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


#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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* 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> /* 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 7-7.

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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* programs for commercial purposes,
* 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 7-5.

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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* 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<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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* 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 <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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K & R C Programs Exercise 7-3.

K and R C, Solution to Exercise 7-3:
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 Revise minprintf to handle more of the other facilities of printf. minprintf walks along the argument list and printf does the actual printing for the facilities supported.
To handle more of the other facilities of printf we collect in local fmt the % and any other characters until an alphabetic character -the format letter.local fmt is the format argument for printf. 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 <stdarg.h>
#include <stdio.h>
#include<ctype.h>
#define LOCALFMT 100

/* minprintf: minimal printf with variable argument list */
void minprintf(char *fmt, ...)
{
va_list ap;
char *p, *sval;
char localfmt[LOCALFMT];
int i;
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 != '%') {
putchar(*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);
printf("%d", ival);
break;
case 'c':
ival = va_arg(ap, int);
putchar(ival);
break;
case 'u':
uval = va_arg(ap, unsigned int);
printf("%u", uval);
break;
case 'o':
uval = va_arg(ap, unsigned int);
printf("%o", uval);
break;
case 'x':
uval = va_arg(ap, unsigned int);
printf("%x", uval);
break;
case 'X':
uval = va_arg(ap, unsigned int);
printf("%X", uval);
break;
case 'e':
dval = va_arg(ap, double);
printf("%e", dval);
break;
case 'f':
dval = va_arg(ap, double);
printf("%f", dval);
break;
case 'g':
dval = va_arg(ap, double);
printf("%g", dval);
break;
case 's':
for (sval = va_arg(ap, char *); *sval; sval++)
putchar(*sval);
break;
default:
putchar(*p);
break;
}
}
va_end(ap);
}




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K & R C Programs Exercise 6-5.

K and R C, Solution to Exercise 6-5:
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 function undef that will remove a name and definition from the table maintained by lookup and install. 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
***********************************************************/
int undef(char * name) {
struct nlist * np1, * np2;

if ((np1 = lookup(name)) == NULL) /* name not found */
return 1;

for ( np1 = np2 = hashtab[hash(name)]; np1 != NULL;
np2 = np1, np1 = np1->next ) {
if ( strcmp(name, np1->name) == 0 ) { /* name found */

/* Remove node from list */

if ( np1 == np2 )
hashtab[hash(name)] = np1->next;
else
np2->next = np1->next;

/* Free memory */

free(np1->name);
free(np1->defn);
free(np1);

return 0;
}
}

return 1; /* name not found */
}


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K & R C Programs Exercise 6-2.

K and R C, Solution to Exercise 6-2:
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 that reads a C program and prints in alphabetical order each group of variable names that are identical in the first 6 characters, but different somewhere thereafter. Don’t count words within strings and comments. Make 6 a parameter that can be from the command line. 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>
#include<ctype.h>
#include<string.h>
#include<stdlib.h>

struct tnode {
char *word;
int match;
struct tnode *left;
struct tnode *right;
};
#define MAXWORD
#define YES 0
#define NO 0

struct tnode *addtreex(struct tnode *, char *,int, int);
void treexprint(struct tnode *);
int getword(char *,int);


main(int argc, char argv[])
{
struct tnode *root;
char word[MAXWORD];
int found = NO;
int num;

num = (--argc && (*++argv)[0] == '-') ? atoi(argv[0] +1) : 6;
root = NULL;
while(getword(word, MAXWORD) != EOF) {
if(isalpha(word[0]) && strlen(word) >= num)
root = addtreex(root, word, num, &found);
found = NO;
}
treexprint(root);
return 0;
}
struct tnode *talloc(void);
int compare(char *,struct tnode *, int, int *);
//addtreex: add a node with w, at or below p
struct tnode *addtreex(struct tnode *p, char *w,int num, int *found)
{
int cond;
if(p == NULL) {
p = talloc();
p->word = strdup(w);
p->match = *found;
p->left = p->right = NULL;
}
else if((cond = compare(w,p,num,found)) < 0)
p->left = addtreex(p->left, w, num, found);
else if(cond > 0)
p->right = addtreex(p->right, w, num, found);
return p;
}

//compare:compare words and update p->match
int compare(char *s,struct tnode *p, int num, int *found)
{
int i;
char *t = p->word;
for(i = 0; *s == *t;i++,s++,t++)
if(*s == '')
return 0;
if(i >= num) {
*found = YES;
p->match = YES;
}
return *s - *t;
}

/*treexprint: in-order print of tree p if p->match == YES */
void treexprint(struct tnode *p)
{
if(p != NULL) {
treexprint(p->left);
if(p->match)
printf("%sn",p->word);
treexprint(p->right);
}
}


//getword: get next word or character from input
int getword(char *word, int lim)
{
int c, d, comment(void), getch(void);
void ungetch(int);
char *w = word;
while(isspace(c = getch()))
;
if(c !=EOF)
*w++ = c;
if(isalpha(c) || c == '-' || c == '#'){
for(; --lim > 0; w++)
if(!isalnum(*w = getch()) && *w != '-') {
ungetch(*w);
break;
}
}
else if(c == ''' || c == '\'){
for(; --lim > 0; w++)
if(((*w = getch()) == '\')
*++w = getch();
else if(*w == c) {
w++;
break;
}else if(*w == EOF)
break;
}else if(c == '/')
if((d = getch()) == '*')
c = comment();
else
ungetch(d);
*w = '';
return c;
}

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K & R C Programs Exercise 6-1.

K and R C, Solution to Exercise 6-1:
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 write a routine that properly handles underscores, string constants, comments, or preprocessor control lines.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<ctype.h>
//getword: get next word or character from input
int getword(char *word, int lim)
{
int c, d, comment(void), getch(void);
void ungetch(int);
char *w = word;
while(isspace(c = getch()))
;
if(c !=EOF)
*w++ = c;
if(isalpha(c) || c == '-' || c == '#'){
for(; --lim > 0; w++)
if(!isalnum(*w = getch()) && *w != '-') {
ungetch(*w);
break;
}
}
else if(c == ''' || c == '\'){
for( ; --lim > 0; w++)
if(((*w = getch()) == '\')
*++w = getch();
else if(*w == c) {
w++;
break;
}else if(*w == EOF)
break;
}else if(c == '/')
if((d = getch()) == '*')
c = comment();
else
ungetch(d);
*w = '';
return c;
}

//comment: skip over comment and return a character
int comment()
{
int c;
while((c = getch()) != EOF)
if(c == '*')
if((c = getch()) == '/')
break;
else
ungetch(c);
return c;
}

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