K & R C Programs Exercise 4-2.

K and R C, Solution to Exercise 4-2:
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 extend the atof function to handle scientific notations of the form 5234.73e-12
atof function converts the intial nptr string to the double. atof means ASCII to float. In this program that atof function handles the scientific notations also like 12.e-3.. 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]
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*
* Happy Coding
***********************************************************/

#include <ctype.h>
#include <limits.h>
#include <float.h>
#include <signal.h>
#include <stdio.h>

int my_atof(char *string, double *pnumber) {
/* Convert char string to double data type. */
double retval;
double one_tenth = 0.1;
double ten = 10.0;
double zero = 0.0;
int found_digits = 0;
int is_negative = 0;
char *num;

/* Check pointers. */
if (pnumber == 0) {
return 0;
}
if (string == 0) {
*pnumber = zero;
return 0;
}
retval = zero;

num = string;

/* Advance past white space. */
while (isspace(*num))
num++;

/* Check for sign. */
if (*num == '+')
num++;
else if (*num == '-') {
is_negative = 1;
num++;
}
/* Calculate the integer part. */
while (isdigit(*num)) {
found_digits = 1;
retval *= ten;
retval += *num - '0';
num++;
}

/* Calculate the fractional part. */
if (*num == '.') {
double scale = one_tenth;
num++;
while (isdigit(*num)) {
found_digits = 1;
retval += scale * (*num - '0');
num++;
scale *= one_tenth;
}
}
/* If this is not a number, return error condition. */
if (!found_digits) {
*pnumber = zero;
return 0;
}
/* If all digits of integer & fractional part are 0, return 0.0 */
if (retval == zero) {
*pnumber = zero;
return 1; /* Not an error condition, and no need to
* continue. */
}
/* Process the exponent (if any) */
if ((*num == 'e') || (*num == 'E')) {
int neg_exponent = 0;
int get_out = 0;
long index;
long exponent = 0;
double getting_too_big = DBL_MAX * one_tenth;
double getting_too_small = DBL_MIN * ten;

num++;
if (*num == '+')
num++;
else if (*num == '-') {
num++;
neg_exponent = 1;
}
/* What if the exponent is empty? Return the current result. */
if (!isdigit(*num)) {
if (is_negative)
retval = -retval;

*pnumber = retval;

return (1);
}
/* Convert char exponent to number <= 2 billion. */
while (isdigit(*num) && (exponent < LONG_MAX / 10)) {
exponent *= 10;
exponent += *num - '0';
num++;
}

/* Compensate for the exponent. */
if (neg_exponent) {
for (index = 1; index <= exponent && !get_out; index++)
if (retval < getting_too_small) {
get_out = 1;
retval = DBL_MIN;
} else
retval *= one_tenth;
} else
for (index = 1; index <= exponent && !get_out; index++) {
if (retval > getting_too_big) {
get_out = 1;
retval = DBL_MAX;
} else
retval *= ten;
}
}
if (is_negative)
retval = -retval;

*pnumber = retval;

return (1);
}

double atof(char *s) {
double d = 0.0;
if (!my_atof(s, &d)) {
#ifdef DEBUG
fputs("Error converting string in [sic] atof()", stderr);
#endif
raise(SIGFPE);
}
return d;
}

#ifdef UNIT_TEST
char *strings[] = {
"1.0e43",
"999.999",
"123.456e-9",
"-1.2e-3",
"1.2e-3",
"-1.2E3",
"-1.2e03",
"cat",
"",
0
};
int main(void)
{
int i = 0;
for (; *strings[i]; i++)
printf("atof(%s) = %gn", strings[i], atof(strings[i]));
return 0;
}
#endif

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K & R C Programs Exercise 4-1.

K and R C, Solution to Exercise 4-1:
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 which returns the position of the occurrence of sub string t in string s, or -1 if there is none.

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

#include<stdio.h>
#include<conio.h>
void main()
{
int flag=0;
char str[80],search[10];
puts("Enter a string:");
gets(str);

puts("Enter search substring:");
gets(search);
flag=strindex(str, search);
if (flag == -1)
{
printf("SEARCH UNSUCCESSFUL! AND POSITION IS:%d",flag);
}
else{
printf("SEARCH SUCCESSFUL! AND POSITION IS:%d",flag);
}
getch();
}

//strindex: returns the right most index of t in s, -1 if none*/
int strindex(char s[], char t[])
{
int k,i,j,pos;
pos = -1;
for(i=0;s[i] != ''; i++)
{
for(j=i, k = 0; t[k] != '' && s[j] == t[k]; j++, k++)
;
if (k > 0 && t[k] == '')
pos=i;
}
return pos;
}
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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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***********************************************************/

#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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K & R C Programs Exercise 3-4.

