Question

I got a almost good working program to count words from standard input.
The word what has to be count is a program argument.

The problem is that I use a white space to see a word but I also must count within th word itself.
Example: if my input is aa aaaa #EOF, and I want to count aa the result should be 4. My code result 2.

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <math.h>


int word_cnt(const char *s, char *argv[])
{
    int cnt = 0;

    while(*s != '\0')
    {
        while(isspace(*s))
        ++s;
        if(*s != '\0')
        {
            if(strncmp(s, argv[1], strlen(argv[1])) == 0)
            ++cnt;

            while(!isspace(*s) && *s != '\0')
            ++s;
        }
     }

    return cnt;
}

int main(int argc, char *argv[])
{
    char buf[1026] = {'\0'};
    char *p="#EOF\n";
    int tellen = 0;

    if (argc != 2)
    {
        printf("Kan het programma niet uitvoeren, er is geen programma argument gevonden\n");
        exit(0);
    }

    while((strcmp(buf, p) !=0))
    {
        fgets (buf, 1025, stdin);
        tellen += word_cnt(buf, argv);
    }

    printf("%d", tellen);

    return 0;
}

Answer 1

int word_cnt(const char *s, char *argv[])
{
    int cnt = 0;
    int len = strlen(argv[1]);
    while(*s)
    {
            if(strncmp(s, argv[1], len) == 0)
              ++cnt;

            ++s;
     }

    return cnt;
}