K & R C Programs Exercise 3-6.

K and R C, Solution to Exercise 3-6:
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 change version of itoa that accepts three arguments instead of two(K and R C Exercise 3-4). The third argument is a minimum field width; the converted number must be paddle with blanks on the left if necessary to make it wide enough. Read more about C Programming Language .

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

#include <stdlib.h>
#include <stdio.h>
#include <limits.h>
#define abs(x) ((x) < 0 ? -(x) : (x))
void itoa(int n, char s[], int w);
void reverse(char s[]);

int main(void) {
char buffer[20];

printf("INT_MIN: %dn", INT_MIN);
itoa(INT_MIN, buffer);
printf("Buffer : %sn", buffer);

return 0;
}
//itoa: convert to n characters in s, w characters wide.
void itoa(int n, char s[], int w)
{
void int i, sign;
void everse(char s[]);
sign = n;
i = 0;
do{
s[i++] = abs(n%10) + '0';
printf("%d %% %d + '0' = %dn", n, 10, s[i-1]);
}while((n/=10)!=0);
if(sign < 0)
s[i++] = '-';
while(i < w)
s[i++] = ' ';
s[i] = '';
reverse(s);
}
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 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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***********************************************************/


#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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C Program to get the system information.

C Program to get the system information. We can use the ‘uname’ function to get the system information, which is defined in the “sys/utsname.h” library.Structure is used to hold information returned by the uname function.
“uname” function members and their use:
sysname returns the name of the operating system in use.
release returns the current release level of the operating system implementation.
version returns the current version level within the release of the operating system.
machine is the description of the type of hardware that is in use.
Read more about C Programming Language.

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


#include<sys/utsname.h> /* Header for 'uname' */

main()
{
struct utsname uname_pointer;

uname(&uname_pointer);

printf("System name - %s n", uname_pointer.sysname);
printf("Nodename - %s n", uname_pointer.nodename);
printf("Release - %s n", uname_pointer.release);
printf("Version - %s n", uname_pointer.version);
printf("Machine - %s n", uname_pointer.machine);
printf("Domain name - %s n", uname_pointer.domainname);
}


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C Program to reverse a Linked List.

Data structures using C,
C Program to reverse a Linked List. Linked list is a data structure in which the objects are arranged in a linear order. Linked List contains group of nodes, in which each node contains two fields, one is data field and another one is the reference field which contains the address of next node. In this program, We reverse the linked list, but without sorting the linked list.Read more about C Programming Language .

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*
* Happy Coding
***********************************************************/


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

#define MAX 100

struct leftnode {
int num;
struct leftnode *next;
};


void linklist_add(struct leftnode **n, int val);

void linklist_reverse(struct leftnode **n);

void linklist_display(struct leftnode *n);

int main(void) {
struct leftnode *new = NULL;
int i = 0;

for(i = 0; i <= MAX; i++)
linklist_add(&new, i);

printf("Linked list before reversal:n");
linklist_display(new);
linklist_reverse(&new);
printf("Linked list after reversal:n");
linklist_display(new);

return 0;
}


void linklist_add(struct leftnode **n, int val) {
struct leftnode *temp = NULL;
temp = malloc(sizeof(struct leftnode));
temp->num = val;
temp->next = *n;
*n = temp;
}


void linklist_reverse(struct leftnode **n) {
struct leftnode *a = NULL;
struct leftnode *b = NULL;
struct leftnode *c = NULL;
a = *n, b = NULL;

while(a != NULL) {
c = b, b = a, a = a->next;
b->next = c;
}

*n = b;
}

void linklist_display(struct leftnode *n) {
while(n != NULL)
printf(" %d", n->num), n = n->next;

printf("n");
}

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C Program to implement Integration.

C Program to implement Integration. Integration is the important concept of calculus, which is the inverse of the differentiation. A definite integral gives the area between the graph of a function and the horizontal axis between vertical lines at the endpoints of an interval. Integration is used to find the area, volume of irregular shapes. In the following code, We find the integration of the defined function between two boundary limits, which are entered by the user. Read more about C Programming Language .

