I'm writing a function that finds the most common alphabetic character in a file. The function should ignore all characters other than alphabetic.
At the moment I have the following:
int most_common(const char *filename)
{
char frequency[26];
int ch = 0;
FILE *fileHandle;
if((fileHandle = fopen(filename, "r")) == NULL){
return -1;
}
for (ch = 0; ch < 26; ch++)
frequency[ch] = 0;
while(1){
ch = fgetc(fileHandle);
if (ch == EOF) break;
if ('a' <= ch && ch <= 'z')
frequency[ch - 'a']++;
else if ('A' <= ch && ch <= 'Z')
frequency[ch - 'A']++;
}
int max = 0;
for (int i = 1; i < 26; ++i)
if (frequency[i] > frequency[max])
max = i;
return max;
}
Now the function returns how many times the most frequent letter occurred, not the character itself. I'm a bit lost, as I'm not sure if that's the way this function should look like at all. Does it make sense and how possibly can I fix the problem?
I would really appreciate your help.
The variable frequency is indexed by the character code. So frequency[0] is 5, if there have been 5 'a's.
In your code you are assigning the count to max, not the character code, so you're returning the count not the actual character.
You need to store both the maximum frequency count and the character code that it referred to.
I would fix this with:
int maxCount = 0;
int maxChar = 0;
// i = A to Z
for (int i = 0; i <= 26; ++i)
{
// if freq of this char is greater than the previous max freq
if (frequency[i] > maxCount)
{
// store the value of the max freq
maxCount = frequency[i];
// store the char that had the max freq
maxChar = i;
}
}
// character codes are zero-based alphabet.
// Add ASCII value of 'A' to turn back into a char code.
return maxChar + 'A';
Note that I changed int i = 1 to int i = 0. Starting at 1 would mean starting at B, which is a subtle bug you might not notice. Also, the loop should terminate at <= 26, otherwise you'd miss out Z too.
Note the braces. Your braces style (no braces for single-statement blocks) comes very highly un-recommended.
Also, i++ is more common than ++i in cases like this. In this context it will make no difference, so would advise i++.
Related
I basically posted this question yesterday, but I'm running into a different type of error. So I'm trying to create a program where the user enters a 12-bit binary hamming code sequence like "100010010001" and it should print out its corresponding ASCII character, which in this case is 'A'.
The problem is it seems to work for single characters like 'A' but if I type the binary sequence 100010010001010110010010010010000011 which represents 'ABC' it prints out random characters unrelated to what should be printed out. I don't understand what's causing this problem?
char charToBin(char usersInput[]) {
char j = 0 ;
for (int i = 0; i <= 12 ; i++) {
if((i == 0) || (i == 1) || (i == 3) || (i == 7)){
continue ;
} ;
usersInput[j] = usersInput[i] ;
j++ ;
} ;
char c = strtol(usersInput, (char **)NULL, 2);
return c;
}
You should try my answer to that question of your yesterday. That solution will work and the first method should be really fast if you trying to break up the array into steps because it is a small array.
Nonetheless, for this code, the main problem that you have with your code is that you are trying to remove 4 indexes from a char array of 12 indexes without doing anything to tell strtol() that. So in your code, you only have to tell strtol() to stop at index 8 by injecting char '\0' at position 8 or the array which would be the 9th char. Also, since you are going to skip index 0 and 1, there is no point starting from index 0 and evaluating whether if it is index 0 or 1. You can start at index 2. Besides that, "i" should be less than 12 in your loop. That is because the last index of a 12 indexes array would be 11, and not 12. Also, j does not need to be a char. That should be an int.
char charToBin(char usersInput[]) {
int j = 0 ;
for (int i = 2; i < 12 ; i++) {
if( (i == 3) || (i == 7) ){
continue ;
} ;
usersInput[j] = usersInput[i] ;
j++ ;
} ;
usersInput[8] = '\0';
char c = strtol(usersInput, (char **)NULL, 2);
return c;
}
You must provide more info about your program like fully functional main(), example input and outputs, how to call the functions and etc.
