C program to generate and print prime numbers in a given range.

Write a C program to generate and print prime numbers in a given range. Also print the number of prime numbers.
Prime number is a whole number and greater than 1, which is divisible by one or itself.
Example: 2, 3, 5, 7, 11, 13………… Read more about C Programming Language .

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


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

void main()
{
int M, N, i, j, flag, temp, count = 0;

clrscr();

printf("Enter the value of M and Nn");
scanf("%d %d", &M,&N);

if(N < 2)
{
printf("There are no primes upto %dn", N);
exit(0);
}
printf("Prime numbers aren");
temp = M;

if ( M % 2 == 0)
{
M++;
}
for (i=M; i<=N; i=i+2)
{
flag = 0;

for (j=2; j<=i/2; j++)
{
if( (i%j) == 0)
{
flag = 1;
break;
}
}
if(flag == 0)
{
printf("%dn",i);
count++;
}
}
printf("Number of primes between %d and %d = %dn",temp,N,count);
}



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C Program to copy and concatenate strings without using standard functions.

C Program to copy and concatenate strings without using standard functions.
In this program , we copy one string from another, and without using the standard library function strcpy from string.h .
Here we appends the one string to another without using the strcat function. Read more about C Programming Language .

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



#include <stdio.h>

void newStrCpy(char *,char *);
void newStrCat(char *,char *);

void main()
{
char str1[80],str2[80];
int opn;
do
{
printf("Press 1- NewStrCpy t 2- NewStrCatt 3- Exitt Your Option?");
scanf("%d",&opn);
switch(opn)
{
case 1: flushall();
printf("n Read the Source String n");
gets(str1);
newStrCpy(str2,str1);
printf(" Copied String: %sn",str2);
break;
case 2: flushall();
printf(" Read the First String n");
gets(str1);
printf(" Read the Second String n");
gets(str2);
newStrCat(str1,str2);
printf("Concatenated String: n");
puts(str1);
break;
case 3: printf(" Exit!! Press a key . . .");
getch();
break;
default: printf(" Invalid Option!!! Try again !!!n");
break;
}
}while(opn != 3);
} /* End of main() */

void newStrCpy(char *d,char *s)
{
while( (*d++ = *s++));
}
void newStrCat(char *s, char *t)
{
while(*s)
s++;
while(*s++ = *t++);
}

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C Program to convert from infix expression into prefix expression

C Program to convert from infix expression into prefix expression.
We use the infix expressions for the mathematical expressions in a program, these expressions will converted into equivalent machine instructions by the compiler using stacks.
Using stacks we can efficiently convert the expressions from infix to postfix, infix to prefix, postfix to infix, and postfix to prefix.
Example: infix to prefix:
infix: x^y^z-m+n+p/q,
postfix: ++-^x^yzmn/pq. Read more about C Programming Language.

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* and browse!
*
* Happy Coding
***********************************************************/


#include<stdio.h>

#include<conio.h>

//Stack precedence function

int F(char symbol)

{

switch(symbol)

{

case ‘+’ :

case ‘-‘ :

return 1;

case ‘*’:

case ‘^’:

return 6:

case ‘)’:

return 0:;

case ‘#’:

return -1;

default:

return 8;

}

}

//Input precedence function

int G(char symbol)

{

switch(symbol)

{

case ‘+’ :

case ‘-‘ :

return 2;

case ‘*’:

return 4;

case ‘^’:

return 5:

case ‘(‘:

return 0;

case ‘)’:

return 9:;

case ‘#’:

return -1;

default:

return 7;

}

}

Void infix_prefix(char infix[], char prefix[])

{

int top, j, i;

char symbol, s[40];

top = -1;

s[++top] = ‘#’;

J = 0;

strrev(infix);

for(i = 0;i < strlen(infix); i++)

{

symbol= infix[i];

while(F(s[top]) > G(symbol))

{

prefix[j] = s[top--];

j++;

}

if(F(s[top]) != G(symbol))

s[++top] = symbol;

else

top--;

}

while(s[top != ‘#’)

{

prefix[j++] = s[top--];

}

prefix[j] = ‘’;

strrev(prefix);

}

void main()

{

char infix[20];

char prefix[20];

printf(“/nEnter a valid infix expressionn”);

scanf(“%s”,infix);

infix_prefix(infix, prefix);

printf(“nnThe prefix expression isn”);

printf(“%sn”,prefix);

}


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C Program to demonstrate isalpha, isdigit, is space.

