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



#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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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 7-2.

K and R C, Solution to Exercise 7-2:
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 program that will print arbitrary input in a sensible way. As a minimum, it should print non-graphic characters in octal or hexadecimal according to local custom, and break long text lines. 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>
#define OCTAL 8
#define HEXADECIMAL 16


void ProcessArgs(int argc, char *argv[], int *output)
{
int i = 0;
while(argc > 1)
{
--argc;
if(argv[argc][0] == '-')
{
i = 1;
while(argv[argc][i] != '')
{
if(argv[argc][i] == 'o')
{
*output = OCTAL;
}
else if(argv[argc][i] == 'x')
{
*output = HEXADECIMAL;
}

++i;
}
}
}
}

int can_print(int ch)
{
char *printable = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890 !"#%&'()*+,-./:;<=>?[\]^_{|}~tfvrn";

char *s;
int found = 0;

for(s = printable; !found && *s; s++)
{
if(*s == ch)
{
found = 1;
}
}

return found;
}

int main(int argc, char *argv[])
{
int split = 80;
int output = HEXADECIMAL;
int ch;
int textrun = 0;
int binaryrun = 0;
char *format;
int width = 0;

ProcessArgs(argc, argv, &output);

if(output == HEXADECIMAL)
{
format = "%02X ";
width = 4;
}
else
{
format = "%3o ";
width = 4;
}

while((ch = getchar()) != EOF)
{
if(can_print(ch))
{
if(binaryrun > 0)
{
putchar('n');
binaryrun = 0;
textrun = 0;
}
putchar(ch);
++textrun;
if(ch == 'n')
{
textrun = 0;
}

if(textrun == split)
{
putchar('n');
textrun = 0;
}
}
else
{
if(textrun > 0 || binaryrun + width >= split)
{
printf("nBinary stream: ");
textrun = 0;
binaryrun = 15;
}
printf(format, ch);
binaryrun += width;
}
}

putchar('n');

return 0;
}

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K & R C Programs Exercise 7-1.

K and R C, Solution to Exercise 7-1:
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.
Write a C program that converts upper case to lower case or lower case to upper, depending on the name it is invoked with, as found in argv[0]. Read more about C Programming Language .

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
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* and browse!
*
* Happy Coding
***********************************************************/


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

//lower: converts upper case to lower case
//upper:converts upper case to lower case

main(int argc, *argv[])
{
int c;
if(strcmp(argv[0],"lower") == 0)
while((c = getchar()) != EOF)
putchar(tolower(c)) != EOF)
else
while(c = getchar()) != EOF)
putchar(toupper(c));
return 0;
}

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K & R C Programs Exercise 6-6.

K and R C, Solution to Exercise 6-6:
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 implement a simple version of the #define processor(i.e , no arguments) suitable for use with C programs, based on the routines of this section, You may also find getch and ungetch helpful.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>

#define MAXWORD 100

struct nlist{
struct nlist *next;
char *name;
char *defn;
};
void error(int, char*);
int getch(void);
void getdef(void);
int getword(char *,int);
struct nlist *install(char *,char *);
struct nlist *lookup(char *);
void skipblanks(void);
void undef(char *);
void ungetch(int);
void ungets(char *);


//simple version of #define processor
main()
{
char w[MAXWORD];
struct nlist *p;
while(getword(w, MAXWORD) != EOF)
if(strcmp(w, "#") == 0)
getdef();
else if(!isalpha(w[0]))
printf("%s",w);
else if ((p == lookup(w)) == NULL)
printf("%s",w);
else
ungets(p->defn);
}

//getdef:get defination and install it
void getdef()
{
int c,i;
char def[MAXWORD], dir[MAXWORD], name[MAXWORD];
skipblanks();
if(!isalpha(getword(dir,MAXWORD)))
error(dir[0],"getdef:expecting a directve after #");
else if(strcmp(dir,"define") == 0) {
skipblanks();
if(!isalpha(getword(name, MAXWORD)))
error(name[0],"getdef:non alpha name expected"):
else{
skipblanks();
for(i = 0;i < MAXWORD-1; i++)
if((def[i] = getch()) == EOF || def[i] == 'n')
break;
def[i] = '';
if(i <= 0)
error('n',"getdef:non alpha in define");
else
install(name, def);
}
}else if(strcmp(dir,"undef") == 0) {
skipblanks();
if(!isalpha(getword(name, MAXWORD)))
error(name[0],"getdef:non alphain undef"):
else
undef(name);
}else
error(dir[0],"getdef:expecting a directive after #");
}

