putchar and printf not behaving as expected

Question

I'm working through some of the exercises in K&R. Exercise 1-6 asks for verification that the expression getchar() != EOF is either 0 or 1. I understand why it is, but the code I wrote to prove it didn't work as expected. I wrote the following two snippets:

Version 1:

int main(void)
{
    int c;

    while (c = getchar() != EOF)
    {
        putchar(c);
    }

    printf("%d at EOF\n", c);

    return 0;
}

Version 2:

int main(void)
{
    int c;

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

    printf("%d at EOF\n", c);

    return 0;
}

My questions:

When I type in a character and hit enter with version one, why do I not see either a 0 or 1 on the screen? Instead, my cursor moves to the first position on next line, which is otherwise empty. I though putchar would send c to stdout.
While the use of printf in the second version does produce a 0 or 1 appropriately, it duplicates the 1 for each non-EOF character (I see the number 1 on two consecutive lines for each character I input). Why?

Many thanks in advance for your thoughts. If there is a reference that you think would help, please send a link.

CLARIFICATION:

I know I'm assigning c a value of either 0 or 1. That's what I want to do, and it's what the exercise wants. That's also why I don't have parentheses around c = getchar(). My question deals more with understanding why the output isn't what I had expected. Sorry for any confusion.

I actually compiled the code with gcc. I'm working in an IDE supplied by a course (CS50), and when I tried to compile with make and clang, I couldn't. Wouldn't those warnings just prevent my code from compiling? — BruceM, Jul 03 '17 at 21:43
"1 for each non-EOF character" --> All characters are non-EOF. End-of_file is a condition, not a character. — chux - Reinstate Monica, Jul 03 '17 at 23:06
The whole point of adding warning options to the compilation is to prevent **faulty** code from compiling. The compiler knows more about C than you do. The compiler only produces warnings when you tell it to, or when it is convinced that what you've written isn't what you meant to write. I don't run code that doesn't compile with `gcc -Wall -Wextra -Werror -O3 -g -Wstrict-prototypes -Wold-style-declarations -Wold-style-definitions -Wmissing-prototypes` (or as many of these as the local GCC will support). I've better things to do with my time than look for bugs the compiler can spot for me. — Jonathan Leffler, Jul 03 '17 at 23:10

dbush · Accepted Answer · 2017-07-04T03:14:10.547

The assignment operator = has lower precedence than the inequality operator !=.

So this:

while (c = getchar() != EOF)

Is parsed as:

while (c = (getchar() != EOF))

So then c is assigned the boolean value 1 if getchar is not EOF and 0 if it does return EOF.

As a result, the first program print the character for the ASCII code 1, which is a non-printable character. That's why you don't see anything. The second program, using the %d format specifier to printf, converts the number 1 to its string representation.

You need parenthesis to have the result of getchar assigned to c:

while ((c = getchar()) != EOF)

EDIT:

To further clarify the output you're getting, in both programs the variable c has the value 1 inside of each while loop. The difference here is that putchar is printing the character with the ASCII value of 1 (an unprintable character), while printf with %d print the textual representation of the value 1, i.e. 1.

If you changed the printf call to this:

printf("%c", c);

You would get the same output as using putchar.

As for the printing of 1 twice for each character, that is because you're actually entering two characters: the key you press, plus the enter key. When reading from the console, the getchar function doesn't return until the enter key is pressed.

You didn't answer the part of the question " it duplicates the 1 for each non-EOF character, why?" (which I guess is due to OP forgetting that he pressed Enter) — M.M, Jul 04 '17 at 00:00
@dbush, can you also please explain why I see the value 1 twice for each non-EOF character I enter? — BruceM, Jul 04 '17 at 02:51
@BruceM See my edit. The extra print is for the newline from the enter key. — dbush, Jul 04 '17 at 03:14

putchar and printf not behaving as expected

1 Answers1