K & R C Programs Exercise 5-1.

K and R C, Solution to Exercise 5-1:
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 get next integer from input into space, so that getint treats a + or – not followed by a digit as a valid representation of zero and fix it to push such a character back on the input. Read more about C Programming Language .

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

include<ctype.h>

int getch(void);
void ungetch(int);

/* getint: get next integer from input into *pn */
int getint(int *pn)
{
int c, sign, sawsign;

while (isspace(c = getch())) /* skip white space */
;
if (!isdigit(c) && c != EOF && c != '+' && c != '-') {
ungetch(c); /* it's not a number */
return 0;
}
sign = (c == '-') ? -1 : 1;
if (sawsign = (c == '+' || c == '-'))
c = getch();
if (!isdigit(c)) {
ungetch(c);
if (sawsign)
ungetch((sign == -1) ? '-' : '+');
return 0;
}
for (*pn = 0; isdigit(c); c = getch())
*pn = 10 * *pn + (c - '0');
*pn *= sign;
if (c != EOF)
ungetch(c);
return c;
}
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K & R C Programs Exercise 4-14.

K and R C, Solution to Exercise 4-14:
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 C Program to swap two arguments using macros.
C Program to swap(t, x, y) that interchanges two arguments of type t using the block structure.

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

#include <stdio.h>

#define swap(t, x, y)
do {
t safe ## x ## y;
safe ## x ## y = x;
x = y;
y = safe ## x ## y;
} while (0)

int main(void) {
int inum1, inum2;
double dnum1, dnum2;
char *ch1, *ch2;
printf("nEnter two Intgers:n");
scanf("%d%d",&inum1,&inum2);
printf("nIntegers before swap:n inum1= %dn inum2= %dn", inum1, inum2);
swap(int, inum1, inum2);
printf("nIntegers after swap:n inum1= %dn inum2= %dn", inum1, inum2);

printf("nEnter two Doubles:n");
scanf("%f%f",&dnum1,&dnum2);
printf("nDoubles before swap:n dnum1= %gn dnum2= %gn", dnum1, dnum2);
swap(double, dnum1, dnum2);
printf("nDoubles after swap:n dnum1= %gn dnum2= %gn", dnum1, dnum2);

printf("nEnter two Strings:n");
scanf("%s%s",ch1,ch2);
printf("n Strings before swap:n ch1= %sn ch2 = %sn", ch1, ch2);
swap(char *, ch1, ch2);
printf("nStrings after swap:n ch1= %sn ch2= %sn", ch1, ch2);

return 0;
}
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K & R C Programs Exercise 4-13.

K and R C, Solution to Exercise 4-13:
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.
Write a C Program to reverse the string using the recursive methods.
In this program reverse determines the length of the string and then calls the reverser, which reverses the string s in place. Read more about C Programming Language .

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*
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***********************************************************/
#include<stdio.h>
/* reverse: reverse the string s in place */
void reverse(char s[])
{

void reverser(char s[], int i, int len);
reverser(s,0,strlen(s));
}

/* reverser: reverse string s in place recursive */
void reverser(char s[], int i, int len)
{
int c, j;
j = len - (i + 1);
if(i < j)
{
c = s[i];
s[i] = s[j];
s[j] = c;
reverser(s, ++i, len);
}
}

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K & R C Programs Exercise 4-12.

K and R C, Solution to Exercise 4-12:
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.
Write a C program to convert an integer into a string by calling a recursive routine.Recursive routine is the programming technique that a routine invoking itself again and again. Read more about C Programming Language .

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

#include<math.h>
void itoa(int n, char s[]);
int main(void) {
char buffer[20];

//for testing!
printf("INT_MIN: %dn", INT_MIN);
itoa(INT_MIN, buffer);
printf("Buffer : %sn", buffer);

return 0;
}
/* itoa: convert n to characters in s; recursive */
void itoa(int n, char s[])
{
static int i;
if(n / 10)
itoa(n /10, s);
else{
i = 0;
if(n < 0)
s[i++] = '-';
}
s[i++] = abs(n) %10 + '0';
s[i] = '';
}

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K & R C Programs Exercise 4-11.

K and R C, Solution to Exercise 4-11:
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 K & R C Programs Exercise 4-3, Modify the gettop function so that it doesn’t need to use ungetch by using internal static variable. Read more about C Programming Language .

