Trying to get scanf to iterate and evaluate each section of the string with isdigit. However it seems to be skipping the first 'block' thus offsetting everything. Recommendations on what I'm doing wrong?
int main (void) {
int icount = 0;
int c = 0;
int compare = 0;
int len = 0;
char s[256] = "";
printf("Enter a string:\n\n");
while (scanf("%s", s) == 1) {
scanf("%255s", s);
len = strlen(s);
printf("the string is %d characters long\n", len);
while (len > 0) {
printf("c is on its s[%d] iteration\n", c);
if (isdigit(s[c])) {
compare = compare + 1;
}
c++;
len--;
}
if (compare == strlen(s)) {
icount = icount + 1;
}
c++;
}
printf("\ni count is %d\n", icount);
return 0;
}
When I run it I keep getting data back like this:
./a
Enter a string:
17 test 17
the string is 4 characters long
c is on its s[0] iteration
c is on its s[1] iteration
c is on its s[2] iteration
c is on its s[3] iteration
the string is 2 characters long
c is on its s[5] iteration
c is on its s[6] iteration
i count is 0
From the comments above, I believe this might be what you are looking for
#include <ctype.h>
#include <stdio.h>
int main (void)
{
int icount;
int index;
char string[256];
printf("Enter a string:\n\n");
icount = 0;
while (scanf("%255s", string) == 1)
{
int isNumber;
isNumber = 1;
for (index = 0 ; ((string[index] != '\0') && (isNumber != 0)) ; ++index)
{
printf("index is on its string[%d] iteration\n", index);
if (isdigit(string[index]) == 0)
isNumber = 0;
}
if (isNumber != 0)
icount += 1;
}
printf("\nicount is %d\n", icount);
return 0;
}
ended up going with this simple code as my knowledge level is... well... simple
. thanks for the help with that iteration and second scanf it was about to drive me over the edge!
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main (void) {
int icount = 0;
int c = 0;
int compare = 0;
char s[256] = "";
printf("Enter a string:\n\n");
while (scanf("%255s", s) == 1) {
compare = 0;
for (c = 0 ; s[c] != '\0' ; c++) {
if (isdigit(s[c])) {
compare = compare + 1;
}
}
if (compare == strlen(s)) {
icount = icount + 1;
}
}
printf("%d integers\n", icount);
return 0;
}
Related
I need to write a function that will count words in a string. For the
purpose of this assignment, a "word" is defined to be a sequence
of non-null, non-whitespace characters, separated from other words by
whitespace.
This is what I have so far:
int words(const char sentence[ ]);
int i, length=0, count=0, last=0;
length= strlen(sentence);
for (i=0, i<length, i++)
if (sentence[i] != ' ')
if (last=0)
count++;
else
last=1;
else
last=0;
return count;
I am not sure if it works or not because I can't test it until my whole program is finished and I am not sure it will work, is there a better way of writing this function?
You needed
int words(const char sentence[])
{
}
(note braces).
For loops go with ; instead of ,.
Without any disclaimer, here's what I'd have written:
See it live http://ideone.com/uNgPL
#include <string.h>
#include <stdio.h>
int words(const char sentence[ ])
{
int counted = 0; // result
// state:
const char* it = sentence;
int inword = 0;
do switch(*it) {
case '\0':
case ' ': case '\t': case '\n': case '\r': // TODO others?
if (inword) { inword = 0; counted++; }
break;
default: inword = 1;
} while(*it++);
return counted;
}
int main(int argc, const char *argv[])
{
printf("%d\n", words(""));
printf("%d\n", words("\t"));
printf("%d\n", words(" a castle "));
printf("%d\n", words("my world is a castle"));
}
See the following example, you can follow the approach : count the whitespace between words .
