I'm working through K&R at the moment and I've found my first exercise that seems to be a good use for my first CodeReview post.

Exercise 1-6 - Verify that the expression getchar() != EOF is 0 or 1.

Exercise 1-7 - Write a program to print the value of EOF.

I've completed both of these together in the same file.

I think I've understood the question here, but let me know if I haven't. I wanted to post here before looking up any sort of "official" answer so that I could learn some stuff instead of just "going to the back of the book".

#include <stdio.h>

void main()

 int c;

 c = getchar();
 while ((c = getchar()) != EOF == (0 

Compiled with:

gcc putchar.c -o putchar

Sending EOF using:

<ctrl-d>

Runs through fine, and when I ctrl-d I get a -1 return code.

This is a nice, clear, short attempt at the exercise.

You've avoided a couple of common problems:

• You're correctly using int for the return value from getchar(). A common mistake is to be misled by the name and to assign directly to a char variable. That doesn't work, because truncating to char prevents EOF from being distinguished from a valid char value.


• You've correctly matched the %d specifier to the type of EOF.

One simple way to improve your code is to get the compiler to point out weaknesses for you (by default, most C compilers are very lenient; you have to specifically ask for more warnings):

gcc -std=c11 -Wall -Wextra -Wwrite-strings -Wno-parentheses -Wpedantic -Warray-bounds -Wconversion putchar.c -o putchar

You can make it easier by writing a simple Makefile that contains just:

CFLAGS = -std=c11 -Wall -Wextra -Wwrite-strings -Wno-parentheses -Wpedantic -Warray-bounds -Wconversion

And you can then build simply with

make putchar

When you've enabled warnings, you'll see:

198374.c:3:6: warning: return type of ‘main’ is not ‘int’ [-Wmain]
 void main()
 ^~~~

The fix for that should be obvious.

Now to the meat of the program. Firstly, before the loop we have a statement

 c = getchar();

This consumes and ignores one character. We probably didn't mean to throw that one away.

Now consider this expression:

 (c = getchar()) != EOF == (0 | 1)

We're actually comparing (after the assignment) c != EOF against 0|1. But | is the bitwise-or operator, so 0|1 evaluates to just 1. When c is not EOF, then c!=EOF is 1, and the condition is satisfied. The last time around the loop (when c becomes EOF), then c!=EOF is 0, and the loop exits.

What we want to do is save the result of c != EOF and compare that to 0 and to 1 separately. Or, even more simply for this exercise, just print it to standard output:

 do 
 c = getchar();
 printf("getchar() != EOF: %dn", c != EOF);
 while (c != EOF);

NoteL given that we haven't been asked to copy the input to output, we shouldn't have the putchar() call.

If we really want to prove that c != EOF is always either 0 or 1, we could test it automatically:

 do 
 c = getchar();
 if ((c != EOF) != 0 && (c != EOF) != 1) 
 fprintf(stderr, "(c != EOF) was neither 0 nor 1n");
 return 1;
 
 while (c != EOF);

You might choose to use a switch statement instead of the if; I think that's clearer:

 c = getchar();
 switch (c != EOF) 
 case 0: /* okay */
 case 1: /* okay */
 break;
 default:
 fprintf(stderr, "(c != EOF) was neither 0 nor 1n");
 return 1;
 

Modified code

#include <stdio.h>

int main()

 int c;

 do 
 c = getchar();
 switch (c != EOF) 
 case 0: /* okay */
 case 1: /* okay */
 break;
 default:
 fprintf(stderr, "(c != EOF) was neither 0 nor 1n");
 return 1;
 
 while (c != EOF);

 printf("%dn", c);