C: counting number of words (need help fixing) - c

Question

I am new to C programming and trying to write a code for counting the number of words in a string.Here is my code for counting the number of codes.
#include<stdio.h>
#include<string.h>
void main()
{
int count=0,i,len;
char str[100];
printf("enter the sentence");
gets(str);
len=strlen(str);
for(i=0;i<=len;i++)
{
if(str[i]==' ')
count++;
}
printf("the number of words are :\t%d",count+1);
}
When my input is:Here is four words it works fine. it gives output
the number of words are : 4
My question is how do I handle "two consecutive spaces" between the word, "space at the beginning" of the input and "space at the last" of the input.

Answer 1

Instead of counting spaces, count the first non-space character of each word.
#include<stdio.h>
int main()
{
int count=0;
char str[100];
printf("enter the sentence");
gets(str);
char *cur= str;
for (;;)
{
while (*cur == ' ')
{
cur++;
}
if (*cur == 0)
{
break;
}
count++;
while (*cur != 0 && *cur != ' ')
{
cur++;
}
}
printf("the number of words are :\t%d",count);
return 0;
}

Answer 2

You can use:
while(str[i]==' '&&str[i]!=EOF)
{
count++;
i++;
}
instead of your if part. You also need to add these code before the for loop to read the beginning spaces.

Answer 3

I think the loop in the current form may not work properly,
It should be as follows,
for(i=0;i<len;i++)
{
if(i!=0)
{
if(str[i]==' ')
count++;
}
}
To check the other criteria change the code as follows,
for(i=0;i<len;i++)
{
if(str[i]==' ')
{
if(i!=0)
{
if(str[i+1]!=' ')
{
count++;
}
}
}

Answer 4

Just ignore spaces at the beginning and spaces directly after other spaces, and +1 if there are no spaces at the last.
#include <stdio.h>
#include <string.h>
// #include <stdlib.h>
int main() // void main is a bad practice
{
int count = 0, i, len, ignoreSpace;
char str[100];
printf("enter the sentence\n");
gets(str);
len = strlen(str);
ignoreSpace = 1; // handle space at the beginning
for(i = 0; i < len; i++) // not i<=len
{
if(str[i] == ' '){
if(!ignoreSpace){
count++;
ignoreSpace = 1; // handle two or more consecutive spaces
}
}else{
ignoreSpace = 0;
}
}
if( !ignoreSpace ) // handle space at the last
count++;
printf("the number of words are :\t%d\n", count); // +1 is moved to previous line
// system("pause");
return 0;
}

Answer 5

use strtok and first call of strtok use strtok(string," ") and for rest of calls use strtok(NULL, " \n")

Answer 6

You should count the transitions from space to non-space characters + a possible non-space character in the beginning itself.
#include<stdio.h>
#include<string.h>
int main()
{
int count=0,i,len, cur_is_spc;
char str[100];
printf("enter the sentence");
gets(str);
len=strlen(str);
cur_is_spc = 0; // 0, if current character is not space. 1, if it is.
for(i=0; str[i]!='\0'; i++)
{
if(str[i] != ' ')
{
switch(cur_is_spc) // currently holding value for previous character
{
case 0: count++; break; //count the spc->non-spc transitions
case 1: break;
default: cout << "Erroneous value"; exit(1);
}
cur_is_spc = 1; //updated for current character.
}
else
{
cur_is_spc = 0; //updated for current character.
}
}
printf("the number of words are :\t%d",count+1);
return 0;
}
Here, I am checking with only spaces. But there can be characters like newline, tab etc. How would your code handle them? Hint: use isspace() function.
/moreover, the transition can be done from non-alphabet characters to alphabet characters if you decide that words are made up of alphabets only. This approach is inherently flexible to suit your needs.

Answer 7

One quick way to do this is use strtok and break everything according to a predicate. This function satisfy all your requirements.
#include<stdio.h>
#include<string.h>
int countSpace(char* str){
int counter = 0;
char * newString;
newString= strtok (str, " "); // now the newString has the str except first word
while (newString != NULL){
counter++; // Put counter here to ignore the newString == NULL
// Or just -1 from the counter on main()
newString= strtok (NULL, " "); //Break the str in to words seperated by spaces
}
return counter;
}
void main(){
int count=0,i,len;
char str[100];
printf("Enter the sentence:\n");
fgets (str , 100 , stdin);
count = countSpace(str);
printf("The number of words are :\t%d\n",count);
return 0;
}
Thank you

Answer 8

Why not use strtok and bypass it altogether:
int main()
{
int num_words = 0;
char str_one[] = "This string has a trailing space ";
char str_two[] = " This string has a preceeding space";
char str_three[] = "This string contains two spaces consecutively twice!";
char delim[] = " ";
char *ret;
/* fgets() for user input as desired... */
if (( ret = strtok(str_one, delim)) != NULL )
{
while ( ret )
{
num_words++;
ret = strtok(NULL, delim);
}
}
else
{
/* no spaces, but might contain a word if the string isn't empty */
if ( str_one[0] != '\0' )
num_words = 1;
}
printf("str_one contains %i words\n", num_words);
num_words = 0;
...
return 0;
}
And by the way: main should ALWAYS return!!!