int words(const char *sentence)
{
int count=0,i,len;
char lastC;
len=strlen(sentence);
if(len > 0)
{
lastC = sentence[0];
}
for(i=0; i<=len; i++)
{
if((sentence[i]==' ' || sentence[i]=='\0') && lastC != ' ')
{
count++;
}
lastC = sentence[i];
}
return count;
}
To test :
int main()
{
char str[30] = "a posse ad esse";
printf("Words = %i\n", words(str));
}
Output :
Words = 4
#include <ctype.h> // isspace()
int
nwords(const char *s) {
if (!s) return -1;
int n = 0;
int inword = 0;
for ( ; *s; ++s) {
if (!isspace(*s)) {
if (inword == 0) { // begin word
inword = 1;
++n;
}
}
else if (inword) { // end word
inword = 0;
}
}
return n;
}
bool isWhiteSpace( char c )
{
if( c == ' ' || c == '\t' || c == '\n' )
return true;
return false;
}
int wordCount( char *string )
{
char *s = string;
bool inWord = false;
int i = 0;
while( *s )
{
if( isWhiteSpace(*s))
{
inWord = false;
while( isWhiteSpace(*s) )
s++;
}
else
{
if( !inWord )
{
inWord = true;
i++;
}
s++;
}
}
return i;
}
Here is one of the solutions. It counts words with multiple spaces or just space or space followed by the word.
#include <stdio.h>
int main()
{
char str[80];
int i, w = 0;
printf("Enter a string: ");
scanf("%[^\n]",str);
for (i = 0; str[i] != '\0'; i++)
{
if((str[i]!=' ' && str[i+1]==' ')||(str[i+1]=='\0' && str[i]!=' '))
{
w++;
}
}
printf("The number of words = %d", w );
return 0;
}
I know this is an old thread, but perhaps someone needs a simple solution, just checks for blank space in ascii and compares current char to that while also makign sure first char is not a space, cheers!
int count_words(string text){
int counter = 1;
int len = strlen(text);
for(int i = 0; i < len; i++){
if(text[i] == 32 && i != 0) {
counter++;
}
}
return counter;}
Here is another solution:
#include <string.h>
int words(const char *s)
{
const char *sep = " \t\n\r\v\f";
int word = 0;
size_t len;
s += strspn(s, sep);
while ((len = strcspn(s, sep)) > 0) {
++word;
s += len;
s += strspn(s, sep);
}
return word;
}
#include<stdio.h>
int main()
{
char str[50];
int i, count=1;
printf("Enter a string:\n");
gets(str);
for (i=0; str[i]!='\0'; i++)
{
if(str[i]==' ')
{
count++;
}
}
printf("%i\n",count);
}
#include<stdio.h>
#include<string.h>
int getN(char *);
int main(){
char str[999];
printf("Enter Sentence: "); gets(str);
printf("there are %d words", getN(str));
}
int getN(char *str){
int i = 0, len, count= 0;
len = strlen(str);
if(str[i] >= 'A' && str[i] <= 'z')
count ++;
for (i = 1; i<len; i++)
if((str[i]==' ' || str[i]=='\t' || str[i]=='\n')&& str[i+1] >= 'A' && str[i+1] <= 'z')
count++;
return count;
}
#include <stdio.h>
int wordcount (char *string){
int n = 0;
char *p = string ;
int flag = 0 ;
while(isspace(*p)) p++;
while(*p){
if(!isspace(*p)){
if(flag == 0){
flag = 1 ;
n++;
}
}
else flag = 0;
p++;
}
return n ;
}
int main(int argc, char **argv){
printf("%d\n" , wordcount(" hello world\nNo matter how many newline and spaces"));
return 1 ;
}
I found the posted question after finishing my function for a C class I'm taking. I saw some good ideas from code people have posted above. Here's what I had come up with for an answer. It certainly is not as concise as other's, but it does work. Maybe this will help someone in the future.
My function receives an array of chars in. I then set a pointer to the array to speed up the function if it was scaled up. Next I found the length of the string to loop over. I then use the length of the string as the max for the 'for' loop.
I then check the pointer which is looking at array[0] to see if it is a valid character or punctuation. If pointer is valid then increment to next array index. The word counter is incremented when the first two tests fail. The function then will increment over any number of spaces until the next valid char is found.
The function ends when null '\0' or a new line '\n' character is found. Function will increment count one last time right before it exit to account for the word preceding null or newline. Function returns count to the calling function.