But before that i think Same code like this may works for you.
p = len of usersInput / 12
char * charToBin(char usersInput[], int p) {
char j = 0 ;
char *c;
for(int n=0;n<p;n++){
for (int i = 0; i <= 12 ; i++) {
if((i == 0) || (i == 1) || (i == 3) || (i == 7)){
continue ;
} ;
usersInput[j+n] = usersInput[i+n] ;
j++ ;
} ;
c[n] = strtol(usersInput, (char **)NULL, 2);
}
return c;
}
Your for() loop should only iterate for 12 times,hence the condition should be i<12 and create a local character array as I did(to_Char[]) of length 9(8+1 for Null at end) and read appropriate bits into it from usersInput[], usersInput[] shouldn't be provided as input to strtol() as its of 32bits in length but our characters are just 8bits in length.
char charToBin(char usersInput[]) {
char j = 0;
char to_Char[9];
for (int i = 0; i < 12; i++) {
if ((i == 0) || (i == 1) || (i == 3) || (i == 7)) {
continue;
};
to_Char[j] = usersInput[i];
j++;
};
to_Char[j] = '\0';
char c = strtol(to_Char, (char**)NULL, 2);
return c;
}
I am reading K&R 2nd Edition and I am having trouble understanding exercise 1-13. The answer is this code
#include <stdio.h>
#define MAXHIST 15
#define MAXWORD 11
#define IN 1
#define OUT 0
main()
{
int c, i, nc, state;
int len;
int maxvalue;
int ovflow;
int wl[MAXWORD];
state = OUT;
nc = 0;
ovflow = 0;
for (i = 0; i < MAXWORD; i++)
wl[i] = 0;
while ((c = getchar()) != EOF)
{
if(c == ' ' || c == '\n' || c == '\t')
{
state = OUT;
if (nc > 0)
{
if (nc < MAXWORD)
++wl[nc];
else
++ovflow;
}
nc = 0;
}
else if (state == OUT)
{
state = IN;
nc = 1;
}
else
++nc;
}
maxvalue = 0;
for (i = 1; i < MAXWORD; ++i)
{
if(wl[i] > maxvalue)
maxvalue = wl[i];
}
for(i = 1; i < MAXWORD; ++i)
{
printf("%5d - %5d : ", i, wl[i]);
if(wl[i] > 0)
{
if((len = wl[i] * MAXHIST / maxvalue) <= 0)
len = 1;
}
else
len = 0;
while(len > 0)
{
putchar('*');
--len;
}
putchar('\n');
}
if (ovflow > 0)
printf("There are %d words >= %d\n", ovflow, MAXWORD);
return 0;
}
At the top, wl is being declared and initialized. What I don't understand is why is it looping through it and setting everything to zero if it just counts the length of words? It doesn't keep track of how many words there are, it just keeps track of the word length so why is everything set to 0?
I know this is unclear it's just been stressing me out for the past 20 minutes and I don't know why.
The ith element of the array wl[] is the number of words of length i that have been found in an input file. The wl[] array needs to be zero-initialized first so that ++wl[nc]; does not cause undefined behavior by attempting to use an uninitialized variable, and so that array elements that represent word lengths that are not present reflect that no such word lengths were found.
Note that ++wl[nc] increments the value wl[nc] when a word of length nc is encountered. If the array were not initialized, the first time the code attempts to increment an array element, it would be attempting to increment an indeterminate value. This attempt would cause undefined behavior.
Further, array indices that represent counts of word lengths that are not found in the input should hold values of zero, but without the zero-initialization, these values would be indeterminate. Even attempting to print these indeterminate values would cause undefined behavior.
The moral: initialize variables to sensible values, or store values in them, before attempting to use them.
It would seem simpler and be more clear to use an array initializer to zero-initialize the wl[] array:
int wl[MAXWORD] = { 0 };
After this, there is no need for the loop that sets the array values to zero (unless the array is used again) for another file. But, the posted code is from The C Answer Book by Tondo and Gimpel. This book provides solutions to the exercises found in the second edition of K&R in the style of K&R, and using only ideas that have been introduced in the book before each exercise. This exercise, 1.13, occurs in "Chapter 1 - A Tutorial Introduction". This is a brief tour of the language lacking many details to be found later in the book. At this point, assignment and arrays have been introduced, but array initializers have not (this has to wait until Chapter 4), and the K&R code that uses arrays has initialized arrays using loops thus far. Don't read too much into code style from the introductory chapter of a book that is 30+ years old.
