How to count the number of integers in a file in C?

I have this code which reads a file from the first argument to the main and counts the number of integers stored in it.

#include

#include 
#include 

int array[100000];
int count = 0;
int main(int argc, char* argv[]){
    FILE* file;
    int i;



    file = fopen(argv[1],"r");

    while(!feof(file)){
        fscanf(file, "%d", &array[count]);
        count++;
    }

    for(i=0; i        printf(" \n a[%d] = %d\n",i,array[i]);
    }

    return 0;
}

The output when I execute this file is

 a[0] = 1

 a[1] = 2


 a[2] = 3

 a[3] = 4

 a[4] = 5

 a[5] = 6

 a[6] = 7


 a[7] = 8

 a[8] = 9

 a[9] = 10

 a[10] = 0

Why is the value of count one greater than expected?

My input file using "./a.out /home/ghost/Desktop/file.txt" is as follows :

1 2 3 4 5 6 7 8 9 10

Answer

while(!feof(file)){
    fscanf(file, "%d", &array[count]);
    count++;

}

Instead of checking for eof, you need to check the returncode of fscanf():

while(fscanf(file, "%d", &array[count]) == 1)
    count++;

But it would be better to build in some safety too, like:

#define NUM_ITEMS 1000

int array[NUM_ITEMS];

int main(int argc, char* argv[]){
{
    FILE* file;
    int i, count = 0;


    file = fopen(argv[1], "r");

    if (!file) {
        printf("Problem with opening file\n");
        return 0; // or some error code
    }

    while(count < NUM_ITEMS && fscanf(file, "%d", &array[count]) == 1)
        count++;


    fclose(file);

    for(i=0; i        printf("a[%d] = %d\n", i, array[i]);
    }

    return 0;
}