C Program to demonstrate the following functions:isalpha, isdigit, isspace.The same principles apply to isalnum, iscntrl, isgraph,islower, isprint, ispunct, isupper, isxdigit.
In the standard library ctype.h all the above functions are declared.
All the subroutines mentioned here, returns non zero(true), If the checked argument is true for the respective subroutines. Read more about C Programming Language .

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* and browse!
*
* Happy Coding
***********************************************************/

#include <stdio.h> /* printf */
#include <ctype.h> /* isalpha isdigit isspace etc */

#define FALSE 0
#define TRUE 1

/* function declarations */
int char_type(char);

main()
{
char ch;
/* get a character from the keyboard */
printf(" Please enter a charcater => ");
ch = getc(stdin);

char_type(ch); /* Figure out the character type */


}

//char_type:decides the character type
int char_type(char ch)
{
/* returns non zero if A-Z or a-z */
if ( isalpha(ch) != FALSE)
printf("%c is an Alpha character.n",ch);

/* returns non zero if 0-9 */
if ( isdigit(ch) != FALSE)
printf("%c is a numeric character.n",ch);

/* returns non zero if a space, CR, Tab, NL FF */
if ( isspace(ch) != FALSE)
printf("%c is white spacen", ch);

}

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K & R C Programs Exercise 7-9.

K and R C, Solution to Exercise 7-9:
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.
Write a C Functions like isupper() can be implemented to save space or to save time. Explore both possibilities. Read more about C Programming Language .
isupper: return 1 (true) if c is an upper case letter
Normal C function:

int  isupper(char c)
{
if (c >= 'A' && c <= 'Z')
return 1;
else
return 0;
}

This simple code tests the character is upper or lower , If the character within the range of ASCII upper case letters it returns 1, otherwise 0.
To save space or to save time using the macros is the best possibility
C Code:

#define isupper(c)  ((c) > = 'A' && (c) <= 'Z') ? 1:0

Macro version of isupper is more efficient because, there is no overhead of the function call and it uses more space because the macro is expanded in line every time it is invoked.

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K & R C Programs Exercise 7-8.

K and R C, Solution to Exercise 7-8:
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 to print a set of files, starting each new one on a new page, with a title and running page for each file 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
***********************************************************/

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

#define MAXBOT 3
#define MAXHDR 5
#define MAXLINE 100
#define MAXPAGE 66

/*print: print files - each new one on a new page */
main(int argc, char *argv[])
{

FILE *fp;
void fileprint(FILE *fp, char *fname);

if(argc == 1)
fileprint(stdin," ");
else
if((fp = fopen(*++argv,"r")) == NULL) {
fprintf(stderr,"find:can't open %sn",*argv);
exit(1);
} else {

fileprint(fp, *argv);
fclose(fp);
}
return 0;
}


//fileprint: print file name
void fileprint(FILE *fp, char *fname)
{
int lineno, pageno = 1;
char line[MAXLINE];
int heading(char *fname, int pageno);
lineno = heading(fname, pageno++);
while(fgets(line, MAXLINE, fp) != NULL) {
if(lineno == 1) {
fprintf(stdout,"f");
lineno = heading(fname, pageno++);
}
fputs(line, stdout);
if(++lineno > MAXPAGE - MAXBOT)
lineno = 1;
}
fprintf(stdout,"f");
}

//heading: put heading and enough blank lines
int heading(char *fname, int pageno)
{
int ln = 3;
fprintf(stdout,"nn");
fprintf(stdout,"%s page %dn", fname, pageno);
while(ln++ < MAXHDR)
fprintf(stdout,"n");
return ln;
}
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K & R C Programs Exercise 7-7.

