How does EOF act in getchar() and scanf() on Windows system?

Question

I have two pieces of codes to test how the two console I/O functions, getchar() & scanf(), handle the EOF. But I still do not have a clear comprehension about the actual operations behind the outputs and their behaviors. Can someone explains that for me? Thanks a lot! (I am using Windows OS)

// 1st piece of Code
#include <stdio.h>
#include <ctype.h>

int main(void)
{
   char ch;

   while ((ch=getchar()) != EOF)
   {
      putchar(toupper(ch));
   }

   return 0;
} 

If I type

abc

or

abc(ctrl+z)

The program will have the same outputs:

ABC

---

// 2nd piece of Code
#include<stdio.h>

int main(void)
{
    int x;
    while(scanf("%d",&x) != EOF)
    {
        /*Ctrl + z + Enter*/
        printf("x=%d\n",x);
    }
    return 0;
}

If I type

123

The program will output:

x=123

Otherwise, if I type

123(ctrl+z)

The program will have an infinite output:

x=123

x=123

x=123

x=123

...

Answer 1

getchar() returns the value of the character converted to unsigned char or EOF in case of error.

The error can be "end of file" or something else; usually (for getchar()) the programmer does not care about the error, just that an error happened.

---

scanf() returns the number of values matched and assigned (basically the number of % in the format string) or EOF in case of error. Note that the number can be less than the number of % in case, for example, of badly formatted input

Just like for getchar() the error can be "end of file" or something else. Particularly, reading less than the number of % is not an error.

So you may prefer to test for the correct number of assignments rather than testing for error in scanf()

#include <stdio.h>

int main(void) {
    int x;
    while (scanf("%d", &x) != 1) {
        /*Ctrl + z + Enter*/
        printf("x=%d\n", x);
    }
    return 0;
}