn");
return 0.0;
}
}

void showTop(void)
{
if(sp > 0)
printf("Top of stack contains: %8gn", val[sp-1]);
else
printf("The stack is empty!n");
}



void duplicate(void)
{
double temp = pop();

push(temp);
push(temp);
}

void swapItems(void)
{
double item1 = pop();
double item2 = pop();

push(item1);
push(item2);
}

void clearStack(void)
{
sp = 0;
}

/*check string s for supported math functions */
void mathfnc(char s[])
{
double op2;

if( 0 == strcmp(s, "sin"))
push(sin(pop()));
else if( 0 == strcmp(s, "cos"))
push(cos(pop()));
else if (0 == strcmp(s, "exp"))
push(exp(pop()));
else if(!strcmp(s, "pow"))
{
op2 = pop();
push(pow(pop(), op2));
}
else
printf("%s Error: is not a supported function.n", s);
}

int getch(void);
void unGetch(int);

/* Getop: get next operator or numeric operand. */
int Getop(char s[])
{
int i = 0;
int c;
int next;


/* Skip whitespace */
while((s[0] = c = getch()) == ' ' || c == 't')
;
s[1] = '';

if(isalpha(c))
{
i = 0;
while(isalpha(s[i++] = c ))
c = getch();
s[i - 1] = '';
if(c != EOF)
unGetch(c);
return IDENTIFIER;
}

/* Not a number but may contain a unary minus. */
if(!isdigit(c) && c != '.' && c != '-')
return c;

if(c == '-')
{
next = getch();
if(!isdigit(next) && next != '.')
{
return c;
}
c = next;
}
else
c = getch();

while(isdigit(s[++i] = c))
c = getch();
if(c == '.') /* Collect fraction part. */
while(isdigit(s[++i] = c = getch()))
;
s[i] = '';
if(c != EOF)
unGetch(c);
return NUMBER;
}

#define BUFSIZE 100

char buf[BUFSIZE];
int bufp = 0;
int getch(void)
{
return (bufp > 0) ? buf[--bufp]: getchar();
}


void unGetch(int c)
{
if(bufp >= BUFSIZE)
printf("nUnGetch: too many charactersn");
else
buf[bufp++] = c;
}


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C Program implement Kruskal’s algorithm.

Write a C Program implement Kruskal’s algorithm.
Kruskal’s algorithm is a greedy algorithm that finds the minimum spanning tree of a graph. Graph should be weighted, connected, and undirected.Minimum spanning tree is a spanning tree with weight less than or equal to the weight of every other spanning tree. Read more about C Programming Language .

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* contact [email protected]
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* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
int i,j,k,a,b,u,v,n,ne=1;
int min,mincost=0,cost[9][9],parent[9];
int find(int);
int uni(int,int);
void main()
{
clrscr();
printf("nntImplementation of Kruskal's algorithmnn");
printf("nEnter the no. of verticesn");
scanf("%d",&n);
printf("nEnter the cost adjacency matrixn");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&cost[i][j]);
if(cost[i][j]==0)
cost[i][j]=999;
}
}
printf("nThe edges of Minimum Cost Spanning Tree arenn");
while(ne<n)
{
for(i=1,min=999;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(cost[i][j]<min)
{
min=cost[i][j];
a=u=i;
b=v=j;
}
}
}
u=find(u);
v=find(v);
if(uni(u,v))
{
printf("n%d edge (%d,%d) =%dn",ne++,a,b,min);
mincost +=min;
}
cost[a][b]=cost[b][a]=999;
}
printf("ntMinimum cost = %dn",mincost);
getch();
}
int find(int i)
{
while(parent[i])
i=parent[i];
return i;
}
int uni(int i,int j)
{
if(i!=j)
{
parent[j]=i;
return 1;
}
return 0;
}
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Dijkstras Algorithm

Learn C Programming

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K & R C Programs Exercise 3-5.

K and R C, Solution to Exercise 3-5:
K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition). You can learn and solve K&R C Programs Exercise.
Write a c program to convert the integer n into a base b character representation in the string s. In particular, itob(n, s, 16) formats n as a hexadecimal integer in s. Read more about C Programming Language .

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* programs for commercial purposes,
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* and browse!
*
* Happy Coding
***********************************************************/


#include <stdlib.h>
#include <stdio.h>

void itob(int n, char s[], int b);
void reverse(char s[]);

int main(void) {
char buffer[10];
int i;

for ( i = 2; i <= 20; ++i ) {
itob(255, buffer, i);
printf("Decimal 255 in base %-2d : %sn", i, buffer);
}
return 0;
}


/* itob: convert n to charecters in s - base b */

void itob(int n, char s[], int b) {

int i, j, sign;
void reverse(char s[]);


if ((sign = n) < 0)
n = -n;
i = 0;
do {
j = n%b;
s[i++] = (j <= 9) ? j+'0' : j+'a'-10;
} while ((n /= b) > 0);
if (sign < 0)
s[i++] = '-';
s[i] = '';
reverse(s);
}


/* Reverses string s[] in place */

void reverse(char s[]) {
int c, i, j;
for ( i = 0, j = strlen(s)-1; i < j; i++, j--) {
c = s[i];
s[i] = s[j];
s[j] = c;
}
}


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K & R C Programs Exercise 2-8.