Much has changed in C since K&R was published, e.g., main() is no longer a valid function signature for the main() function. Note that the function signature must be one of int main(void) or int main(int argc, char *argv[]) (or alternatively int main(int argc, char **argv)), with a caveat for implementation-defined signatures for main().
Everything is set to 0 because if you dont initialize the array, the array will be initialize with random number in it. Random number will cause error in your program. Instead of looping in every position of your array you could do this int wl[MAXWORD] = {0}; at the place of int wl[MAXWORD]; this will put 0 at every position in your array so you dont hava to do the loop.
I edited your code and put some comments in as I was working through it, to explain what's going on. I also changed some of your histogram calculations because they didn't seem to make sense to me.
Bottom line: It's using a primitive "state machine" to count up the letters in each group of characters that isn't white space. It stores this in wl[] such that wl[i] contains an integer that tells you how many groups of characters (sometimes called "tokens") has a word length of i. Because this is done by incrementing the appropriate element of w[], each element must be initialized to zero. Failing to do so would lead to undefined behavior, but probably would result in nonsensical and absurdly large counts in each element of w[].
Additionally, any token with a length that can't be reflected in w[] will be tallied in the ovflow variable, so at the end there will be an accounting of every token.
#include <stdio.h>
#define MAXHIST 15
#define MAXWORD 11
#define IN 1
#define OUT 0
int main(void) {
int c, i, nc, state;
int len;
int maxvalue;
int ovflow;
int wl[MAXWORD];
// Initializations
state = OUT; //Start off not assuming we're IN a word
nc = 0; //Start off with a character count of 0 for current word
ovflow = 0; //Start off not assuming any words > MAXWORD length
// Start off with our counters of words at each length at zero
for (i = 0; i < MAXWORD; i++) {
wl[i] = 0;
}
// Main loop to count characters in each 'word'
// state keeps track of whether we are IN a word or OUTside of one
// For each character in the input stream...
// - If it's whitespace, set our state to being OUTside of a word
// and, if we have a character count in nc (meaning we've just left
// a word), increment the counter in the wl (word length) array.
// For example, if we've just counted five characters, increment
// wl[5], to reflect that we now know there is one more word with
// a length of five. If we've exceeded the maximum word length,
// then increment our overflow counter. Either way, since we're
// currently looking at a whitespace character, reset the character
// counter so that we can start counting characters with our next
// word.
// - If we encounter something other than whitespace, and we were
// until now OUTside of a word, change our state to being IN a word
// and start the character counter off at 1.
// - If we encounter something other than whitespace, and we are
// still in a word (not OUTside of a word), then just increment
// the character counter.
while ((c = getchar()) != EOF) {
if (c == ' ' || c == '\n' || c == '\t') {
state = OUT;
if (nc > 0) {
if (nc < MAXWORD) ++wl[nc];
else ++ovflow;
}
nc = 0;
} else if (state == OUT) {
state = IN;
nc = 1;
} else {
++nc;
}
}
// Find out which length has the most number of words in it by looping
// through the word length array.
maxvalue = 0;
for (i = 1; i < MAXWORD; ++i) {
if(wl[i] > maxvalue) maxvalue = wl[i];
}
// Print out our histogram
for (i = 1; i < MAXWORD; ++i) {
// Print the word length - then the number of words with that length
printf("%5d - %5d : ", i, wl[i]);
if (wl[i] > 0) {
len = wl[i] * MAXHIST / maxvalue;
if (len <= 0) len = 1;
} else {
len = 0;
}
// This is confusing and unnecessary. It's integer division, with no
// negative numbers. What we want to have happen is that the length
// of the bar will be 0 if wl[i] is zero; that the bar will have length
// 1 if the bar is otherwise too small to represent; and that it will be
// expressed as some fraction of MAXHIST otherwise.
