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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K & R C Programs Exercise 5-10.

K and R C, Solution to Exercise 5-10:
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 Program to evaluate reverse Polish expression from the command line, where each operator or operand is a separate argument.
For example:expr 2 3 4 + * evaluates 2 X (3 + 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>

#define MAX 1000

double stack[MAX];
int stack_height = 0;

void panic(const char *msg) {
fprintf(stderr, "%sn", msg);
exit(EXIT_FAILURE);
}

void push(double value) {
if (stack_height == MAX)
panic("stack is too high!");
stack[stack_height] = value;
++stack_height;
}

double pop(void) {
if (stack_height == 0)
panic("stack is empty!");
return stack[--stack_height];
}

int main(int argc, char **argv) {
int i;
double value;

for (i = 1; i < argc; ++i) {
switch (argv[i][0]) {
case '':
panic("empty command line argument");
break;
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
push(atof(argv[i]));
break;
case '+':
push(pop() + pop());
break;
case '-':
value = pop();
push(pop() - value);
break;
case '*':
push(pop() * pop());
break;
case '/':
value = pop();
push(pop() / value);
break;
default:
panic("unknown operator");
break;
}
}

printf("%gn", pop());
return 0;
}

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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 .

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

#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 .

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

#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 5-9.

K and R C, Solution to Exercise 5-9:
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.
C program to , Rewrite the routines of K & R C Programs Exercise 5-8 day_of_year and month_day with pointers instead of indexing.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
***********************************************************/

#include <stdio.h>

static char daytab[2][13] = {
{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
};



int day_of_year(int year, int month, int day)
{
int i, leap;
char *p;

leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;

/* Set `p' to point at first month in the correct row. */
p = &daytab[leap][1];

/* Move `p' along the row, to each successive month. */
for (i = 1; i < month; i++) {
day += *p;
++p;
}
return day;
}

void month_day(int year, int yearday, int *pmonth, int *pday)
{
int i, leap;
char *p;

leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;
p = &daytab[leap][1];
for (i = 1; yearday > *p; i++) {
yearday -= *p;
++p;
}
*pmonth = i;
*pday = yearday;
}


int main(void)
{
int year, month, day, yearday;

year = 2012;
month = 7;
day = 9;
printf("The date is: %d-%02d-%02dn", year, month, day);
printf("day_of_year: %dn", day_of_year(year, month, day));



yearday = 61; /* 2012-03-01 */
printf("Yearday is %dn", yearday);
month_day(year, yearday, &month, &day);
printf("month_day_pointer: %d %dn", month, day);

return 0;
}


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K & R C Programs Exercise 5-8.

K and R C, Solution to Exercise 5-8:
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.
There is no error-checking in the function day_of_year or month_day. Remedy this defect.
In day_of_year we check for reasonable values in month and day. If month is less than one or greater than twelve, day_of_year returns -1. If day is less than one or day exceeds the number of days for the month, the function returns -1.
In month_day we first check for negative year. You may want to add this kind of check in day_of_year also. Then we proceed to decrement year day while we check that the month does not exceed 12. 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>

static char daytab[2][13] = {
{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
};

/* day_of_year: set day of year from month & day */
int day_of_year(int year, int month, int day)
{
int i, leap;

if (year < 1752 || month < 1 || month > 12 || day < 1)
return -1;

leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;
if (day > daytab[leap][month])
return -1;

for (i = 1; i < month; i++)
day += daytab[leap][i];
return day;
}

/* month_day: set month, day from day of year */
int month_day(int year, int yearday, int *pmonth, int *pday)
{
int i, leap;

if (year < 1752 || yearday < 1)
return -1;

leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;
if ((leap && yearday > 366) || (!leap && yearday > 365))
return -1;

for (i = 1; yearday > daytab[leap][i]; i++)
yearday -= daytab[leap][i];
*pmonth = i;
*pday = yearday;

return 0;
}


/* main: test day_of_year and month_day */
int main(void)
{
int year, month, day, yearday;

for (year = 1970; year <= 2000; ++year) {
for (yearday = 1; yearday < 366; ++yearday) {
if (month_day(year, yearday, &month, &day) == -1) {
printf("month_day failed: %d %dn",
year, yearday);
} else if (day_of_year(year, month, day) != yearday) {
printf("bad result: %d %dn", year, yearday);
printf("month = %d, day = %dn", month, day);
}
}
}

return 0;
}

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K & R C Programs Exercise 5-7.

K and R C, Solution to Exercise 5-7:
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 Program to rewrite the readline function to store lines in the arraysupplied by main, rather than calling alloc to maintain storage. How much faster is the program?
The readlile function is slightly faster than the original version in the book K and R C Program page 109.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]
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* and browse!
*
* Happy Coding
***********************************************************/


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

#define TRUE 1
#define FALSE 0

#define MAXLINES 5000 /* maximum number of lines */
#define MAXLEN 1000 /* maximum length of a line */
char *lineptr[MAXLINES];
char lines[MAXLINES][MAXLEN];


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;
}


int readlines(char *lineptr[], int maxlines)
{
int len, nlines;
char *p, line[MAXLEN];

nlines = 0;
while ((len = getline(line, MAXLEN)) > 0)
if (nlines >= maxlines || (p = malloc(len)) == NULL)
return -1;
else {
line[len - 1] = ''; /* delete the newline */
strcpy(p, line);
lineptr[nlines++] = p;
}
return nlines;
}

int readlines2(char lines[][MAXLEN], int maxlines)
{
int len, nlines;

nlines = 0;
while ((len = getline(lines[nlines], MAXLEN)) > 0)
if (nlines >= maxlines)
return -1;
else
lines[nlines++][len - 1] = '';
return nlines;
}

int main(int argc, char *argv[])
{

readlines2(lines, MAXLINES);

if (argc > 1 && *argv[1] == '2') {
puts("readlines2()");
readlines2(lines, MAXLINES);
} else {
puts("readlines()");
readlines(lineptr, MAXLINES);
}

return 0;
}

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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 5-5.

K and R C, Solution to Exercise 5-5:
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.
C Program to write the versions of the library functions strncpy, strncat, and strncmp, which operate on at most the first n characters of their argument strings.
For example, strncpy(s, t, n) copies at most n characters of t to s. 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
***********************************************************/

/*strncpy: copy n characters from t to s*/
void strncpy(char *s, char *t, int n)
{
while(*t && n--> 0)
*s++ = *t++;
while( n--> 0)
*s++ = '';
}

/*strncat: concatenate n characters of t to end of s*/
void strncat(char *s, char *t, int n)
{
void strncpy(char *s, char *t, int n);
int strlen(char *);
strncpy(s+strlen(s), t, n);
}

/*strncmp: comapre at most n characters of t with s */
int strncmp(char *s, char *t, int n)
{
for( ; *s == *t; s++, t++)
if(*s == '' || --n <= 0)
{
return 0;
}
return *s - *t;
}
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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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