Why does the statement getchar() != EOF always return 1 in my code?

Question

In my C programming book, if I do not input a real char, the statement getchar() != EOF will return 0. But when I practiced this principle, I found it did not make sense according to my result. My code always return 1 no matter what character I give. Is there anything wrong with my code?

Here is it:

#include <stdio.h>

main()
{   
    printf("%d\n", getchar() != EOF);
}

@user3121023 Thank you, it runs correctly, but isn't space equal to end of file? — code-life balance, Jan 31 '20 at 11:29
No, files can have spaces in them, so it can't be used to indicate EOF. — HolyBlackCat, Jan 31 '20 at 11:29
This is a definitive answer you are looking for: https://stackoverflow.com/a/10720881/147407 — Andrejs Cainikovs, Jan 31 '20 at 11:33

score 3 · Answer 1 · answered Jan 31 '20 at 11:35

3

My code always return 1 no matter what character I give.

That’s the expected behaviour: getchar() should only return EOF if no character was read. In all other cases it should return the character read, with a value ≠ EOF.

answered Jan 31 '20 at 11:35

Konrad Rudolph

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... and that is why the function returns an `int` rather than a `char`. – John Bollinger Jan 31 '20 at 13:04
@JohnBollinger Maybe. Unfortunately C has quite a lot of functions that operate on `char`s but accept/return `int` instead, even where this doesn’t make sense. – Konrad Rudolph Jan 31 '20 at 13:32

Why does the statement getchar() != EOF always return 1 in my code?

1 Answers1