K and R C, Solution to Exercise 7-7:
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 modify the pattern-finding program of chapter 5(C Programming Language  2nd Edition, page no 117.) to take its input from a set of named files or, if no files are named as arguments, from the standard input. Should the file name be printed when a matching line is found.Read more about C Programming Language .

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* contact [email protected]
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* and browse!
*
* Happy Coding
***********************************************************/

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

#define MAXLINE 1000

/*print lines that match pattern from 1st argument */
main(int argc, char *argv[])
{
char pattern[MAXLINE];
int c, excpet = 0,number = 0;
FILE *fp;
void fpat(FILE *fp, char *fname, char *pattern, int except, int number);

while(--argc > 0 && (*++argv)[0] == '-')
while(c = *++argv[0])
switch(c) {
case 'x':
except = 1;
break;
case 'n':
number = 1;
break;
default:
printf("find:illigal option %cn",c);
argc = 0;
break;
}
if(argc >= 1)
strcpy(pattern, *argv);
else{
printf("Usage:find[-x] [-n] pattern [file....]n");
exit(1);
}
if(argc == 1)
fpat(stdin,"",pattern,except,number);
else
while(--argc > 0)
if((fp = fopen(*++argv,"r")) == NULL) {
fprintf(stderr,"find:can't open %sn",*argv);
exit(1);
} else {
fpat(fp, *argv, pattern,except,number);
fclose(fp);}
return 0;
}

/*fpat: find pattern*/
void fpat(FILE *fp, char *fname, char *pattern, int except, int number)
{
char line[MAXLINE];
long loneno = 0;

while(fgets(line, MAXLINE, fp) != NULL){
++lineno;
if((strstr(line,pattern) != NULL) !=except) {
if(*fname)
printf("%s -",fname);
if(number)
printf("%ld: ",lineno);
printf("%s",line);
}
}
}



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K & R C Programs Exercise 7-6.

K and R C, Solution to Exercise 7-6:
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.
Write a C Program to compare two files, printing the first line where they differ.
Read more about C Programming Language .

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* contact [email protected]
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*
* Happy Coding
***********************************************************/

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

#define MAXLINE 100
/*comp: compare two file, printing the first line where they differ*/
main(int argc, char *argv[])
{
FILE *fp1, *fp2;
void filecomp(FILE *fp1, FILE *fp2);
if(argc != 3){
fprintf(stderr,"comp:need two file namesn");
exit(1);
} else {
if((fp1 = fopen(*++argv, "r")) == NULL) {
fprintf(stderr, "comp:can't open %sn",*argv);
exit(1);
} else if((fp2 = fopen(*++argv, "r")) == NULL) {
fprintf(stderr, "comp:can't open %sn",*argv);
exit(1);
}else {
filecomp(fp1,fp2);
fclose(fp1);
fclose(fp2);
exit(0);

}
}
}

//filecomp: compare two files -a line at a time
void filecomp (FILE *fp1, FILE *fp2)
{
char line1[MAXLINE], line2[MAXLINE];
char *lp1 = *lp2;
do{
lp1 = fgets(line1, MAXLINE, fp1);
lp2 = fgets(line2, MAXLINE, fp2);
if(lp1 == line1 && lp2 == line2){

if(strcmp(line1,line2) !=0) {
printf("First difference in linen%sn",line1);
lp1 = lp2 = NULL;
}
} else if(lp1 != line1 && lp2 == line2)
printf("end of first file at linen%sn",line2);
else if(lp1 == line1 && lp2 == line2)
printf("end of second file at line n%sn",line1);

}while(lp1 == line1 && lp2 == line2);
}



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K & R C Programs Exercise 7-5.

K and R C, Solution to Exercise 7-5:
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.
Rewrite the postfix calculator of chapter 4 to use scanf and/or sscanf to do the input number conversion.
Read more about C Programming Language .