//undef:remove a name and defination from the table
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 */
}

//error:print error message and skip the rest of the line
void error(int c, char *s)
{
printf("error:%sn",s);
while(c != EOF && c != 'n')
c = getch();
}

//skipblanks:skip blan and tab characters
void skipblanks()
{
int c;
while((c = getch()) == ' ' ||c == 't')
;
ungetch(c);
}

/*ungets: push string back onto the input*/
void ungets(char s[])
{
int len=strlen(s);
void ungetch(int);
while (len >0)
ungetch(s[--len]);
}
/*note: modify the getword function to return
spaces so that the output resembles the input data.*/
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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,
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*
* 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 6-4.

K and R C, Solution to Exercise 6-4:
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 C Program that prints the distinct words in its input sorted into decreasing order of frequency of occurrence. Precede each word by its count. 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<ctype.h>

#define MAXWORD 100
#define NDISTINCT 1000

struct tnode(
char *word;
int count;
struct tnode *addtree(struct tnode *, char *);
int getword(char *, int);
void sortlist(void);
void treestore(struct tnode *);
struct tnode *list[DISTINCT];
int ntn = 0;

/* print distinct words sorted in decreasing order of freq.*/
main()
{
struct tnode *root;
char word[MAXWORD]
int i;

root = NULL;
while(getword(word, MAXWORD) != EOF)
if(isalpha(word[0]))
root = addtree(root, word);
treestore(root);
sortlist();
for(i = 0;i < ntn; i++)
printf("%2d:%20sn",list[i]->count,list[i]->word);
return 0;
}

//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;
}

//TREESTORE:STORE IN LIST[] POINTERS TO TREE NODES

void treestore(struct tnode *p)
{
if(p != NULL) {
treestore(p->left);
if(ntn < NDISTINCT)
list[ntn++] = p;
treestore(p->right);
}
}

//sortlist: sort list of pointers to tree nodes
void sortlist()
{
int gap, i,j;
struct tnode *temp;

for(gap = ntn/2;gap > 0;gap /= 2)
for(i = gap; i < ntn;i++)
for(j = i - gap;j >= 0;j -= gap) {
if((list[j]->count) >= (list[j+gap]->count))
break;
temp = list[j];
list[j] = list[j+gap];
list[j+gap] = temp;
}
}
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K & R C Programs Exercise 6-3

K and R C, Solution to Exercise 6-3:
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 Write a cross-referencer program that prints a list of all words in a document, and, for each word, a list of the line numbers on which it occurs. Remove noise words like “the”, “and,” and so on. 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!
* Original URL: http://www.c-program-example.com/2012/02/k-r-c-programs-exercise-6-3.html
* GitHub URL: https://github.com/snadahalli/cprograms/blob/master/knrc/knrc6_3.c
***********************************************************/

/** Write a cross-referencer program that prints a list of all words in a
* document, and, for each word, a list of the line numbers on which it
* occurs. Remove noise words like "the", "and," and so on.
**/

#include
#include
#include
#include

char *dupstr(char *s) {
char *p = NULL;
if(s != NULL){
p = malloc(strlen(s) + 1);
if(p) {
strcpy(p, s);
}
}
return p;
}

int i_strcmp(const char *s, const char *t) {
int diff = 0;
char cs = 0;
char ct = 0;

while(diff == 0 && *s != '' && *t != '') {
cs = tolower((unsigned char)*s);
ct = tolower((unsigned char)*t);
if(cs < ct) {
diff = -1;
}
else if(cs > ct) {
diff = 1;
}
++s;
++t;
}

if(diff == 0 && *s != *t) {
if(*s == '') {
diff = -1;
}
else {
diff = 1;
}
}

return diff;
}

struct linelist {
struct linelist *next;
int line;
};

struct wordtree {
char *word;
struct linelist *firstline;
struct wordtree *left;
struct wordtree *right;
};

void printlist(struct linelist *list) {
if(list != NULL) {
printlist(list->next);
printf("%6d ", list->line);
}
}

void printtree(struct wordtree *node) {
if(node != NULL) {
printtree(node->left);
printf("%18s ", node->word);
printlist(node->firstline);
printf("n");
printtree(node->right);
}
}

struct linelist *addlink(int line) {
struct linelist *new = malloc(sizeof *new);
if(new != NULL) {
new->line = line;
new->next = NULL;
}
return new;
}