#include <ctype.h>
char wordCount(char array[]) {
char *pointer; //Declare pointer type char
pointer = &array[0]; //Pointer to array
int count; //Holder for word count
count = 0; //Initialize to 0.
long len; //Holder for length of passed sentence
len = strlen(array); //Set len to length of string
for (int i = 0; i < len; i++){
//Is char punctuation?
if (ispunct(*(pointer)) == 1) {
pointer += 1;
continue;
}
//Is the char a valid character?
if (isalpha(*(pointer)) == 1) {
pointer += 1;
continue;
}
//Not a valid char. Increment counter.
count++;
//Look out for those empty spaces. Don't count previous
//word until hitting the end of the spaces.
if (*(pointer) == ' ') {
do {
pointer += 1;
} while (*(pointer) == ' ');
}
//Important, check for end of the string
//or newline characters.
if (*pointer == '\0' || *pointer == '\n') {
count++;
return(count);
}
}
//Redundent return statement.
count++;
return(count);
}
I had this as an assignment...so i know this works.
The function gives you the number of words, average word length, number of lines and number of characters.
To count words, you have to use isspace() to check for whitespaces. if isspace is 0 you know you're not reading whitespace. wordCounter is a just a way to keep track of consecutive letters. Once you get to a whitespace, you reset that counter and increment wordCount. My code below:
Use isspace(c) to
#include <stdio.h>
#include <ctype.h>
int main() {
int lineCount = 0;
double wordCount = 0;
double avgWordLength = 0;
int numLines = 0;
int wordCounter = 0;
double nonSpaceChars = 0;
int numChars = 0;
printf("Please enter text. Use an empty line to stop.\n");
while (1) {
int ic = getchar();
if (ic < 0) //EOF encountered
break;
char c = (char) ic;
if (isspace(c) == 0 ){
wordCounter++;
nonSpaceChars++;
}
if (isspace(c) && wordCounter > 0){
wordCount++;
wordCounter =0;
}
if (c == '\n' && lineCount == 0) //Empty line
{
break;
}
numChars ++;
if (c == '\n') {
numLines ++;
lineCount = 0;
}
else{
lineCount ++;
}
}
avgWordLength = nonSpaceChars/wordCount;
printf("%f\n", nonSpaceChars);
printf("Your text has %d characters and %d lines.\nYour text has %f words, with an average length of %3.2f ", numChars, numLines, wordCount, avgWordLength);
}
Here is one solution. This one will count words correctly even if there are multiple spaces between words, no spaces around interpuncion symbols, etc. For example: I am,My mother is. Elephants ,fly away.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int countWords(char*);
int main() {
char string[1000];
int wordsNum;
printf("Unesi nisku: ");
gets(string); /*dont use this function lightly*/
wordsNum = countWords(string);
printf("Broj reci: %d\n", wordsNum);
return EXIT_SUCCESS;
}
int countWords(char string[]) {
int inWord = 0,
n,
i,
nOfWords = 0;
n = strlen(string);
for (i = 0; i <= n; i++) {
if (isalnum(string[i]))
inWord = 1;
else
if (inWord) {
inWord = 0;
nOfWords++;
}
}
return nOfWords;
}
this is a simpler function to calculate the number of words
int counter_words(char* a){`
// go through chars in a
// if ' ' new word
int words=1;
int i;
for(i=0;i<strlen(a);++i)
{
if(a[i]==' ' && a[i+1] !=0)
{
++words;
}
}
return words;}
This code gets user input then 256 % length of string is used. If the result is 3 and the input is abc the output is bcd. This works fine. However if the input is for example "hey whatsup" the length is 11 and it should be 10 because the space shouldn't be included for the length.
How can I programm this code so it dosen't count space to the length?
Is it even possible to implement it while using fgets?
Thank you in advance.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
char array[20];
int length = 0;
int i;
int key = 256;
printf("input: ");
fgets(array, 20, stdin);
length = strlen(array) - 1;
key = key % length;
if (key > 0) {
for (i = 0; i < length; i++) {
if (array[i] == ' ') {
printf("%c", array[i]);
continue;
}
array[i] = array[i] + key;
printf("%c", array[i]);
}
}
return 0;
}
1) If you don't want to include the spaces in the calculation of the key, you have to make your own function to calculate the number of spaces.
2) The code length = strlen(array) - 1; seems to "take care" of a '\n' in the end of the string. However, you can't be sure that there is a '\n'. You need to check for that first.
