Question

For reference, I’m copying this example (nearly) letter for letter from page 18 in The C Programming Language, Second Edition.

#include <stdio.h>
/*count characters in input, 2nd version*/
main(){
    double n;
    for (n = 0; getchar() != EOF; ++n)
        ;
    printf("%.0f\n", n); /*this never prints*/
}

I wasn’t sure if it was my version of gcc (I’m a noob):

% gcc --version
gcc (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5
Copyright (C) 2010 Free Software Foundation, Inc.

Because

for (n = 0; getchar() != EOF; ++n)
    printf("%.0f",n); /*returns 0123456...n*/

I tried printing “foo” instead of value n, just to see. It still isn’t printing.

What am I missing here? I don’t like moving ahead while ignoring little problems like these.

=====================================

EDIT

So the end result should be:

% gcc ./counter.c -o ./counter
% ./counter
foo
3

Right now, this is the output from the first snippet:

% ./counter
foo
0123

And the second one:

% ./counter
foo

^C
%

Answer 1

If you are on an OSX or linux box, you need to type Ctrl+D on its own line to generate an EOF character. On Windows, Ctrl+Z on its own line. Don’t type Ctrl+Z on a unix box because that will just send your proces to the background.

You are typing Ctrl+C which is break, and will send a SIGTERM to your program.

On my mac, I get:

$ ./foo
hello
6D

Or if you don’t want to signal the EOF condition, use echo and a pipe:

$ echo "hello" | ./foo 
6

Note that the EOF has to be on its own line. a ^D is printed, and then 6 overwrites the ^, so it looks like the output is 6D.

Of course, in the two above examples, the characters being counted are h e l l o \n. If you don’t want a newline, do this:

$ echo -n "hello" | ./foo 
5