我没有得到所需的期望输出

Question

The output of this program is to get the number of words in a sentence irrespective of number of whitespaces or any of the symbols as mentioned in the question.

Here's the question:

A line of English text will be given, where words are separated by one of the following symbols:

' ', '\\t', '.', ',' and ';'

Each word may be separated from the next and the previous by one or more of the following symbols. You have to count the number of words in the sentence.

Note that to read the input, you have to read until the EOF symbol is read, as in the following example.

int main()
{
    int c;

    c = getchar();
    while ( c != EOF ) {
       c = getchar();
    }
    return 0;
}

Note: it is possible to solve this question without arrays, so the maximum length of the line is not important.

Input

A line of English text with words separated from one another by one or more occurrences of the symbols

' ', '\\t', '.', ',' and ';'

Output

The number of words in the line.

Sample Input

This is a sentence, it has words separated by spaces and fullstops.

Sample Output

12

I know I have to use the format given in the question, but I didn't know how to use it.

I only know the program to find number of words in a sentence when the words are separated by only single space or a single \\t. But, I didn't get any idea to solve this question.

Answer 1

[one of delimiters]...[one of delimiters][not one of delimiters]... 
                                        ^^ count this edge.

like this

#include <stdio.h>
#include <stdbool.h>

bool IsDelimiter(char ch, const char *delimiters){
    while(*delimiters)
        if(*delimiters++ == ch)
            return true;
    return false;
}

int main(void){
    bool prev_is_delmiter = true;
    int ch, wc = 0;

    while((ch = getchar()) != EOF && ch != '\n'){
        bool current_is_delmiter = IsDelimiter(ch, " \t,.;");
        if(prev_is_delmiter && !current_is_delmiter){
            ++wc;
        }
        prev_is_delmiter = current_is_delmiter;
    }
    printf("%d\n", wc);
}

Answer 2

This is a simple application of four elements.

1. while()
2. getchar()
3. switch()
4. printf()

the following code shows one way to implement the function:

#include <stdio.h>  // getchar(), EOF, printf()

int main( void )
{
    int wordCount = 0;
    int ch;
    int inWord = 0;

    while( (ch = getchar()) != EOF )
    {
        switch( ch )
        {
            case ' ':
            case '\t':
            case '.':
            case ',':
            case ';':
                inWord = 0;
                break;

            default:
                if ( !inWord )
                {
                    wordCount++;
                    inWord = 1;
                }
                break;
        } // end switch
    } // end while

    printf( "\n%d\n", wordCount );
    return 0;
}