3) Doing key % 0 will be "bad" so check for that as well
The code could look something like:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int cnt_spaces(char* arr)
{
int res = 0;
while(*arr)
{
if (*arr == ' ')
{
++res;
}
++arr;
}
return res;
}
int main() {
char array[20];
int length = 0;
int i;
int key = 256;
printf("input: ");
fgets(array, 20, stdin);
length = strlen(array);
if (strlen(array) == 0) return 0; // or add error handling
// Remove \n if present
if (array[length-1] == '\n')
{
array[length-1] = '\0';
--length;
}
printf("len = %d\n", length);
int spaces = cnt_spaces(array);
printf("spaces = %d\n", spaces);
if (length == spaces) return 0; // or add error handling
key = key % (length - spaces);
printf("key = %d\n", key);
if (key > 0) {
for (i = 0; i < length; i++) {
if (array[i] != ' ') {
array[i] = array[i] + key;
}
printf("%c", array[i]);
}
}
return 0;
}
Example:
input: a b c
len = 5
spaces = 2
key = 1
b c d
I have school task. To reverse each word in sentence, so example :
Input: Fried chicken, fried duck.
Output: deirF nekcihc, deirf kcud.
So except dot and comma it's not reversed.
The first code
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
int i, n, titik = 0, coma = 0;
char s[5001];
char c[5001];
char *tok;
scanf("%[^\n]s", s);
if (s[0] == ' ')
printf(" ");
tok = strtok(s, " ");
while (tok != NULL) {
strcpy(c, tok);
n = strlen(c);
for (i = n; i >= 0; i--) {
if (c[i] == ',') {
coma = 1;
} else
if (c[i] == '.') {
titik = 1;
} else
printf("%c", c[i]);
}
if (coma) {
printf(",");
coma = 0;
} else
if (titik){
printf(".");
titik = 0;
}
tok = strtok(NULL," ");
if (tok == NULL)
printf("\n");
else
printf(" ");
}
}
Second code is
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main() {
int i, j, n, prana = 0, titik = 0, coma = 0, end = 0;
char s[5001];
scanf("%[^\n]s", s);
n = strlen(s);
for (i = 0; i <= n; i++) {
if (isspace(s[i]) || iscntrl(s[i])) {
if (iscntrl(s[i]))
end = 1;
for (j = i - 1; j >= prana; j--) {
if (s[j] == '.') {
titik = 1;
} else
if (s[j] == ',') {
coma = 1;
} else
printf("%c", s[j]);
}
prana = i + 1;
if (titik) {
titik = 0;
if (end)
printf(".");
else
printf(". ");
} else
if (coma) {
coma = 0;
if (end)
printf(",");
else
printf(", ");
} else {
if (end)
printf("");
else
printf(" ");
}
}
}
printf("\n");
return 0;
}
Why the second code is accepted in test case?, but first code is not.
I tested the result it's same. Really identical in md5 hash.
The output of the two codes id different, because you print the terminating null character for each token in the first code. This loop:
for (i = n; i >=0 ; i--) ...
will have i == n in its first iteration. For a C string of length n, s[n] is the terminating null. This character may not show in the console, but it is part of the output.
To fix the loop, you could start with i = n - 1, but C uses inclusive lower bounds and exclusive upper bounds, and a more idomatic loop syntax is:
i = n;
while (i-- > 0) ...
Not related to your question at hand, but your codes are rather complicated, because they rely on many assumptions: words separated by spaces; only punctuation is comma or stop; repeated punctuation marks are ignored, special case for last word.
Here's a solution that treats all chunks of alphabetic characters plus the apostrophe as words and reverses them in place:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
void reverse(char *str, int i, int j)
{
while (i < j) {
int c = str[--j];
str[j] = str[i];
str[i++] = c;
}
}
int main()
{
char str[512];
int begin = -1;
int i;
if (fgets(str, sizeof(str), stdin) == NULL) return -1;
for (i = 0; str[i]; i++) {
if (isalpha((unsigned char) str[i]) || str[i] == '\'') {
if (begin == -1) begin = i;
} else {
if (begin != -1) {
reverse(str, begin, i);
begin = -1;
}
}
}
printf("%s", str);
return 0;
}
So I have an assignment where I should delete a character if it has duplicates in a string. Right now it does that but also prints out trash values at the end. Im not sure why it does that, so any help would be nice.
Also im not sure how I should print out the length of the new string.