K and R C, Solution to Exercise 3-4:
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 represent the binary numbers in a negative sign.There are a number of ways of representing signed integers in binary, for example, signed-magnitude, excess-M, one’s complement and two’s complement. In this program , we handle the problem that is, the value of n equal to -(2 to the power (wordsize – 1))(C Programming Language (2nd Edition) by Brian W.Kernighan & Dennis M.Richie page no 64) by changing the (n /= 10) > 0 to (n /= 10) != 0. Read more about C Programming Language .

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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[]);
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 - modified.
void itoa(int n, char s[])
{
void int i, sign;
void everse(char s[]);
sign = n;
i = 0;
do{
s[i++] = abs(n%10) + '0';
}while((n/=10)!=0);
if(sign < 0)
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 3-3.

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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* contact [email protected]
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* and browse!
*
* 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 3-2.

K and R C, Solution to Exercise 3-2:
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 converts the characters like newline and tab into visible escape sequences like n and t as it copies the string t to s. Use a Switch case, and Write the function for the other direction as well, converting escape sequences into the real characters. Read more about C Programming Language .

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

#include <stdio.h>

void escape(char * s, char * t);
void unescape(char * s, char * t);

int main(void) {
char text1[50] = "aHello,ntWorld! Mistakeeb was "Extra 'e'"!n";
char text2[51];

printf("Original string:n%sn", text1);

escape(text2, text1);
printf("Escaped string:n%sn", text2);

unescape(text1, text2);
printf("Unescaped string:n%sn", text1);

return 0;
}

/*escape: expand newline and tab and others tabs into visible sequences using switch*/

void escape(char * s, char * t) {
int i, j;
i = j = 0;

while ( t[i] ) {



switch( t[i] ) {
case 'n':
s[j++] = '\';
s[j] = 'n';
break;

case 't':
s[j++] = '\';
s[j] = 't';
break;

case 'a':
s[j++] = '\';
s[j] = 'a';
break;

case 'b':
s[j++] = '\';
s[j] = 'b';
break;

case 'f':
s[j++] = '\';
s[j] = 'f';
break;

case 'r':
s[j++] = '\';
s[j] = 'r';
break;

case 'v':
s[j++] = '\';
s[j] = 'v';
break;

case '\':
s[j++] = '\';
s[j] = '\';
break;

case '"':
s[j++] = '\';
s[j] = '"';
break;

default:



s[j] = t[i];
break;
}
++i;
++j;
}
s[j] = t[i];
}


/* unescape: convert escape sequences into real charecters while copying the string t to s */

void unescape(char * s, char * t) {
int i, j;
i = j = 0;

while ( t[i] ) {
switch ( t[i] ) {
case '\':


switch( t[++i] ) {
case 'n':
s[j] = 'n';
break;

case 't':
s[j] = 't';
break;

case 'a':
s[j] = 'a';
break;

case 'b':
s[j] = 'b';
break;

case 'f':
s[j] = 'f';
break;

case 'r':
s[j] = 'r';
break;

case 'v':
s[j] = 'v';
break;

case '\':
s[j] = '\';
break;

case '"':
s[j] = '"';
break;

default:



s[j++] = '\';
s[j] = t[i];
}
break;

default:



s[j] = t[i];
}
++i;
++j;
}
s[j] = t[i];
}


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K & R C Programs Exercise 3-1.

K and R C, Solution to Exercise 3-1:
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 implement binary Search. Binary search technique is simple searching technique which can be applied if the items to be compared are either in ascending order or descending order. The general idea used in binary search is similar to the way we search for the telephone number of a person in the telephone directory. Binary search is the divide and conquer strategy. Read more about C Programming Language .

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

/*C Program for Binary search */

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

int n, a[30], item, i, j, mid, top, bottom;
printf("Enter how many elements you want:n");
scanf("%d", &n);
printf("Enter the %d elements in ascending ordern", n);
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
printf("nEnter the item to searchn");
scanf("%d", &item);
bottom = 1;
top = n;

do {
mid = (bottom + top) / 2;
if (item < a[mid])
top = mid - 1;
else if (item > a[mid])
bottom = mid + 1;
} while (item != a[mid] && bottom <= top);

if (item == a[mid]) {
printf("Binary search successfull!!n");
printf("n %d found in position: %dn", item, mid + 1);
} else {
printf("n Search failedn %d not foundn", item);
}
return 0;
}
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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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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
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* and browse!
*
* Happy Coding
***********************************************************/

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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