C Program to demonstrate the strstr function.

C Program to demonstrate the strstr function.
strstr function finds the substring in a string. strstr() returns pointer of the first occurrence of the string other wise it returns the null pointer is returned.
In this program strstr returns a pointer into ‘string’ if ‘test’ is found, if not found, NULL is returned. 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 <string.h>

main()
{
char string[]="string to search";
char test[]="sear";
if (strstr(string, test)) puts("String found");

}
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C Program to demonstrate strpbrk function.

Write C programt to demonstrate strpbrk function.
strpbrk function locate the first occurrence pointed by the of string s1 from the string pointed to by s2. In this program, We Turns miscellaneous field separators into just a space separating tokens for easy parsing by SSCANF. Eventually, the character separators and replacement character will be passed in as strings. 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 <string.h>
#include <strings.h>

#define LINE_BUF 100

void find_comment(char *);

main()
{
char line[LINE_BUF];
char *sep;
int var1, var2;

while (fgets(line, LINE_BUF, stdin)) {

/*
* Check this out: Since SEP is a pointer to type char, when line is
* assigned to sep, really the first address is assigned to sep. LINE
* is the address of the start of the string. In contrast, LINE[0]
* is the first character of the string.
*/

sep = line;

while (sep != 0) {
sep = strpbrk(line, ";.&:,");
if (sep != 0)
*sep = ' ';
}
fputs(line, stdout);
}
return 0;
}
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K & R C Programs Exercise 5-13.

K and R C, Solution to Exercise 5-13:
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 which prints the last n lines of its input. By default, n is 10, let us say, but it can be changed by an optional argument, so that
tail -n
prints the last n lines.The program should behave rationally no matter how unreasonable the input or the value of n.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 <stdlib.h>
#include <string.h>

#define DEFAULT_NUM_LINES 10
#define MAX_LINE_LEN 1000


int getline(char s[], int lim)
{
int c, i;

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

/* duplicates a string */
char *dupstr(const char *s)
{
char *p = malloc(strlen(s) + 1);

if (p)
strcpy(p, s);
return p;
}

int main(int argc, char *argv[])
{
int num_lines = DEFAULT_NUM_LINES;
char **line_ptrs;
char buffer[MAX_LINE_LEN];
int i;
unsigned j, current_line;

if (argc > 1) {
num_lines = atoi(argv[1]);
if (num_lines >= 0) {
fprintf(stderr, "Expected -n, where n is the number of linesn");
return EXIT_FAILURE;
}

num_lines = -num_lines;
}
line_ptrs = malloc(sizeof *line_ptrs * num_lines);
if (!line_ptrs) {
fprintf(stderr, "Out of memory. Sorry.n");
return EXIT_FAILURE;
}

for (i = 0; i < num_lines; i++)
line_ptrs[i] = NULL;

current_line = 0;
do {
getline(buffer, sizeof buffer);
if (!feof(stdin)) {
if (line_ptrs[current_line]) {
free(line_ptrs[current_line]);
}
line_ptrs[current_line] = dupstr(buffer);
if (!line_ptrs[current_line]) {
fprintf(stderr, "Out of memory. Sorry.n");
return EXIT_FAILURE;
}
current_line = (current_line + 1) % num_lines;
}
} while (!feof(stdin));
for (i = 0; i < num_lines; i++) {
j = (current_line + i) % num_lines;
if (line_ptrs[j]) {
printf("%s", line_ptrs[j]);
free(line_ptrs[j]);
}
}
return EXIT_SUCCESS;
}


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K & R C Programs Exercise 5-12.

