counting words with arguments, fgets(), strncmp() - c

Question

i would like to write a code that count how often the argument occurs in the input.
those are the requirements:
It may be assumed
that the lines in the input do not exceed 1024 characters. The string #EOF on the beginning of a line indicates the end of the input. It it not necessary to consider word
boundaries, and overlapping words must be counted as well: with an input of baaaab,
the word aa shall be counted three times. Also, the program must be case sensitive.
i already wrote a code, but i seem to have made some mistakes. Does anyone have an idea?
int main(int argc, char *argv[])
{
char buf[1026]="start";
int count=0;
while (strncmp(buf,"#EOF",4)!=0)
{
fgets(buf, 1025, stdin);
if (strncmp(buf, argv[1], strlen(argv[1]))==0)
{
count++;
}
}
if(argc==1)
printf("Please specify a program argument.");
if(argc>=2)
printf("%d", count);
return 0;
}
this is the program input with the argument let:
Let it be, let it be, let it be, let it be.
Whisper words of wisdom, let it be.
#EOF
and there is no output while it should be 4
this is the program input with argument aa:
aa aaaaa aa
aa aaa
#EOF
and the output is 2 while it should be 9
this is the program input with argument EOF:
De volgende EOF behoort ook tot de invoer: EOF
# Net als deze #EOF. Maar hieronder niet meer.
#EOF
and there is no input while it should be 3
thanks in advance

Answer 1

strncmp() tests for exact equality of the first n characters of each string provided. However, what you want is to count each occurrence, not just if the start of the line matches. For example, if you're looking for "let" in "Let it be, let it be, let it be, let it be.", you're only ever testing "Let" against "let". No match, no count. You never test further down the string.
So what you want to do is to loop over the result of fgets(), like so:
fgets(buf, 1025, stdin);
for (char *p = buf; *p; ++p) {
if (strncmp(p, argv[1], strlen(argv[1])) == 0)
{
count++;
}
}
This will test "let" against "Let", then "et ", then "t i", etc. until you've checked the whole line and counted the matches.
If you were to use strstr() instead of strncmp(), the loop would look like this:
for (char *p = buf; (p = strstr(p, argv[1])); ++p)
{
count++;
}

Answer 2

Your code only counts the first occurrence of the word in each line of input. You need to iterate through each input string to find ALL occurrences. Try something like this:
int main(int argc,char *argv[])
{
char buf[1026] = "start";
int len, matches = 0;
if (argc < 2) {
printf("Please specify a program argument.");
exit(1);
}
len = strlen(argv[1]);
while (strncmp(buf,"#EOF",4) != 0) {
fgets(buf,1025,stdin);
int buflen = strlen(buf);
for (int i = 0; i <= buflen - len; ++i) {
if (strncmp(&buf[i],argv[1],len) == 0)
++matches;
}
}
printf("'%s' found %d times\n",argv[1],matches);
return 0;
}

Answer 3

This is a functional and correct code written based on the answer given by Fred Larson
Big thanks to him.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
char buf[1026]="start";
int N;
int count=0;
char *p;
if(argc==1)
{
printf("Please specify a program argument.\n");
return(1);
}
N=strlen(argv[1]);
while (strncmp(buf,"#EOF",4)!=0)
{
fgets(buf, 1025, stdin);
for (p = buf;*p;p++)
{
if (strncmp(p, argv[1], N)==0)
{
if (strncmp(buf,"#EOF",4)!=0)
count++;
}
}
}
if(argc>=2)
printf("%d\n", count);
return 0;
}