Counting words in an string

[Imara]

My program has to count the amount of times the word given as an argument to main is given by the user. This all worked untill I added the bit where the program has to exit when EOF is the input.

Here is my file

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

void swap(int *p, int *q);

int main(int argc, char *argv[])
{
    char buf[1026] = {0};
    char *p = buf;
    int cnt = 0, i;
    const char *word = argv[1];
    int length_arg = strlen(argv[1]);

    while (fgets (buf, 1025, stdin) != "EOF")
    {
        for(i = 0; fgets (buf, 1025, stdin) != NULL; i++)
        {
            for (i = 0; i < 1025; i++)
            {
                if (strncmp(p + i, word, length_arg) == 0)
                    cnt++;
            }
        }
        printf("%d", cnt);
    }

    printf("%d", cnt);

    return 0;
}

the input is:
Let it be, let it be, let it be, let it be.
Whisper words of wisdom, let it be.
EOF

The outcome should be (my argument to main is "let")
4

Can someone help me out?

[Imara]

No worries, I got it

[jimblumberg]

In the following snippet:

while (fgets (buf, 1025, stdin) != "EOF")

The fgets() either returns your buf, or NULL if it failed. However you can not use the operator!= with C-Strings, you must use strcmp() to compare C-Strings.

Jim