void deletelist(struct linelist *listnode) {
if(listnode != NULL) {
deletelist(listnode->next);
free(listnode);
}
}

void deleteword(struct wordtree **node) {
struct wordtree *temp = NULL;
if(node != NULL) {
if(*node != '') {
if((*node)->right != NULL) {
temp = *node;
deleteword(&temp->right);
}
if((*node)->left != NULL) {
temp = *node;
deleteword(&temp->left);
}
if((*node)->word != NULL) {
free((*node)->word);
}
if((*node)->firstline != NULL) {
deletelist((*node)->firstline);
}
free(*node);
*node = NULL;
}
}
}

struct wordtree *addword(struct wordtree **node, char *word, int line) {
struct wordtree *wordloc = NULL;
struct linelist *newline = NULL;
struct wordtree *temp = NULL;
int diff = 0;

if(node != NULL && word != NULL) {
if(NULL == *node) {
node = malloc(sizeof **node);
if(NULL != *node) {
(*node)->left = NULL;
(*node)->right = NULL;
(*node)->word = dupstr(word);
if((*node)->word != NULL) {
(*node)->firstline = addlink(line);
if((*node)->firstline != NULL) {
wordloc = *node;
}
}
}
}
else {
diff = i_strcmp((*node)->word, word);
if(0 == diff) {

newline = addlink(line);
if(newline != NULL) {
wordloc = *node;
newline->next = (*node)->firstline;
(*node)->firstline = newline;
}
}
else if(0 < diff) {
temp = *node;
wordloc = addword(&temp->left, word, line);
}
else {
temp = *node;
wordloc = addword(&temp->right, word, line);
}
}
}

if(wordloc == NULL) {
deleteword(node);
}

return wordloc;
}


char *char_in_string(char *s, int c) {
char *p = NULL;

/* if there's no data, we'll stop */
if(s != NULL) {
if(c != '') {
while(*s != '' && *s != c) {
++s;
}
if(*s == c) {
p = s;
}
}
}
return p;
}

char *tokenise(char **s, char *delims) {
char *p = NULL;
char *q = NULL;

if(s != NULL && *s != '' && delims != NULL) {
/* pass over leading delimiters */
while(NULL != char_in_string(delims, **s)) {
++*s;
}
if(**s != '') {
q = *s + 1;
p = *s;
while(*q != '' && NULL == char_in_string(delims, *q)) {
++q;
}

*s = q + (*q != '');
*q = '';
}
}

return p;
}

int NoiseWord(char *s)
{
int found = 0;
int giveup = 0;

char *list[] = {
"a",
"an",
"and",
"be",
"but",
"by",
"he",
"I",
"is",
"it",
"off",
"on",
"she",
"so",
"the",
"they",
"you"
};

int top = sizeof list / sizeof list[0] - 1;
int bottom = 0;
int guess = top / 2;

int diff = 0;

if(s != NULL) {
while(!found && !giveup) {
diff = i_strcmp(list[guess], s);
if(0 == diff) {
found = 1;
}
else if(0 < diff) {
top = guess - 1;
}
else {
bottom = guess + 1;
}
if(top < bottom) {
giveup = 1;
}
else {
guess = (top + bottom) / 2;
}
}
}

return found;
}

char *GetLine(char *s, int n, FILE *fp) {
int c = 0;
int done = 0;
char *p = s;

while(!done && --n > 0 && (c = getc(fp)) != EOF) {
if((*p++ = c) == 'n') {
done = 1;
}
}

*p = '';
if(EOF == c && p == s) {
p = NULL;
}
else {
p = s;
}
return p;
}

#define MAXLINE 8192

int main(void) {
static char buffer[MAXLINE] = {0};
char *s = NULL;
char *word = NULL;
int line = 0;
int giveup = 0;
struct wordtree *tree = NULL;

char *delims = " tnrafv!"%^&*()_=+{}[]\|/,.<>:;#~?";

while(!giveup && GetLine(buffer, sizeof buffer, stdin) != NULL) {
++line;
s = buffer;
while(!giveup && (word = tokenise(&s, delims)) != NULL) {
if(!NoiseWord(word)) {
if(NULL == addword(&tree, word, line)) {
printf("Error adding data into memory. Giving up.n");
giveup = 1;
}
}
}
}

if(!giveup) {
printf("%18s Line Numbersn", "Word");
printtree(tree);
}

deleteword(&tree);
return 0;
}
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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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* programs for commercial purposes,
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* 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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