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*
* Happy Coding
***********************************************************/

#include <stdio.h>
/* Propram to perform definite integration of a given function between
two boundary limits input by user. Feel free to use and modify it, but
please do not remove this comment.
source: C for Engineering, http://c4engineering.hypermart.net */

#define N 1000

void main(void) {
float i, a, b, sum = 0;
printf(
"nThis program will integrate a function between two boundary limits.");
printf("nnEnter the first boundary limit:");
scanf("%f", &a);
printf("nEnter the second boundary limit:");
scanf("%f", &b);
if (a > b) {
i = a;
a = b;
b = i;
}

for (i = a; i < b; i += (b - a) / N) {
/* Define your function below, and include the suitable header files */
y = x * x + 2 * x - 4;
sum += y * (b - a) / N;
}

printf("nnValue of integration is:%.3f", sum);
getch();
return;
}

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K & R C Programs Exercise 3-3.

K and R C, Solution to Exercise 3-3:
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 expand short hand notations like a-z in the string s1 into the equivalent complete list abc—xyz in s2. Allowed for letters of either case and digits, and be prepared to handle cases like a-b-c and a-z0-9 and -a-z. Arranged that a leading or trailing – is taken literally. Read more about C Programming Language .

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*
* Happy Coding
***********************************************************/

#include
#include

void expand(char * s1, char * s2);

int main(void) {
char *s[] = { "a-z-", "z-a-", "-1-6-",
"a-ee-a", "a-R-L", "1-9-1",
"5-5", NULL };
char result[100];
int i = 0;

while ( s[i] ) {

expand(result, s[i]);
printf("Unexpanded: %sn", s[i]);
printf("Expanded : %sn", result);
++i;
}

return 0;
}



//expand: expand short hand notations in s1 into string s2
void expand(char * s1, char * s2) {
static char upper_alph[27] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
static char lower_alph[27] = "abcdefghijklmnopqrstuvwxyz";
static char digits[11] = "0123456789";

char * start, * end, * p;
int i = 0;
int j = 0;


/* Loop through characters in s2 */

while ( s2[i] ) {
switch( s2[i] ) {
case '-':
if ( i == 0 || s2[i+1] == '' ) {

s1[j++] = '-';
++i;
break;
}
else {


if ( (start = strchr(upper_alph, s2[i-1])) &&
(end = strchr(upper_alph, s2[i+1])) )
;
else if ( (start = strchr(lower_alph, s2[i-1])) &&
(end = strchr(lower_alph, s2[i+1])) )
;
else if ( (start = strchr(digits, s2[i-1])) &&
(end = strchr(digits, s2[i+1])) )
;
else {

fprintf(stderr, "EX3_3: Mismatched operands '%c-%c'n",
s2[i-1], s2[i+1]);
s1[j++] = s2[i-1];
s1[j++] = s2[i++];
break;
}


p = start;
while ( p != end ) {
s1[j++] = *p;
if ( end > start )
++p;
else
--p;
}
s1[j++] = *p;
i += 2;
}
break;

default:
if ( s2[i+1] == '-' && s2[i+2] != '' ) {

++i;
}
else {

s1[j++] = s2[i++];
}
break;
}
}
s1[j] = s2[i];
}


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K & R C Programs Exercise 2-9.

K and R C, Solution to Exercise 2-9:
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.
In C two complement number system, x &= (x-1) deletes the rightmost one bit in x.
We take (x-1) and add 1 to it to produce x. The rightmost 0-bit of x-1 changes to 1 in the result x. Therefore, the rightmost 1-bit of x has a corresponding 0-bit in x-1. This is why x & (x-1), in a two’s complement number system, will delete the rightmost 1-bit in x. Read more about C Programming Language .

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***********************************************************/
int bitcount(unsigned x)
{
int b;
for(b = 0;x != 0;x &= x-1)
++b;
return b;
}
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C Program to add two polynomials using structures.

C Program to add two polynomials using structures. Structure is a c composite data type, in which we can define all the data types under the same name or object. Size of the Structure is the size of all data types, plus any internal padding. the key word “struct” is used to declare the structure. A Polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. Read more about C Programming Language .

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*
* Happy Coding
***********************************************************/

#include < stdio.h >
#include < conio.h >
#define MAX 20

struct addpolynomial {
int exp, coef;
};

void main() {
struct addpolynomial p1[MAX], p2[MAX], p3[MAX];
int max1, max2, max3;
clrscr();
printf("nEnter first addpolynomial : ");
max1 = read_addpolynomial(p1);
printf("nEnter second addpolynomial : ");
max2 = read_addpolynomial(p2);
max3 = add_addpolynomial(p1, p2, p3, max1, max2);
printf("nFirst addpolynomial is ");
print_addpolynomial(p1, max1);
printf("nSecond addpolynomial is ");
print_addpolynomial(p2, max2);
printf("n The resultant addpolynomial after addition is");
print_addpolynomial(p3, max3);
}