This is my main.c file:
#include <stdio.h>
#include <string.h>
#include "functions.h"
int main() {
char string[256];
int length;
printf("Enter char array size of string(counting with backslash 0): \n");
/*
Example: The word aabc will get a size of 5.
a = 0
a = 1
b = 2
c = 3
/0 = 4
Total 5 slots to allocate */
scanf("%d", &length);
printf("Enter string you wish to remove duplicates from: \n");
for (int i = 0; i < length; i++)
{
scanf("%c", &string[i]);
}
deleteDuplicates(string, length);
//String output after removing duplicates. Prints out trash values!
for (int i = 0; i < length; i++) {
printf("%c", string[i]);
}
//Length of new string. The length is also wrong!
printf("\tLength: %d\n", length);
printf("\n\n");
getchar();
return 0;
}
The output from the printf("%c", string[i]); prints out trash values at the end of the string which is not correct.
The deleteDuplicates function looks like this in the functions.c file:
void deleteDuplicates(char string[], int length)
{
for (int i = 0; i < length; i++)
{
for (int j = i + 1; j < length;)
{
if (string[j] == string[i])
{
for (int k = j; k < length; k++)
{
string[k] = string[k + 1];
}
length--;
}
else
{
j++;
}
}
}
}
There is a more efficent and secure way to do the exercise:
#include <stdio.h>
#include <string.h>
void deleteDuplicates(char string[], int *length)
{
int p = 1; //current
int f = 0; //flag found
for (int i = 1; i < *length; i++)
{
f = 0;
for (int j = 0; j < i; j++)
{
if (string[j] == string[i])
{
f = 1;
break;
}
}
if (!f)
string[p++] = string[i];
}
string[p] = '\0';
*length = p;
}
int main() {
char aux[100] = "asdñkzzcvjhasdkljjh";
int l = strlen(aux);
deleteDuplicates(aux, &l);
printf("result: %s -> %d", aux, l);
}
You can see the results here:
http://codepad.org/wECjIonL
Or even a more refined way can be found here:
http://codepad.org/BXksElIG
Functions in C are pass by value by default, not pass by reference. So your deleteDuplicates function is not modifying the length in your main function. If you modify your function to pass by reference, your length will be modified.
Here's an example using your code.
The function call would be:
deleteDuplicates(string, &length);
The function would be:
void deleteDuplicates(char string[], int *length)
{
for (int i = 0; i < *length; i++)
{
for (int j = i + 1; j < *length;)
{
if (string[j] == string[i])
{
for (int k = j; k < *length; k++)
{
string[k] = string[k + 1];
}
*length--;
}
else
{
j++;
}
}
}
}
You can achieve an O(n) solution by hashing the characters in an array.
However, the other answers posted will help you solve your current problem in your code. I decided to show you a more efficient way to do this.
You can create a hash array like this:
int hashing[256] = {0};
Which sets all the values to be 0 in the array. Then you can check if the slot has a 0, which means that the character has not been visited. Everytime 0 is found, add the character to the string, and mark that slot as 1. This guarantees that no duplicate characters can be added, as they are only added if a 0 is found.
This is a common algorithm that is used everywhere, and it will help make your code more efficient.
Also it is better to use fgets for reading input from user, instead of scanf().
Here is some modified code I wrote a while ago which shows this idea of hashing:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define NUMCHAR 256
char *remove_dups(char *string);
int main(void) {
char string[NUMCHAR], temp;
char *result;
size_t len, i;
int ch;
printf("Enter char array size of string(counting with backslash 0): \n");
if (scanf("%zu", &len) != 1) {
printf("invalid length entered\n");
exit(EXIT_FAILURE);
}
ch = getchar();
while (ch != '\n' && ch != EOF);
if (len >= NUMCHAR) {
printf("Length specified is longer than buffer size of %d\n", NUMCHAR);
exit(EXIT_FAILURE);
}
printf("Enter string you wish to remove duplicates from: \n");
for (i = 0; i < len; i++) {
if (scanf("%c", &temp) != 1) {
printf("invalid character entered\n");
exit(EXIT_FAILURE);
}
if (isspace(temp)) {
break;
}
string[i] = temp;
}
string[i] = '\0';
printf("Original string: %s Length: %zu\n", string, strlen(string));
result = remove_dups(string);
printf("Duplicates removed: %s Length: %zu\n", result, strlen(result));
return 0;
}
char *remove_dups(char *str) {
int hash[NUMCHAR] = {0};
size_t count = 0, i;
char temp;
for (i = 0; str[i]; i++) {
temp = str[i];
if (hash[(unsigned char)temp] == 0) {
hash[(unsigned char)temp] = 1;
str[count++] = str[i];
}
}
str[count] = '\0';
return str;
}
Example input:
Enter char array size of string(counting with backslash 0):
20
Enter string you wish to remove duplicates from:
hellotherefriend
Output:
Original string: hellotherefriend Length: 16
Duplicates removed: helotrfind Length: 10
For example, the user shall put the input like that, "ABC123," but not "ABC 123" or "A BC123."
