C语言中的单词计数程序的数量超过了应有的

Question

OK guys, so I wrote this program

#include <stdio.h>

/* count words */

main ()
{

    int c, c2;
    long count = 0;

    while ((c = getchar()) != EOF)
    {
        switch(c)
        {
        case ' ':
        case '\n':
        case '\t':
            switch(c2)
            {
            case ' ':
            case '\n':
            case '\t':
                break;
            default:
                ++count;
            }
        }
        c2 = c;
    }
    printf("Word count: %ld\n", count);
}

It counts words from an input, as you can see. So i wrote a file called a-text that only has

a text

and i wrote in the ubuntu prompt

./cw < a-text

and it wrote

Word count: 2

So, what the heck? Shouldn't it just count 1, because after the second word there's no tab nor new line nor space, only EOF. Why does this happen?

Answer 1

Why not count words rather than spaces? Then you don't have a problem when the input ends with space.

#include <ctype.h>
#include <stdio.h>

int main(int argc, char**argv) {
    int was_space = 1;
    int c;
    int count = 0;
    while ((c = getchar()) != EOF) {
        count += !isspace(c) && was_space;
        was_space = isspace(c);
    }
    printf("Word count: %d\n", count);
    return 0;
}

Answer 2

lets see what happens with "a text"

after the first iteration, c2 == 'a', count remains 0
now comes c == ' ' c2 is still 'a', so count == 1, c2 becomes == ' '
now comes c == 't' c2 is still ' '. so count remains == 1
...
now comes c == '\n' c2 is the last 't'. count becomes == 2

IOW

"a text\n"
  ^----^-------- count == 1
       |
       +-------- count == 2