//function to read polynomial
int read_addpolynomial(struct addpolynomial p[]) {
int i, texp;
i = 0;
printf("nEnter exp ( use -1 to exit) : ");
scanf("%d", &texp);
while (texp != -1) {
p[i].exp = texp;
printf("nEnter coef : ");
scanf("%d", &p[i].coef);
i++;
printf("nEnter exp ( use -1 to exit) : ");
scanf("%d", &texp);
}
return (i);
}

//function to print polynomial
int print_addpolynomial(struct addpolynomial p[], intMAX1) {
int i;
for (i = 0; i < max1; i++)
printf("%+dX%d ", p[i].coef, p[i].exp);
return;
}

//function to ad polynomials
int add_addpolynomial( p1, p2, p3, max1, max2)
struct addpolynomial p1[], p2[], p3[];
intMAX1, max2;
{
int i,j,k;
i = j = k = 0;
while ( i <max1 && j <max2)
{
if( p1[i].exp > p2[j].exp)
{
p3[k] = p1[i];
k++;
i++;
}
else
if( p1[i].exp < p2[j].exp)
{
p3[k] = p2[j];
k++;
j++;
}
else
{
p3[k].exp = p1[i].exp;
p3[k].coef = p1[i].coef + p2[j].coef;
i++;
j++;
k++;
}
}
while( i <max1 )
{
p3[k] = p1[i];
k++;
i++;
}
while( j <max2 )
{
p3[k] = p2[j];
k++;
j++;
}
return(k);
}

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C Program for Simple DSC order Priority QUEUE Implementation

Data structures using C,Priority QUEUE is a abstract data type in which the objects are inserted with respect to certain priority. In this program, we created the simple descending order priority queue, here items are inserted in descending order. Read more about C Programming Language .

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*
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***********************************************************/
#define SIZE 5 /* Size of Queue */
int PQ[SIZE],f=0,r=-1; /* Global declarations */

PQinsert(int elem)
{
int i; /* Function for Insert operation */
if( Qfull()) printf("nn Overflow!!!!nn");
else
{
i=r;
++r;
while(PQ[i] <= elem && i >= 0) /* Find location for new elem */
{
PQ[i+1]=PQ[i];
i--;
}
PQ[i+1]=elem;
}
}

int PQdelete()
{ /* Function for Delete operation */
int elem;
if(Qempty()){ printf("nnUnderflow!!!!nn");
return(-1); }
else
{
elem=PQ[f];
f=f+1;
return(elem);
}
}

int Qfull()
{ /* Function to Check Queue Full */
if(r==SIZE-1) return 1;
return 0;
}

int Qempty()
{ /* Function to Check Queue Empty */
if(f > r) return 1;
return 0;
}

display()
{ /* Function to display status of Queue */
int i;
if(Qempty()) printf(" n Empty Queuen");
else
{
printf("Front->");
for(i=f;i<=r;i++)
printf("%d ",PQ[i]);
printf("<-Rear");
}
}

main()
{ /* Main Program */
int opn,elem;
do
{
clrscr();
printf("n ### Priority Queue Operations(DSC order) ### nn");
printf("n Press 1-Insert, 2-Delete,3-Display,4-Exitn");
printf("n Your option ? ");
scanf("%d",&opn);
switch(opn)
{
case 1: printf("nnRead the element to be Inserted ?");
scanf("%d",&elem);
PQinsert(elem); break;
case 2: elem=PQdelete();
if( elem != -1)
printf("nnDeleted Element is %d n",elem);
break;
case 3: printf("nnStatus of Queuenn");
display(); break;
case 4: printf("nn Terminating nn"); break;
default: printf("nnInvalid Option !!! Try Again !! nn");
break;
}
printf("nnnn Press a Key to Continue . . . ");
getch();
}while(opn != 4);
}


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K & R C Programs Exercise 2-2.

K and R C, Solution to Exercise 2-2:
C Program to demonstrate for and while loops. K and R C Program. 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 .

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

/*Write the equivalent to the below for loop without using && or ||.*/

// Original

for(i=0; i<lim-1 && (c=getchar()) != 'n' && c != EOF; ++i)
{
s[i] = c;
}
// Equivalent:


while (i < (lim - 1))
{
c = getchar();

if (c == EOF)
break;
else if (c == 'n')
break;

s[i++] = c;
}

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