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* and browse!
*
* Happy Coding
***********************************************************/



#include<stdio.h>
#include<math.h> /*for atof()*/
#define MAXTOP 100 /* max size of operand or operator */
#define NUMBER '0' /* SIGNAL THAT A NUMBER WAS FOUND */
#define MAXVAL 100
int gettop(char []);
void push(double);
double pop(void);
int sp = 0; /* Next free stack position. */
double val[MAXVAL];


//reverse Polish calulator
int main(void)
{
int type;
double op2;
char s[MAXOP];

while((type = getop(s)) != EOF)
{
switch(type)
{
case NUMBER:
push(atof(s));
break;
case '+':
push(pop() + pop());
break;
case '*':
push(pop() * pop());
break;

case '-':
op2=pop();
push(pop() - op2);
break;
case '-':
op2=pop();
if(op2 != 0.0)
push(pop() / op2);
else printf("nError: Zero divissorn");
break;
case '%':
op2 = pop();
if(op2)
push(fmod(pop(), op2));
else
printf("nError: Division by zero!");
break;
case 'n':
printf("t%.8gn",pop());
break;
default:
printf("error: unknown command %sn", s);
break;

}
}
return 0;
}




/* Getop: get next operator or numeric operand. */
int getop(char s[])
{

int i = 0;
int c;
int rc;
static char lastc[] = " ";
sscanf(lastc,"%c", &c);
lastc[0] = ' ';

/* Skip whitespace */
while((s[0] = c ) == ' ' || c == 't')
if(scanf("%",&c) == EOF)
c = EOF;
s[1] = '';

/* Not a number but may contain a unary minus. */
if(!isdigit(c) && c != '.' )
return c;


if(isdigit(c))
do{
rc = scanf("%c",&c);
if(!isdigit(s[++i] = c))
break;
}while(rc != EOF)

if(c == '.')
do{
rc = scanf("%c",&c);
if(!isdigit(s[++i] = c))
break;
}while(rc != EOF)

s[i] = '';
if(rc != EOF)
lastc[0] = c;
return NUMBER;

}



/* push: push f onto stack. */
void push(double f)
{
if(sp < MAXVAL)
val[sp++] = f;
else
printf("nError: stack full can't push %gn", f);
}

/*pop: pop and return top value from stack.*/
double pop(void)
{
if(sp > 0)
return val[--sp];
else
{
printf("nError: stack emptyn");
return 0.0;
}
}

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K & R C Programs Exercise 7-4.

K and R C, Solution to Exercise 7-4:
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 private version of scanf analogous to minprintf from the previous section.
minscanf is similar to minprintf. This function collects characters from the format string until it finds an alphabetic character after a %. That is the localfmt passsed to scanf along with the appropriate pointer.
The arguments to scanf are pointers: a pointer to a format string and a pointer to the variable that receives the value from scanf.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
***********************************************************/

#include <stdarg.h>
#include <stdio.h>
#include <ctype.h>
#define LOCALFMT 100

/* minscanf: minimal scanf with variable argument list */
void minscanf(char *fmt, ...)
{
va_list ap;
char *p, *sval;
char localfmt[LOCALFMT];
int i,c;
int *ival;
double *dval;
unsigned *uval;

va_start(ap, fmt); /* make ap point to the first unnamed arg */
for (p = fmt; *p; p++) {
if (*p != '%') {
localfmt[i++] = *p;
continue;
}
i = 0;
localfmt[i++] = '%';
while(*p(p+1) && !isalpha(*(p+1)))
localfmt[i++] = *++p;
localfmt[i++] = *(p+1);
localfmt[i] = '/0';
switch (*++p) {
case 'd':
case 'i':
ival = va_arg(ap, int *);
scanf(localfmt, ival);
break;

case 'u':

case 'o':

case 'x':

case 'X':


case 'f':
dval = va_arg(ap, double);
scanf(localfmt, dval);
break;

case 's':
sval = va_arg(ap, char *);
scanf(localfmt, sval);
break;
default:
scanf(localfmt);
break;
}
i = 0;
}
va_end(ap);
}


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