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

#include<stdio.h>
#include<ctype.h>
#define NUMBER '0'

int getch(void);

/* getop: get next operator or numeric operand. */
int getop(char s[])
{
int i ;
int c;
static int lastc = 0;

if(lastc == 0)
c = getch();
else{
c = lastc;
lastc = 0;
}

while((s[0] = c) == ' ' || c == 't')
c = getch();
s[1] = '';

if(!isdigit(c) && c != '.')
return c;
i = 0;
if(isdigit(c))
while(isdigit(s[++i] = c = getch()))
;
if(c == '.')

while(isdigit(s[++i] = c = getch()))
;

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

}
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K & R C Programs Exercise 4-10.

K and R C, Solution to Exercise 4-10:
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 modify the K & R C Programs Exercise 4-5, An alternate organization uses getline to read an entire input line; this makes getch and ungetch unnecessary. revise the caluculater to use this aproach. Read more about C Programming Language .

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

#include<stdio.h>
#include<ctype.h>

#define MAXLINE 100
#define NUMBER '0' //SIGNAL THAT A NUMBER WAS FOUND
int getline(char line[], int limit);
int li = 0;
char line[MAXLINE];
/* Getop: get next operator or numeric operand. */
int Getop(char s[])
{
int i;
int c;

if(line[li] == '')
if(getline(line, MAXLINE) == 0)
return EOF;
else li = 0;
/* Skip whitespace */
while((s[0] = c = line[li++]) == ' ' || c == 't')
{
;
}
s[1] = '';

/* Not a number but may contain a unary minus. */
if(!isdigit(c) && c != '.' )
return c;
i = 0;

if(isdigit(c))
while(isdigit(s[++i] = c =line[i++]))
;

if(c == '.')
while(isdigit(s[++i] = c =line[i++]))
;
s[i] = '';
li--;
return NUMBER;
}

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K & R C Programs Exercise 4-9.

K and R C, Solution to Exercise 4-9:
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 modify the K & R C Programs Exercise 4-8, getch and ungetch do not handlea pushed back EOF correctly. Decide what their properties ought to be if an EOF is pushed back, then implement the design
Read more about C Programming Language .

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

#include<stdio.h>
#define BUFSIZE 100
int buf[BUFSIZE];
int bufp = 0;

//getch: get a (possibly pushed back) character
int getch(void)
{
return (bufp > 0) ? buf[--bufp] : getchar();
}
//ungetch: push a character back onto the input
void ungetch(int c)
{
if(bufp >= BUFSIZE)
printf("ungetch: too many characters!");
else
buf[bufp++] = c;
}
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K & R C Programs Exercise 4-8.

K and R C, Solution to Exercise 4-8:
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 modify the K & R C Programs Exercise 4-7, Suppose there will never be more than one character of pushback. Modify getch and ungetch accordingly. Read more about C Programming Language .

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

#include <stdio.h>
char buf = 0;

//getch: get a (possible pushed character back) character
int getch(void)
{
 int c;
 if(buf != 0)
  c = buf;
 else
  c getchar();
 buf = =0;
 return c;
}

/*ungetch: push string back onto the input*/
void ungetch(int c)
{
 if (buf != 0)
  printf("ungetch: too many characters!n");
 else
  buf = c;
}
int main(void)
{
 int c;

 while ((c = getch()) != EOF) {
  if (c == '/') {
   putchar(c);
   if ((c = getch()) == '*') { 
    ungetch('!');
   }         
  } 
  putchar(c);               
 }
 return 0;
}
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K & R C Programs Exercise 4-7.

K and R C, Solution to Exercise 4-7:
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.
Write C program that will push back an entire string onto the the input.
ungets calls the routine ungetch len times, each time pushing back a character from the string s onto the input. ungets pushes the string back in reverse order.
The routine ungets does not need to worried about the buf and bufp. The routine ungetch handles buf, bufp, and error handling. Read more about C Programming Language .

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

#include<stdio.h>
/*ungets: push string back onto the input*/

void ungets(char s[])
{
int len=strlen(s);
void ungetch(int);
while (len >0)
ungetch(s[--len]);
}
int main(void)
{
char *s = "oh! this is for testing!";
int c;

ungets(s);
while ((c = getch()) != EOF)
putchar(c);
return 0;
}
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K & R C Programs Exercise 4-6.

K and R C, Solution to Exercise 4-6:
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 Add commands for handling variables.(It’s easy to provide twenty-six variables with single-letter names.). Add a variable for the most recently printed value.