K and R C, Solution to Exercise 2-8:
 K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition) You can learn and solve K&R C Programs Exercise.
C Program that returns the value of the integer x rotated to the right by n bit positions. Read more about C Programming Language .

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

unsigned rightroot(unsigned x, int n)
{
int wordlength(void);
int rbit; //rightmost bit

while(n-->0){
rbit=(x & 1)<<(wordlength()-1);
x=x>>1;
x=x|rbit;
}
return x;
//word length computes the wor lengt of machine
int wordlength()
{
int i;
unsigned v = (unsigned~0;
for(i = 1;(v = v >> 1) > 0;i++)
;
return i;
}
//main function to test the program,, you can try in different ways!
#include <stdio.h>

int main(void)
{
unsigned x;
int n;

for(x = 0; x < 200; x += 25)
for(n = 1; n < 8; n++)
printf("%u, %d: %un", x, n, rightrot(x, n));
return 0;
}

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C Program to implement Linear regression algorithm.

Linear Regression  is the predicting the value of one scalar variable(y) using the explanatory another variable(x). Linear regression  is represented by the equation Y = a + bX, where X is the explanatory variable and Y is the scalar variable. The slope of the line is b, and a is the intercept. For linear list square modeling, Linear regression is very helpful. Read more about C Programming Language.

/************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/



#include "stdio.h"
#include "conio.h"
#include "math.h"
#include "string.h"

float mean(float *a, int n);
void deviation(float *a, float mean, int n, float *d, float *S);

void main() {
float a[20], b[20], dx[20], dy[20];
float sy = 0, sx = 0, mean_x = 0, mean_y = 0, sum_xy = 0;
float corr_coff = 0, reg_coff_xy = 0, reg_coff_yx = 0;
char type_coff[7];
int n = 0, i = 0;

clrscr();

printf("Enter the value of n: ");
scanf("%d", &n);
printf("Enter the values of x and y:n");
for (i = 0; i < n; i++)
scanf("%f%f", &a[i], &b[i]);
mean_x = mean(a, n);
mean_y = mean(b, n);
deviation(a, mean_x, n, dx, &sx);
deviation(b, mean_y, n, dy, &sy);

for (i = 0; i < n; i++)
sum_xy = sum_xy + dx[i] * dy[i];
corr_coff = sum_xy / (n * sx * sy);
printf("Enter the type of regression coefficient as 'x on y' or 'y on x': ");
fflush(stdin);
gets(type_coff);

if (strcmp(type_coff, "x on y") == 1) {
reg_coff_xy = corr_coff * (sx / sy);
printf("nThe value of linear regression coefficient is %f",
reg_coff_xy);
} else if (strcmp(type_coff, "y on x") == 1) {
reg_coff_yx = corr_coff * (sy / sx);
printf("nThe value of linear regression coefficient is %f",
reg_coff_yx);
} else
printf("nEnter the correct type of regression coefficient.");
getch();
}

float mean(float *a, int n) {
float sum = 0, i = 0;
for (i = 0; i < n; i++)
sum = sum + a[i];
sum = sum / n;
return (sum);
}

void deviation(float *a, float mean, int n, float *d, float *s) {
float sum = 0, t = 0;
int i = 0;
for (i = 0; i < n; i++) {
d[i] = a[i] - mean;
t = d[i] * d[i];
sum = sum + t;
}
sum = sum / n;
*s = sqrt(sum);
}

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K&R C Program Exercise 1-16

K and R solutions to Revise the the main routine of the longest-line program so it will correctly print the length of arbitrarily long input lines, and as much as possible of the text. K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language, second addition, by Brian W.Keringhan and Dennis M.Ritchie(Prentice Hall,1988). You can learn and solve K&R C Programs Exercise. Read more about C Programming Language .

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include "stdio.h"

#define MAXINLINE 1000 /* maximum input line size */

int getline(char line[], int MAXINLINE);
void copy(char to[], char from[]);

/* print longest input line */
int main(void)
{
int len; /* current line length */
int max; /* maximum length seen so far */
char line[MAXINLINE]; /* current input line */
char longest[MAXINLINE]; /* longest line saved here */

max = 0;

while((len = getline(line, MAXINLINE)) > 0)
{
printf("%d: %s", len, line);

if(len > max)
{
max = len;
copy(longest, line);
}
}
if(max > 0)
{
printf("Longest LINE IN THE PROGRAM is %d characters:n%s", max, longest);
}
printf("n");
return 0;
}

/* getline: read a line into s, return length */
int getline(char s[], int line)
{
int c, i, j;

for(i = 0, j = 0; (c = getchar())!=EOF && c != 'n'; ++i)
{
if(i < line - 1)
{
s[j++] = c;
}
}
if(c == 'n')
{
if(i <= line - 1)
{
s[j++] = c;
}
++i;
}
s[j] = '';
return i;
}

/* copy: copy 'from' into 'to'; assume 'to' is big enough */
void copy(char to[], char from[])
{
int k;

k = 0;
while((to[k] = from[k]) != '')
{
++k;
}
}


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