CODEWITHSUNDEEP
c
Tamar Kravitz
Tamar Kravitz

Reputation: 11

count how many word in sentence- what's wrong

I wrote a code that counts how many words are there in a sentence, but it does not work in cases like this for example:

"hello    world."

It needs to return that there are 2 words, but it returns 4 because of the spaces. It's only good for the case of one space between each word. This is my code:

int counthowmanywordsinasentence(char sentence[])// help forfunc7
{
    int count = 0, i;

    for (i = 0;sentence[i] != '\0';i++)
    {
        if (sentence[i] == ' ')
            count++;    
    }
    return (count+1);
}

Upvotes: 0

Views: 143

Answers (6)

João Neto
João Neto

Reputation: 1840

You'll have to decide what is a word first :) let's assume a word is any sequence of characters with at least one alphabetic character (A-Za-z). then you can follow @Abhilash's suggestion to complete your code.

int wordcount(char *sentence) {
  int count = 0;
  int is_word = 0;

  int i;
  for(i=0; sentence[i]!='\0'; i++) {
    if(isalpha(sentence[i])) {
      is_word = 1;
    }
    if(sentence[i] == ' ' && is_word) {
      count++;
      is_word = 0;
    }
  }

  return count + is_word;
}

Upvotes: 0

Jonathan Mee
Jonathan Mee

Reputation: 38909

So sscanf already does what you need it will eat any number of whitespaces before a string including tabs. This algorithm is safe with leading or trailing spaces.

int countHowManyWordsInASentence(char* sentence){
    int result = 0;
    int i = 0;

    while(sscanf(sentence, "%*s%n", &i) != EOF){
        sentence += i;
        result++;
    }
    return result;
}

sscanf is extremely versatile you can easily read out each word as follows:

int countHowManyWordsInASentence(char* sentence){
    int result = 0;
    int size = strlen(sentence);

    if(size > 0){
        char* word = (char*)malloc((size + 1) * sizeof(char));

        for(int i = 0; sscanf(sentence, "%s%n", word, &i) > 0; sentence += i){
            result++;
        }
        free(word);
    }
    return result;
}

Upvotes: 1

BLUEPIXY
BLUEPIXY

Reputation: 40145

int counthowmanywordsinasentence(char sentence[])
{
    int count = 0, i;
    char ch, pre = ' ';

    for (i = 0; (ch=sentence[i]) != '\0'; i++, pre = ch)
    {
        if (pre == ' ' && ch != ' ')//reckon the rise
            count++;    
    }
    return count;
}

Upvotes: 0

DWilches
DWilches

Reputation: 23016

You can safely replace your if by this new version:

if (sentence[i] == ' ' && sentence[i+1] != ' ')

Which means you will be only counting the last space in each space sequence. So in your case of 4 contiguous spaces, you will count only the last one.

You will still need to decide what to do in these two cases:

" hello world."
"hello world "

As you need to know if these should count as 2 or 3 words in both cases.

Upvotes: 1

Bhargav Rao
Bhargav Rao

Reputation: 52071

This is the simplest of all answers, just add 2 lines

#include <stdio.h>
int counthowmanywordsinasentence(char sentence[])// help forfunc7
{
    int count = 0, i;

    for (i = 0;sentence[i] != '\0';i++)
    {
        if (sentence[i] == ' ')
            count++;  
        while (sentence[i] == ' ')
            i++;     
    }
    return (count+1);
}

Upvotes: 1

agent420
agent420

Reputation: 3511

Use a flag. If you encounter a space & flag is not set, set the flag and increment count. If space is encountered & flag is set , just ignore that case. And if flag is set & char(i.e. sentence[i]) is not space, reset flag.

Upvotes: 2

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