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

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

#define MAXOP 100
#define NUMBER 0
#define IDENTIFIER 1
#define ENDSTRING 2
#define TRUE 1
#define FALSE 0
#define MAX_ID_LEN 32
#define MAXVARS 30


struct varType{
char name[MAX_ID_LEN];
double val;
};


int Getop(char s[]);
void push(double val);
double pop(void);
void showTop(void);
void duplicate(void);
void swapItems(void);
void clearStacks(struct varType var[]);
void dealWithName(char s[], struct varType var[]);
void dealWithVar(char s[], struct varType var[]);

int pos = 0;
struct varType last;

int main(void)
{
int type;
double op2;
char s[MAXOP];
struct varType var[MAXVARS];

clearStacks(var);

while((type = Getop(s)) != EOF)
{
switch(type)
{
case NUMBER:
push(atof(s));
break;
case IDENTIFIER:
dealWithName(s, var);
break;
case '+':
push(pop() + pop());
break;
case '*':
push(pop() * pop());
break;
case '-':
op2 = pop();
push(pop()- op2);
break;
case '/':
op2 = pop();
if(op2)
push(pop() / op2);
else
printf("nError: division by zero!");
break;
case '%':
op2 = pop();
if(op2)
push(fmod(pop(), op2));
else
printf("nError: division by zero!");
break;
case '?':
showTop();
break;
case '#':
duplicate();
break;
case '~':
swapItems();
break;
case '!':
clearStacks(var);
break;
case 'n':
printf("nt%.8gn", pop());
break;
case ENDSTRING:
break;
case '=':
pop();
var[pos].val = pop();
last.val = var[pos].val;
push(last.val);
break;
case '<':
printf("The last variable used was: %s (value == %g)n",
last.name, last.val);
break;

default:
printf("nError: unknown command %s.n", s);
break;
}
}
return EXIT_SUCCESS;
}

#define MAXVAL 100

int sp = 0;
double val[MAXVAL];

/* 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;
}
}

void showTop(void)
{
if(sp > 0)
printf("Top of stack contains: %8gn", val[sp-1]);
else
printf("The stack is empty!n");
}


void duplicate(void)
{
double temp = pop();

push(temp);
push(temp);
}

void swapItems(void)
{
double item1 = pop();
double item2 = pop();

push(item1);
push(item2);
}


void clearStacks(struct varType var[])
{
int i;
sp = 0;

for( i = 0; i < MAXVARS; ++i)
{
var[i].name[0] = '';
var[i].val = 0.0;
}
}

void dealWithName(char s[], struct varType var[])
{
double op2;

if(!strcmp(s, "sin"))
push(sin(pop()));
else if(!strcmp(s, "cos"))
push(cos(pop()));
else if (!strcmp(s, "exp"))
push(exp(pop()));
else if(!strcmp(s, "pow"))
{
op2 = pop();
push(pow(pop(), op2));
}

else
{
dealWithVar(s, var);
}
}


void dealWithVar(char s[], struct varType var[])
{
int i = 0;

while(var[i].name[0] != '' && i < MAXVARS-1)
{
if(!strcmp(s, var[i].name))
{
strcpy(last.name, s);
last.val = var[i].val;
push(var[i].val);
pos = i;
return;
}
i++;
}

/* variable name not found so add it */
strcpy(var[i].name, s);
/* And save it to the last variable */
strcpy(last.name, s);
push(var[i].val);
pos = i;
}

int getch(void);
void unGetch(int);

/* Getop: get next operator or numeric operand. */
int Getop(char s[])
{
int i = 0;
int c;
int next;

/* Skip whitespace */
while((s[0] = c = getch()) == ' ' || c == 't')
{
;
}
s[1] = '';

if(isalpha(c))
{
i = 0;
while(isalpha(s[i++] = c ))
{
c = getch();
}
s[i - 1] = '';
if(c != EOF)
unGetch(c);
return IDENTIFIER;
}

/* Not a number but may contain a unary minus. */
if(!isdigit(c) && c != '.' && c != '-')
{
/* 4-6 Deal with assigning a variable. */
if('=' == c && 'n' == (next = getch()))
{
unGetch('');
return c;
}
if('' == c)
return ENDSTRING;

return c;
}

if(c == '-')
{
next = getch();
if(!isdigit(next) && next != '.')
{
return c;
}
c = next;
}
else
{
c = getch();
}

while(isdigit(s[++i] = c))
{
c = getch();
}
if(c == '.') /* Collect fraction part. */
{
while(isdigit(s[++i] = c = getch()))
;
}
s[i] = '';
if(c != EOF)
unGetch(c);
return NUMBER;
}

#define BUFSIZE 100

int buf[BUFSIZE];
int bufp = 0;

/* Getch: get a ( possibly pushed back) character. */
int getch(void)
{
return (bufp > 0) ? buf[--bufp]: getchar();
}

/* unGetch: push character back on input. */
void unGetch(int c)
{
if(bufp >= BUFSIZE)
printf("nUnGetch: too many charactersn");
else
buf[bufp++] = c;
}



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