[SOLVED] K&R Exercise 1-6: What is this asking me to do?

anon033 · 05-23-2019, 05:06 PM

I am working through K&R and have gotten to exercise 1-6 which asks you to

Code:

Verify that the expression getchar() != EOFis 0 or 1.

I am confused about what exactly this is asking. I may be over thinking this or just not understanding what they mean. Do they mean do this?

Code:

#include <stdio.h>

int main()
{
        int c;

        if ((c = getchar()) != EOF) {
                putchar(c);
                return 0;
        }

        else if ((c = getchar()) == EOF) {
                putchar(c);
                return 1;
        }
}

If someone could elaborate on what they are asking for that would be amazing! I would look it up, but I am afraid I'll find the answer which isn't what I want.

#Solution:

I got confused by directions, but luckily my code was correct. However, as a bonus if you look at this code it can be trimmed. I.E.

Code:

#include <stdio.h>

int main()
{
        int c;

        if ((c = getchar()) != EOF) {
                putchar(c);
                return 0;
        }
}

Mechanikx · 05-23-2019, 05:29 PM

They are asking you to test whether the expression "getchar() != EOF" evaluates to 1 or 0, or true or false. When you test an expression in C and it's true it evaluates to 1 and if not it evaluates to zero.

You can verify in different ways. In general:

If this condition is true then do this, else do this.

anon033 · 05-23-2019, 05:34 PM

I apologize, this is just really confusing to me. So I went ahead and did the next one

Code:

#include <stdio.h>

int main()
{
        int c;

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

        if (c == EOF) {
                printf("%d\n", EOF);
        }
}

but this one is just confusing me. I believe what I did does as they asked, but are they looking for soemthing like this

Code:

#include <stdio.h>

int main()
{

        if (EOF == 0) {
                printf("EOF is true\n");
        }

        else if (EOF == 1) {
                printf("EOF is false\n");
        }
}

edit: This seems like it makes no sense when I read the code out loud haha. I just wanted to be sure my initial try was correct, the directions just got me confused.

Mechanikx · 05-23-2019, 05:42 PM

No need to apologize. In C, EOF = -1. This is to set it apart from the other values getchar returns (characters). In ASCII 0-255. This is why you're storing the return value in "int c" and not "char c".

Since the expression they reference is "getchar() != EOF" I believe this is the expression whose value they want you to examine. Besides like I said EOF is -1 and it's constant.

anon033 · 05-23-2019, 05:46 PM

Quote:

Originally Posted by Mechanikx

No need to apologize. In C, EOF = -1. This is to set it apart from the other values getchar returns (characters). In ASCII 0-255. This is why you're storing the return value in "int c" and not "char c".

Since the expression they reference is "getchar() != EOF" I believe this is expression whose return value they want you to examine. Besides like I said EOF is -1 and it's constant.

Aha! Thank you so much

Mechanikx · 05-23-2019, 06:00 PM

You're quite welcome.