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 4-12.

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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*
* 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 4-6.

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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