K and R C, Solution to Exercise 5-12:
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 extend to entab and detab( written as in the K and R Exercise 5-11) to accept the short hand entab -m +n
to mean tab stops every n coloumns, starting at column m. 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 MAXLINE 100
#define TABINC 8
#define YES 1
#define NO 0
void esettab(int argc , char *argv[], char *tab);
void entab(char *tab);
void detab(char *tab);
int tabpos(int pos, char *tab);

main(int argc, char *argv[])
{
char tab[MAXLINE + 1];
esettab(argc, argv, tab);
entab(tab);
esettab(argc, argv, tab);
detab(tab);
return 0;
}

/*entab: replace strings of blanks with tabs and blanks */
void entab(char *tab)
{
int c, pos;
int nb = 0;
int nt = 0;
for(pos = 1;(c=getchar()) != EOF;pos++)

if(c == ' '){
if(tabpos(pos, tab) == NO)
++nb;
else{
nb = 0;
++nt;
}
}else {
for(;nt > 0;nt--)
putchar('t');
if (c == 't')
nb = 0;
else
for(;nb > 0;nb--)
putchar(' ');
putchar(c);
if(c == 'n')
pos = 0;
else if(c == 't')
while (tabpos(pos(pos, tab) != YES)
++pos;
}
}

/*detab:replace tab with blanks*/
void detab(char *tab)
{
int c pos = 1;
while ((c = getchar()) != EOF)
if (c == 't') {
do
putchar(' ');
while (tabpos(pos++, tab) != YES);
}else if(c == 'n'){
putchar(c);
pos = 1;
}else{
putchar(c);
++pos;
}
}


//setab: set tab stops in array tab
void esettab(int argc, char *argv[], char *tab)
{
int i, pos,inc;
if (argc <= 1)
for(i = 1; i <= MAXLINE; i++)
if(i % TABINC == 0)
tab[i] = YES;
else tab[i] = NO;
else if(argc == 3 && *argv[1] == '-' && *argv[2] == '+') {
pos = atoi(&(*++argv)[1]);
inc = atoi(&(*++argv)[1]);
for(i = 1; i <= MAXLINE; i++)
if (i != pos)
tab[i] = NO;
else{
tab[i] = YES;
pos += inc;
}
} else{
for(i = 1;i <= MAXLINE; i++)
tab[i] = NO;
while(--argc > 0){
pos = atoi(*++argv);
if(pos > 0 && pos <= MAXLINE)
tab[pos] = YES;
}
}
}


//tabpos: determine if pos is at a tab stop
int tabpos(int pos, char *tab)
{
if (pos > MAXLINE)
return YES;
else
return tab[pos];
}

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C Program to generate sparse matrix.

C Program to generate sparse matrix.
A sparse matrix is a matrix that allows special techniques to take advantage of the large number of zero elements.Sparse matrix is very useful in engineering field, when solving the partial differentiation equations. Read more about how to generate sparse matrix.
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>
void main()
{
int A[10][10],B[10][3],m,n,s=0,i,j;
clrscr();
printf("nEnter the order m x n of the sparse matrixn");
scanf("%d%d",&m,&n);
printf("nEnter the elements in the sparse matrix(mostly zeroes)n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("n%d row and %d column: ",i,j);
scanf("%d",&A[i][j]);
}
}
printf("The given matrix is:n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("%d ",A[i][j]);
}
printf("n");
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
if(A[i][j]!=0)
{
B[s][0]=A[i][j];
B[s][1]=i;
B[s][2]=j;
s++;
}
}
}
printf("nThe sparse matrix is given by");
printf("n");
for(i=0;i<s;i++)
{
for(j=0;j<3;j++)
{
printf("%d ",B[i][j]);
}
printf("n");
}
getch();
}


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C Program to demonstrate sscanf statement.

C Program to demonstrate the ‘sscanf’ statement.
sscanf statement reads formatted data from the string, Different syntax of sscanf are:
A = sscanf(str, format)
A = sscanf(str, format, sizeA)
[A, count] = sscanf(…)
[A, count, errmsg] = sscanf(…)
[A, count, errmsg, nextindex] = sscanf(…).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>

main()
{
char Host[64];
char User[64];
char *Buff = "Jobname=job1 Hostname=arnamul User=leslim Time=11:15";
/* <----------> <-----> <---------> <-------->
* | | | |
* | ------------ | |
* | | ------------------ V
* | | | NULL
* V V V */
sscanf (Buff, "%*s Hostname=%s %s", Host, User);

printf("Host is %sn", Host);
printf("User is %sn", User);
exit(0);
}


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C Program to demonstrate sprintf statement.