//if(wl[i] > 0)
// {
// if((len = wl[i] * MAXHIST / maxvalue) <= 0)
// len = 1;
// }
// else
// len = 0;
// Multiply MAXHIST (our histogram maximum length) times the relative
// fraction, i.e., we're using a histogram bar length of MAXHIST for
// our statistical mode, and interpolating everything else.
len = ((double)wl[i] / maxvalue) * MAXHIST;
// Our one special case might be if maxvalue is huge, a word length
// with just one occurrence might be rounded down to zero. We can fix
// that manually instead of using a weird logic structure.
if ((len == 0) && (wl[i] > 0)) len = 1;
while (len > 0) {
putchar('*');
--len;
}
putchar('\n');
}
// If any words exceeded the maximum word length, say how many there were.
if (ovflow > 0) printf("There are %d words >= %d\n", ovflow, MAXWORD);
return 0;
}
From ANSI C programming by K&R (page69), there is a strindex function, which will return a position of the fisrt search-for string in the source string:
#include <stdio.h>
#define MAXLINE 1000 //max input length
int getline(char line[], int max);
int Strindex(char source[], char searchfor[]);
char pattern[] = "ould";
int main()
{
char line[MAXLINE];
int found = 0;
while (getline(line, MAXLINE) > 0)
if (Strindex(line, pattern) >= 0) {
printf("%s", line);
found++;
}
return found;
} // end of main function
int getline(char s[], int lim)
{
int c, i;
i = 0;
while (--lim > 0 && (c = getchar()) != EOF && c != '\n')
s[i++] = c;
if (c == '\n')
s[i++] = c;
s[i] = '\0';
return i;
}
int Strindex(char s[], char t[])
{
int i, j, k;
for (i = 0; s[i] != '\0'; i++) {
for (j = i, k = 0; s[j] == t[k]; j++, k++)
;
if (k > 0 && t[k] == '\0') //here is the k
return i;
}
return -1;
}
My question is:
when j = i, k = 0; s[j] != t[k] (if the t[] and s[] are not empty strings), It seems t[0] will never get a value of \0? Then, what does this k>0 do in the last if statement?: if (k > 0 && t[k] == '\0')
If I get it right the Strindex() function checks if the string t is inside the string s.
So the if (k > 0 && t[k] == '\0') statement means that every character so far of the string t is the same with the string s and we reached the end of string t, so we got it mached and it returns the index of string s where the string t begins.
Well simply it is discarding all the 0 length targets. This is needed if you want to discard 0 length targets.(For this code. But there are better ways to do that).
Suppose s is "CODE" and t is an empty string containing \0. So in this case it enters for loop and the inner for loop breaks. And then k=0 and if that condition (k>0) is omitted if( t[k]=='\0') is used then it will be true and it is returned. But again one may argue that it should match any indices if empty target is passed.
Otherwise if t is empty then this function will return 0 as a string index. It seems to be a waste to put this condition into cycle though.
seems as it's not needed, but before touching any code from K&R, please, do full thoroughly checking.... probably it's needed. :) (edit:) indeed, that checks the case of calling strindex(3) with a null string "" second parameter to fail, instead of succeeding at the first case.
the next code always succeeds at index 0 if you call:
strindex("something", "");
without that test.
Indeed, i maintains the source string index, j is the index of the being checked char, and k is the index in the second string of the checked char (the matched length). the k > 0 test makes it fail in case the second string index is 0 as then, the null string has been tried, and it always succeeds at length 0.
The check of null string second parameter is more general, and as you can get that second string from the result of another calculation, it seems better to consider that case than to disallow it.
This is what I came up with, but I always get a Run-Time Check Failure #2 - Stack around the variable 'h' was corrupted.
int mostCommonLetter(char s[]) {
int i=0, h[26],k=0, max=0, number=0;
while ( k < 26){
h[k] = '0';
k++;
}
while(s[i] != '\0'){
h[whichLetter(s[i])] = h[whichLetter(s[i])]+1;
i++;
}
h[26] = '\0';
for(i=0;h[i]!='\0';i++){
if(h[i] > max)
number=i;
}
return number;
}
You cannot do h[26] = '\0'; - h has 26 elements indexed 0..25. As you know the length of h you don't need to 0-terminate it, simply do for (i=0; i < 26; ++i)
Also, are you certain whichLetter always returns a value in the 0..25 range? What does it do if it e.g. encounters a space?
This writes past the end of the array:
h[26] = '\0';
Make the for loop depend on the length rather than the last character:
for(i=0;i<26;i++){
if(h[i] > max)
number=i;
}
So pray tell, how would I go about getting the largest contiguous string of letters out of a string of garbage in C? Here's an example:
char *s = "(2034HEY!!11 th[]thisiswhatwewant44";
Would return...
thisiswhatwewant
I had this on a quiz the other day...and it drove me nuts (still is) trying to figure it out!
