[C Program] How to count the numbers of digit from a file ?

Question

It's error. What's wrong of my codes?

#include "stdafx.h"
#include "stdlib.h"
#include "ctype.h"

int _tmain(int argc, _TCHAR* argv[])
{

FILE* input;
int num;
int numCount = 0;

    input = fopen("123.txt", "r");

    if (!input)
    {
      printf("No file \a\n");
      exit (101);
    }


    while ((fscanf(input, "%d", &num)) == 1)
     printf("%d", num);

    if (isdigit(input))
        numCount++;


    printf("number count: %d", numCount); 

    return 0;
}

Answer 1

while ((fscanf(input, "%d", &num)) == 1)
    printf("%d", num);

if (isdigit(input))
    numCount++;

You are only checking input once. And you shoul be checking num instead, input is a FILE.

while ((fscanf(input, "%d", &num)) == 1){
    printf("%d", num);
    if (isdigit(num))
        numCount++;
}

Answer 2

I modified your code and its working now. You did not use command line arguments in your code, so argc and argv were not needed(however that would not cause any problem).

#include <stdlib.h>
#include <ctype.h>
#include <stdio.h>
int main(void)
{

   FILE* input;
   int num;
   int numCount = 0;

   input = fopen("123.txt", "r");

   if (!input)
  {
    printf("No file \a\n");
    exit (101);
  }


  while ((fscanf(input, "%d", &num)) == 1)
    {
      printf("%d\n", num);
      if (isdigit(num))
       numCount++;
    }
    /* The function isdigit() returns non-zero if its argument is a digit between 0 and  9. Otherwise, zero is returned.Hence the value of numCount is 0 if number is not between 0 and 9*/



  printf("number count: %d", numCount);

  return 0;

}

Answer 3

Your logic is completely wrong. You should read individual characters using fgetc() and then test them with isdigit(). The loop should terminate when fgetc() returns EOF.

Answer 4

isdigit() tests a character, not a FILE*.

And if you are going to count something, numCount++ should be inside a loop, and not once in the whole program