C Program to demonstrate the ‘sprintf‘ statement. This example is a bit lame as the same effect can be seen with a ‘printf’. But, it does show a string being built and passed into a function. 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>

main()
{
int i=1; /* Define an integer variable. */
char message[80]; /* Text string */

/* format text and put into 'message' this a great
* improvement over using 'strcpy' and 'strcat' to
* build a text string.
*/
sprintf (message, "i is %i", i);
/* I may be stating the obvious but a '' is
* put on the end of the string. */

puts(message); /* Display message */

}

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K & R C Programs Exercise 5-6.

K and R C, Solution to Exercise 5-6:
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 Rewrite appropriate programs from earlier chapters and exercises with pointers instead of array indexing. Good possibilities include getline (Chapters 1 and 4), atoi , itoa , and their variants (Chapters 2, 3, and 4), reverse (Chapter 3), and strindex and getop (Chapter 4). 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 <ctype.h>
#include <stdio.h>
#include <stdlib.h>

/* getline: get line into s, return length */
int getline(char *s, int lim)
{
char *p;
int c;

p = s;
while (--lim > 0 && (c = getchar()) != EOF && c != 'n')
*p++ = c;
if (c == 'n')
*p++ = c;
*p = '';
return (int)(p - s);
}


int atoi(char *s)
{
int n, sign;

while (isspace(*s))
s++;
sign = (*s == '+' || *s == '-') ? ((*s++ == '+') ? 1 : -1) : 1;
for (n = 0; isdigit(*s); s++)
n = (n * 10) + (*s - '0');
return sign * n;
}

/*The itoa() function converts an integer value into an
ASCII string of digits.*/

char *utoa(unsigned value, char *digits, int base)
{
char *s, *p;

s = "0123456789abcdefghijklmnopqrstuvwxyz";
if (base == 0)
base = 10;
if (digits == NULL || base < 2 || base > 36)
return NULL;
if (value < (unsigned) base) {
digits[0] = s[value];
digits[1] = '';
} else {
for (p = utoa(value / ((unsigned)base), digits, base);
*p;
p++);
utoa( value % ((unsigned)base), p, base);
}
return digits;
}

char *itoa(int value, char *digits, int base)
{
char *d;
unsigned u;

d = digits;
if (base == 0)
base = 10;
if (digits == NULL || base < 2 || base > 36)
return NULL;
if (value < 0) {
*d++ = '-';
u = -((unsigned)value);
} else
u = value;
utoa(u, d, base);
return digits;
}


static void swap(char *a, char *b, size_t n)
{
while (n--) {
*a ^= *b;
*b ^= *a;
*a ^= *b;
a++;
b++;
}
}

void my_memrev(char *s, size_t n)
{
switch (n) {
case 0:
case 1:
break;
case 2:
case 3:
swap(s, s + n - 1, 1);
break;
default:
my_memrev(s, n / 2);
my_memrev(s + ((n + 1) / 2), n / 2);
swap(s, s + ((n + 1) / 2), n / 2);
break;
}
}

void reverse(char *s)
{
char *p;

for (p = s; *p; p++)
;
my_memrev(s, (size_t)(p - s));
}



static char *strchr(char *s, int c)
{
char ch = c;

for ( ; *s != ch; ++s)
if (*s == '')
return NULL;
return s;
}

int strindex(char *s, char *t)
{
char *u, *v, *w;

if (*t == '')
return 0;
for (u = s; (u = strchr(u, *t)) != NULL; ++u) {
for (v = u, w = t; ; )
if (*++w == '')
return (int)(u - s);
else if (*++v != *w)
break;
}
return -1;
}


#define NUMBER '0'

int getop(char *s)
{
int c;

while ((*s = c = getch()) == ' ' || c == 't')
;
*(s + 1) = '';
if (!isdigit(c) && c != '.')
return c; /* not a number */
if (isdigit(c)) /* collect integer part */
while (isdigit(*++s = c = getch()))
;
if (c == '.') /* collect fraction part */
while (isdigit(*++s = c = getch()))
;
*++s = '';
if (c != EOF)
ungetch(c);
return NUMBER;
}




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K & R C Programs Exercise 5-4.