UPDATE:
My fault guys, I forgot to include the fact that the only function you are allowed to use is the strlen function. Thus making it harder...
Uae strtok() to split your string into tokens, using all non-letter characters as delimiters, and find the longest token.
To find the longest token you will need to organise some storage for tokens - I'd use linked list.
As simple as this.
EDIT
Ok, if strlen() is the only function allowed, you can first find the length of your source string, then loop through it and replace all non-letter characters with NULL - basically that's what strtok() does.
Then you need to go through your modified source string second time, advancing one token at a time, and find the longest one, using strlen().
This sounds similar to the standard UNIX 'strings' utility.
Keep track of the longest run of printable characters terminated by a NULL.
Walk through the bytes until you hit a printable character. Start counting. If you hit a non-printable character stop counting and throw away the starting point. If you hit a NULL, check to see if the length of the current run is greater then the previous record holder. If so record it, and start looking for the next string.
What defines the "good" substrings compared to the many others -- being lowercase alphas only? (i.e., no spaces, digits, punctuation, uppercase, &c)?
Whatever the predicate P that checks for a character being "good", a single pass over s applying P to each character lets you easily identify the start and end of each "run of good characters", and remember and pick the longest. In pseudocode:
longest_run_length = 0
longest_run_start = longest_run_end = null
status = bad
for i in (all indices over s):
if P(s[i]): # current char is good
if status == bad: # previous one was bad
current_run_start = current_run_end = i
status = good
else: # previous one was also good
current_run_end = i
else: # current char is bad
if status == good: # previous one was good -> end of run
current_run_length = current_run_end - current_run_start + 1
if current_run_length > longest_run_length:
longest_run_start = current_run_start
longest_run_end = current_run_end
longest_run_length = current_run_length
status = bad
# if a good run ends with end-of-string:
if status == good: # previous one was good -> end of run
current_run_length = current_run_end - current_run_start + 1
if current_run_length > longest_run_length:
longest_run_start = current_run_start
longest_run_end = current_run_end
longest_run_length = current_run_length
Why use strlen() at all?
Here's my version which uses no function whatsoever.
#ifdef UNIT_TEST
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#endif
/*
// largest_letter_sequence()
// Returns a pointer to the beginning of the largest letter
// sequence (including trailing characters which are not letters)
// or NULL if no letters are found in s
// Passing NULL in `s` causes undefined behaviour
// If the string has two or more sequences with the same number of letters
// the return value is a pointer to the first sequence.
// The parameter `len`, if not NULL, will have the size of the letter sequence
//
// This function assumes an ASCII-like character set
// ('z' > 'a'; 'z' - 'a' == 25; ('a' <= each of {abc...xyz} <= 'z'))
// and the same for uppercase letters
// Of course, ASCII works for the assumptions :)
*/
const char *largest_letter_sequence(const char *s, size_t *len) {
const char *p = NULL;
const char *pp = NULL;
size_t curlen = 0;
size_t maxlen = 0;
while (*s) {
if ((('a' <= *s) && (*s <= 'z')) || (('A' <= *s) && (*s <= 'Z'))) {
if (p == NULL) p = s;
curlen++;
if (curlen > maxlen) {
maxlen = curlen;
pp = p;
}
} else {
curlen = 0;
p = NULL;
}
s++;
}
if (len != NULL) *len = maxlen;
return pp;
}
#ifdef UNIT_TEST
void fxtest(const char *s) {
char *test;
const char *p;
size_t len;
p = largest_letter_sequence(s, &len);
if (len && (len < 999)) {
test = malloc(len + 1);
if (!test) {
fprintf(stderr, "No memory.\n");
return;
}
strncpy(test, p, len);
test[len] = 0;
printf("%s ==> %s\n", s, test);
free(test);
} else {
if (len == 0) {
printf("no letters found in \"%s\"\n", s);
} else {
fprintf(stderr, "ERROR: string too large\n");
}
}
}
int main(void) {
fxtest("(2034HEY!!11 th[]thisiswhatwewant44");
fxtest("123456789");
fxtest("");
fxtest("aaa%ggg");
return 0;
}
#endif
While I waited for you to post this as a question I coded something up.
This code iterates through a string passed to a "longest" function, and when it finds the first of a sequence of letters it sets a pointer to it and starts counting the length of it. If it is the longest sequence of letters yet seen, it sets another pointer (the 'maxStringStart' pointer) to the beginning of that sequence until it finds a longer one.