Here is my code:
unsigned int convert_to_num(char * string) {
unsigned result = 0;
char ch;
//printf("check this one %s\n", string);
while(ch =*string++) result = result * 26 + ch - 'A' + 1;
return result;
}
int main()
{
char input_string[100];
char arr_col[100] = {'\0'};
char arr_row[100] = {'\0'};
int raiseflag;
int started_w_alpha =0;
int digitflag = 0;
while(scanf("%s", &input_string) != EOF) {
int i = 0, j = 0, digarr = 0;
while (i <=5) {
if (input_string[i] == '\0') {printf("space found!");}
if ((input_string[i] >= 'A' && input_string[i] <= 'Z') && (digitflag == 0)) {
started_w_alpha = 1;
arr_col[j] = input_string[i]; j++;
}
//printf("something wrong here %s and %d and j %d\n", arr_holder, i, j);
if (started_w_alpha == 1) {
if (input_string[i] >=48 && input_string[i]<=57){ digitflag = 1; arr_row[digarr] =input_string[i]; digarr++; }
}
i++; if (i == 5) { raiseflag =1; }
}
printf(" => [%d,%s]\n", convert_to_num(arr_col), arr_row);
if (raiseflag == 1) { raiseflag = 0; memset(arr_col, 0, 5); memset(input_string, 0, 5); memset(arr_row, 0, 5); digitflag = 0; started_w_alpha = 0; }
}
return 0;
}
Apparently, \0 doesn't work in my case because I have an array of 5 and user can put 2 chars. I want to exit the loop whenever a space is found in between the characters.
This is the whole code. I added {'\0'} my array because of the extra characters I get when there is less than 5 characters.
Thanks!
Since the index is starting from 0 and input_string[5]; array size is 5, the only valid indexes are from 0 to 4.
but your loop while (i <=5) { go till 5, it is mean you exceed the array.
If you insert 5 characters to the string, the terminating null is the 6th.
Since you exceed the array it written over some other variable. but you still can find it when you check input_string[5]
So if you want to insert 5 characters you array size should be at least 6
char input_string[6];
if you want to check only the first 5 elements you'll have to change the loop to:
while (i < 5) {
and as I wrote in the comment if you find the terminating null, no use to continue the loop, since it contain garbage or leftover from the previous iteration.
Therefor you should break if it found, like this:
if (input_string[i] == '\0') {printf("space found!"); break;}
EDIT
check this program: it use fgets to read the whole input, then search for white spaces.
Note it doesn't trim the input, means it won't remove spaces when thay appear at the beginning or at the end of the input.
#include <ctype.h>
#include <string.h>
#include <stdio.h>
int main()
{
int i ,size;
char input_string[100];
fgets(input_string,100,stdin);
i=0;
size = strlen(input_string);
while (i<size-1){ //enter is also count
if (isspace(input_string[i]))
{
printf("space found!");
break;
}
i++;
}
return 0;
}
EDIT2
Now with a trim, so it will remove leading and ending spaces:
#include <ctype.h>
#include <string.h>
#include <stdio.h>
char* trim(char *input_string)
{
int i=0;
char *retVal = input_string;
i = strlen(input_string)-1;
while( i>=0 && isspace(input_string[i]) ){
input_string[i] = 0;
i--;
}
i=0;
while(*retVal && isspace(retVal[0]) ){
retVal ++;
}
return retVal;
}
int main()
{
int i ,size;
char input_string[100],*ptr;
fgets(input_string,100,stdin);
ptr = trim(input_string);
i=0;
size = strlen(ptr);
while (i<size){
if (isspace(ptr[i]))
{
printf("space found!");
break;
}
i++;
}
return 0;
}