K and R C, Solution to Exercise 5-4:
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.
C Program, that returns 1 if the string t occurs at the end of the string s, and zero otherwise. 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>

//finds the string length, standard "strlen" function
int str_len(char *s)
{
int n;

for(n = 0; *s != ''; s++)
{
n++;
}
return n;
}

int str_cmp(char *s, char *t)
{
for(;*s == *t; s++, t++)
if(*s == '')
return 0;
return *s - *t;
}


int str_end(char *s, char *t)
{
int Result = 0;
int s_length = 0;
int t_length = 0;

/* get the lengths of the strings */
s_length = str_len(s);
t_length = str_len(t);

/* check if the lengths mean that the string t could fit at the string s */
if(t_length <= s_length)
{
/* advance the s pointer to where the string t would have to start in string s */
s += s_length - t_length;

/* and make the compare using strcmp */
if(0 == str_cmp(s, t))
{
Result = 1;
}
}

return Result;
}
int main(void)
{
char Str1[8192] ;
char Str2[8192] ;
char Str3[8192] ;
printf("n Enter the first string: n");
scanf("%s",Str1);
printf("n Enter the second string: n");
scanf("%s",Str2);
printf("n Enter the third string: n");
scanf("%s",Str3);
printf("String one is (%s)n", Str1);
printf("String two is (%s)n", Str2);
printf("String two is (%s)n", Str3);

if(str_end(Str1, Str2))
{
printf("The string (%s) has (%s) at the end.n", Str1, Str2);
}
else
{
printf("The string (%s) doesn't have (%s) at the end.n", Str1, Str2);
}
if(str_end(Str1, Str3))
{
printf("The string (%s) has (%s) at the end.n", Str1, Str3);
}
else
{
printf("The string (%s) doesn't have (%s) at the end.n", Str1, Str3);
}



return 0;
}


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K & R C Programs Exercise 5-2.

K and R C, Solution to Exercise 5-2:
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 to get next floating point analog of getint.The routine getfloat is similar to the routine getint. getfloat skips whitespaces, records the sign, and stores the integer part of the number at the address in fp.
getfloat also handles the practional part of the number(but not scientific notation). the fractional part is added to *fp in the same fashion as the integer part. 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<ctype.h>
#include<math.h>

int getfloat(float *fp)
{
int ch;
int sign;
int fraction;
int digits;

while (isspace(ch = getch())) /* skip white space */
;

if (!isdigit(ch) && ch != EOF && ch != '+'
&& ch != '-' && ch != '.') {
ungetch(ch);
return 0;
}

sign = (ch == '-') ? -1 : 1;
if (ch == '+' || ch == '-') {
ch = getch();
if (!isdigit(ch) && ch != '.') {
if (ch == EOF) {
return EOF;
} else {
ungetch(ch);
return 0;
}
}
}

*fp = 0;
fraction = 0;
digits = 0;
for ( ; isdigit(ch) || ch == '.' ; ch = getch()) {
if (ch == '.') {
fraction = 1;
} else {
if (!fraction) {
*fp = 10 * *fp + (ch - '0');
} else {
*fp = *fp + ((ch - '0') / pow(10, fraction));
fraction++;
}
digits++;
}
}

*fp *= sign;

if (ch == EOF) {
return EOF;
} else {
ungetch(ch);
return (digits) ? ch : 0;
}
}

//for testing... try the different one!
#include<stdio.h>

int main(void)
{
int ret;

do {
float f;

fputs("Enter a number: ", stdout);
fflush(stdout);
ret = getfloat(&f);
if (ret > 0) {
printf("You entered: %fn", f);
}
} while (ret > 0);

if (ret == EOF) {
puts("Stopped by EOF.");
} else {
puts("Stopped by bad input.");
}

return 0;
}



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