K & R C Programs Exercise 2-10.

K and R C, Solution to Exercise 2-10:
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 Function lower, which converts the lower case, with a conditional expression instead of if-else.Read more about C Programming Language .

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int lower(int c)
{
return c >= 'A' && c<= 'Z' ? c + 'a' - 'A': c;
}
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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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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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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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K & R C Programs Exercise 2-7.

K and R C, Solution to Exercise 2-7:
 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 function invert(x, p, n) that returns x with the n bits that begin at position p inverted(i.e, 1 changed into 0 and vice versa), leaving the others unchanged.Read more about C Programming Language .

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

unsigned invert(unsigned x, int p, int n)
{
return x ^ (~(~0U << n) << p);
}

/* Program for testing */

#include <stdio.h>

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

for(x = 0; x < 500; x += 49)
for(n = 1; n < 8; n++)
for(p = 1; p < 8; p++)
printf("%u, %d, %d: %un", x, n, p, invert(x, n, p));
return 0;
}

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

#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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K & R C Programs Exercise 2-6.

K and R C, Solution to Exercise 2-6:
 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 Function setbits(x, p, n, y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.Read more about C Programming Language .

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

#include <stdio.h>

unsigned setbits(unsigned x, int p, int n, unsigned y)
{
return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n))))) | ((y & ~(~0 << n)) << (p + 1 - n));
}

int main(void)
{
unsigned i;
unsigned j;
unsigned k;
int p;
int n;

for(i = 0; i < 30000; i += 511)
{
for(j = 0; j < 1000; j += 37)
{
for(p = 0; p < 16; p++)
{
for(n = 1; n <= p + 1; n++)
{
k = setbits(i, p, n, j);
printf("setbits(%u, %d, %d, %u) = %un", i, p, n, j, k);
}
}
}
}
return 0;
}

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K & R C Programs Exercise 2-5.

K and R C, Solution to Exercise 2-5:
 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.
Write the C function any(s1, s2), which returns the first location in the string s1 where any character from the string s2 occurs, or -1 if s1 contains no characters from s2.The standard library function strpbrk does the same job but returns a pointer to the location.Read more about C Programming Language .

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int any(char s1[], chars2[])
{
int i, j;
for(i=0;s1[i]!='';i++)

for(j=0;s2[[j]!='' ;j++)

if(s1[i]==s2[j])
return i;

return -1;
}
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K & R C Programs Exercise 2-4.

K and R C, Solution to Exercise 2-4:
 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.
Write an alternate version of squeeze(s1,s2) that deletes each character in s1 that matches any character in the string s2.
Compare ,First string with the second string, if any character matched,  delete the matched character of the first string. Read more about C Programming Language .

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/*squeeze: delete each char in s1 which is in s2*/
void squeeze(char s1[], chars2[])
{
int i, j, k;
for(i=k=0;s1[i]!='';i++)
{
for(j=0;s2[[j]!='' && s2[j]!=s1[i];j++)
;
if(s2[j]=='')
s1[k++]=s1[i];
}
s1[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 2-3.

K and R C, Solution to Exercise 2-3:
C Function htoi() which converts a string of hexa decimal digits into its equivalent integer value. 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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***********************************************************/

#define YES 1
#define NO 0
//HTOI: CONVERT HEXADECIMAL STRING S TO INTEGER
int htoi(char s[])
{
int hexdigit, i, inhex, n,i=0;
if(s[i]=='0'){
++i;
if(s[i]=='x' || s[i]=='x')
++i;
}
n=0;
inhex=YES;
for(;inhex==YES;++i){
if(s[i]>='0' && s[i]<='9')
hexdigit=s[i]-'0';
else if(s[i]>='a' && s[i]<='f')
hexdigit=s[i]-'a'+10;
else if (s[i]>='A' && s[i]<='F')
hexdigit=s[i]-'A'+10;
else
inhex=NO;
if(inhex==YES)
n=16*n+hexdigit;
}
return n;
}


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