At the end, it allocates enough room for the new string and returns a pointer to it.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int isLetter(char c){
return ( (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') );
}
char *longest(char *s) {
char *newString = 0;
int maxLength = 0;
char *maxStringStart = 0;
int curLength = 0;
char *curStringStart = 0;
do {
//reset the current string length and skip this
//iteration if it's not a letter
if( ! isLetter(*s)) {
curLength = 0;
continue;
}
//increase the current sequence length. If the length before
//incrementing is zero, then it's the first letter of the sequence:
//set the pointer to the beginning of the sequence of letters
if(curLength++ == 0) curStringStart = s;
//if this is the longest sequence so far, set the
//maxStringStart pointer to the beginning of it
//and start increasing the max length.
if(curLength > maxLength) {
maxStringStart = curStringStart;
maxLength++;
}
} while(*s++);
//return null pointer if there were no letters in the string,
//or if we can't allocate any memory.
if(maxLength == 0) return NULL;
if( ! (newString = malloc(maxLength + 1)) ) return NULL;
//copy the longest string into our newly allocated block of
//memory (see my update for the strlen() only requirement)
//and null-terminate the string by putting 0 at the end of it.
memcpy(newString, maxStringStart, maxLength);
newString[maxLength + 1] = 0;
return newString;
}
int main(int argc, char *argv[]) {
int i;
for(i = 1; i < argc; i++) {
printf("longest all-letter string in argument %d:\n", i);
printf(" argument: \"%s\"\n", argv[i]);
printf(" longest: \"%s\"\n\n", longest(argv[i]));
}
return 0;
}
This is my solution in simple C, without any data structures.
I can run it in my terminal like this:
~/c/t $ ./longest "hello there, My name is Carson Myers." "abc123defg4567hijklmnop890"
longest all-letter string in argument 1:
argument: "hello there, My name is Carson Myers."
longest: "Carson"
longest all-letter string in argument 2:
argument: "abc123defg4567hijklmnop890"
longest: "hijklmnop"
~/c/t $
the criteria for what constitutes a letter could be changed in the isLetter() function easily. For example:
return (
(c >= 'a' && c <= 'z') ||
(c >= 'A' && c <= 'Z') ||
(c == '.') ||
(c == ' ') ||
(c == ',') );
would count periods, commas and spaces as 'letters' also.
as per your update:
replace memcpy(newString, maxStringStart, maxLength); with:
int i;
for(i = 0; i < maxLength; i++)
newString[i] = maxStringStart[i];
however, this problem would be much more easily solved with the use of the C standard library:
char *longest(char *s) {
int longest = 0;
int curLength = 0;
char *curString = 0;
char *longestString = 0;
char *tokens = " ,.!?'\"()#$%\r\n;:+-*/\\";
curString = strtok(s, tokens);
do {
curLength = strlen(curString);
if( curLength > longest ) {
longest = curLength;
longestString = curString;
}
} while( curString = strtok(NULL, tokens) );
char *newString = 0;
if( longest == 0 ) return NULL;
if( ! (newString = malloc(longest + 1)) ) return NULL;
strcpy(newString, longestString);
return newString;
}
First, define "string" and define "garbage". What do you consider a valid, non-garbage string? Write down a concrete definition you can program - this is how programming specs get written. Is it a sequence of alphanumeric characters? Should it start with a letter and not a digit?
Once you get that figured out, it's very simple to program. Start with a naive method of looping over the "garbage" looking for what you need. Once you have that, look up useful C library functions (like strtok) to make the code leaner.
Another variant.
#include <stdio.h>
#include <string.h>
int main(void)
{
char s[] = "(2034HEY!!11 th[]thisiswhatwewant44";
int len = strlen(s);
int i = 0;
int biggest = 0;
char* p = s;
while (p[0])
{
if (!((p[0] >= 'A' && p[0] <= 'Z') || (p[0] >= 'a' && p[0] <= 'z')))
{
p[0] = '\0';
}
p++;
}
for (; i < len; i++)
{
if (s[i] && strlen(&s[i]) > biggest)
{
biggest = strlen(&s[i]);
p = &s[i];
}
}
printf("%s\n", p);
return 0;
}