How to count total number of words without using string? - c

Question

I implemented this code for getting no. of words count. It works fine for all the single characters, i.e. if I type "q w r " as input, it gives me 3 words, but when I type "qwe ed df " as input, it displays 2.
#include<stdio.h>
int main()
{
int c=getchar();
int words=0;
while(c!=EOF)
{
if(c==' ' || c=='\n')
{
c=getchar();
}
else if(c>='a' && c<='z')
{
c=getchar();
if(c==' ')
{
words=words+1;
c=getchar();
}
else
{
c=getchar();
}
}
}
printf("%i\n",words);
}

Answer 1

The trick is to count the boundaries.
Your original code has a bug around ...
if(c==' ')
{
words=words+1;
c=getchar();
}
The code only counts words if a pair of reads results in 'a' => 'z' followed by ' '.
The trick is to simplify the loop into a read one character at a time loop. (This ensure strange boundaries don't occur), and a state machine which models whether you are eating words, or spaces.
enum Mode { word = 1, spaces = 2 };
int c=getchar();
enum Mode currentMode = spaces;
int words=0;
while(c!=EOF)
{
if(c==' ' || c=='\n')
{
if( currentMode == word ) {
words=words+1;
}
currentMode = spaces;
}
else if(c>='a' && c<='z')
{
currentMode = word;
}
c=getchar();
}
// count the last word...
if( currentMode == word ) {
words=words+1;
}
The code would now ignore any non-alpha character, as neither a word, nor a space.
EDIT : Fixed incorrect enum usage

Answer 2

The problem with your code is your scattered calls to getchar which are hiding the bug which you are seeing.
You need something that checks if every character is a space and if every character is a letter.
Something like:
while ((c = getchar()) != EOF) {
if (c == ' ' || c == '\n') {
} else if (c >= 'a' && c <= 'z')
} else {
}
}
Now you just need to work out the logic inside those groupings. You have the logic right (if the previous letter was a character and you get a space then increment the word, otherwise just keep going). But your multiple calls to getchar means you can miss that condition.
Fill in the gaps in the above and it should work. (And then test with upper case letters!).

Answer 3

You are reading the characters at too many places in your while loop. This is the reason getting wrong output. Try to use this loop :
while(c!=EOF)
{
while(c == ' ' || c == '\n') //consume multiple spaces
{
c = getchar();
}
for( int i=0 ;c >= 'a' && c <= 'z'; i++) //consuming a word
{
if(i == 0) //increment word count once at the start of each word
{
words++;
}
c = getchar();
}
}

Answer 4

your code have a problem with strings that have
Even number of characters I think
the logic of the program isn't totally fine you have to add this condition inside the last else :
if(c==' ')
{
words=words+1;
c=getchar();
}
the program will be like this:
#include <stdio.h>
int main()
{
int c=getchar();
int words=0;
while(c!=EOF)
{
if(c==' ' || c=='\n')
{
c=getchar();
}
else if(c>='a' && c<='z')
{
c=getchar();
if(c==' ')
{
words=words+1;
c=getchar();
}
else
{
c=getchar();
if(c==' ')
{
words=words+1;
c=getchar();
}
}
}
}
printf("%d\n",words);
return 0;
}
because when you enter string of two characters and add a space, this last space will be read by the last getchar inside the else statement which don't increment the number of words.
check this results.
I hope this helps!

Answer 5

Every time a letter follows a space (white-space) or is the first letter, increment your word count. No need to limit count using narrow integers, file can be very long.
// Useful is...() functions declared here.
#include <ctype.h>
unsigned long long words = 0;
int previous = ' ';
int ch;
while ((ch = getchar()) != EOF) {
if (isspace(previous) && isalpha(ch)) words++;
previous = ch;
}
printf("%llu\n",words);
OP may want to adjust code to cope with digits, punctuation, etc.