Count words from a string with multiple empty spaces - c

Question

I need code that can count the words from a string without counting multiple spaces between them.
I could code a program that counts the words with only 1 empty space between them, but I don't know how I should code it when it would be more than 1 empty space. I thought something like a for loop that checks if the char before it is a space, but I don't know how to do that. And I want to mention that I'm a beginner in C.
#include <stdio.h>
#include <string.h>
int main()
{
char s[200];
int count = 0, i;
printf("enter the string: ");
fgets(s,200,stdin);
for (i = 0;s[i] != '\0';i++)
{
if (s[i] == ' ')
count++;
}
printf("number of words in given string are: %d\n", count+ 1);
return(0);
}

Answer 1

You can introduce a flag to tell whether the previous character was a space. Something like:
#include <stdio.h>
#include <string.h>
int main()
{
char s[200];
int count = 0, i;
int last_was_space = 1;
printf("enter the string: ");
fgets(s,200,stdin);
for (i = 0;s[i] != '\0';i++)
{
if (s[i] == ' ')
{
if (!last_was_space)
{
count++; // Only count when last char wasn't a space
last_was_space = 1;
}
}
else
{
// Update flag (unless this char is a newline)
if (s[i] != '\n') last_was_space = 0;
}
}
if (!last_was_space) ++count; // Count the last word if there wasn't a space before
printf("number of words in given string are: %d\n", count);
return(0);
}

Answer 2

Framing the problem in general terms helps here. Don't think of it as "counting words", or "counting spaces". Think of it as counting "transitions from separator to non-separator". Defining our terms:
Separator: start of string, or a single ASCII space
Non-separator: everything else
Examples (^ is start of string, _ is a literal space, $ is end of string):
^a_quick_brown_fox_jumps$
^ ^ ^ ^ ^ 5 transitions
^_a__quick___brownfox_jumps___$
^ ^ ^ ^ 4 transitions
^$
0 transitions
^___$
0 transitions
^__x$
^ 1 transition
Now in psuedo code:
def is_separator(char x):
return (x == NULL or x == ' ')
def is_non_separator(char x):
return (! is_separator(x))
let count = 0, last_char = NULL
while current_char = read_char():
if (is_non_separator(current_char) and is_separator(last_char)):
count++
From here, you can translate into specific languages or change the meaning of separators without affecting the logic of counting.

Answer 3

A bit more universal
size_t wcount(const char *s, const char *del, int countempty)
{
char *token;
size_t count = 0;
char *str = strdup(s);
if(str)
{
token = strtok(str, del);
while( token != NULL )
{
if(!strlen(token))
{
if(countempty)
{
count++;
}
}
else
{
count++;
}
token = strtok(NULL, del);
}
}
free(str);
return count;
}
int main ()
{
char str[] = "something to count ,., , . !! Stack overflow ";
printf("With empty %zu, Without empty%zu\n", wcount(str," ", 1), wcount(str," .,", 0));
}

Answer 4

count the words from a string without counting multiple spaces between them
Set a flag to determine if the beginning of a word is possible. Fewer special cases than looking for the end of the word.
Typically the requirement for "spaces" implies any white-space, then the task is easily coded:
#include <ctype.h>
#include <stdbool.h>
#include <stdio.h>
#include <string.h>
int main(void) {
char s[200];
printf("enter the string: ");
fgets(s, sizeof s, stdin);
int count = 0;
bool beginning_of_word_possible = true;
for (const char *p = s; *p; p++) {
if (isspace((unsigned char ) *p)) {
beginning_of_word_possible = true;
} else {
if (beginning_of_word_possible) {
count++;
}
beginning_of_word_possible = false;
}
}
printf("number of words in given string are: %d\n", count);
return (0);
}
#P__J__ offered a good idea that passes in a list of delimiters. Below is a similar and short solution that does not allocate memory nor change the supplied string.
#include <string.h>
size_t word_count(const char *s, const char *delimiters) {
size_t count = 0;
while (*(s += strspn(s, delimiters))) { // Advance s by the matching delimiters.
count++;
s += strcspn(s, delimiters); // Advance s by the non-matching delimiters.
}
return count;
}
Test
int main(void) {
const char *de = " \n";
printf("%zu\n", word_count("", de));
printf("%zu\n", word_count("\n", de));
printf("%zu\n", word_count(" ", de));
printf("%zu\n", word_count("abc", de));
printf("%zu\n", word_count(" abc", de));
printf("%zu\n", word_count(" abc \n", de));
printf("%zu\n", word_count("abc xyz", de));
printf("%zu\n", word_count(" abc xyz", de));
printf("%zu\n", word_count(" abc xyz \n", de));
}
Output
0
0
0
1
1
1
2
2
2

Answer 5

Short and Simple Version:
#include <stdio.h>
int main(void) {
char str[] = " Hello, This is a test of a word counter";
int i = 0;
for(char* s=str; strtok(s," "); i++) s = NULL;
printf("number of words in given string are: %d\n", i);
return 0;
}
Output
Success #stdin #stdout 0s 9424KB
number of words in given string are: 9