Thread: Super simple - Opening/Reading a file; Don't get it

I don't know what the condition for my while loop needs to be. I just want it to read over a file of numbers (800+) and print those numbers out to the user.

Code:

#include <stdio.h>
#define ALL_NUMS 800

main(void)
{
  FILE *fptr;
  char c;
  char file_name[20];
  int i, j;

  printf("Please enter the name of the file containing the set of numbers you wish to sort.\n");
  scanf("%s",file_name);

  fptr=fopen(file_name,"r");

  if(fptr == NULL){
    fprintf(stderr, "Can't open input file.\n");
    exit(1);
  }

  while(){
    c=fgetc(fptr);
    printf("%c",c);
  }

  fclose(fptr);

  return 0;
}

FAQ > Definition of EOF and how to use it effectively - Cprogramming.com

Does putchar automatically cast the int to a char?

Code:

int c;
  
while ((c = fgetc(fp)) != EOF)
{
  putchar (c);
}

Originally Posted by new2C-

Does putchar automatically cast the int to a char?

What makes you think putchar expects a char? (Well, besides the name.)

I was thinking it did due to the name. I noticed in the prototype it takes an int parameter, I also noticed the examples in main have some parameters passed as a char, shouldn't that create some warning from the compiler of incompatible data types?

Example:

Code:

#include <stdio.h>

int main ()
{
   char ch;

   for(ch = 'A' ; ch <= 'Z' ; ch++) {
      putchar(ch);
   }
   
   return(0);
}

C has a notion of automatic type promotion, which means that (for instance) passing a char to a function that expects an int will automatically (and silently) convert the char to an int (of the same value). (char and short can be promoted to int, and I believe all of those can be promoted to long int; similarly float can be promoted to double).

As an extra fun fact that you didn't want to know about C, you might get warnings from the above, but on line 7 because 'A' (and any character constant) is of type int, not of type char.