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Write a C Program to Count Number of Words in a Sentence By Taking Input From User


C Program to Count Number of Words In a String


Logic

1. Take input string from user.

2.  Run for loop . Whenever you find space increase the count by 1.

3. Once the loop is complete print the count.



#include <stdio.h>

void main()
{

   char str[100];  // Declare string with a size of 100

   int wordcount=0,i;

   printf("Enter a string");
   fgets(str,100,stdin);

  // Word count logic begins

  for(i=0;str[i]!='\0';i++)
  {
     /* Check for consecutive space .*/

    if(str[i]==' ' && str[i+1]!=' ')
         wordcount++;
   }

/* Check whether, we don't press enter without entering any value*/
if(wordcount > 1){

  printf(" Total number of words is %d",wordcount+1);
 }else{
  printf("Please enter valid string");
 }

}



Output:

Enter a string : cquestions dot in

Total number of words is 3

6 comments:

  1. If we dont give any input in your program, i.e., just press enter... The number of words will be counted as 1 not 0. Is there any way to resolve this bug?

    ReplyDelete
    Replies
    1. Thanks for pointing error, i have corrected it.

      Delete
  2. if consecutive spaces found the words are count without a word .......
    check it and give a new solution.......

    ReplyDelete
  3. Thanks. Solve the issue of word count when consecutive spaces are there.

    ReplyDelete
  4. How do you resolve the issue without complexing the program when the input sentence starts with a blank space?

    ReplyDelete
  5. Check this out!! Here's also a problem for first line's words count but resolved newline problem :)

    #include
    #include

    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':
    if (inword) { inword = 0; counted++; }
    break;
    default: inword = 1;
    } while(*it++);

    return counted;
    }

    int main(int argc, const char *argv[])
    {
    int t,m, count;
    char str[1001];
    scanf("%d",&t);
    for(m=0;m<=t;m++)
    {

    if(m==1) continue;
    else{
    gets(str);
    printf("%d\n", words(str));
    }

    }
    }

    ReplyDelete