Counting chars in C on Mac OS X

I am running an example from The C Programming Language by Kernighan and Ritchie.

#include <stdio.h>

main() {
    long nc;
    nc = 0;
    while (getchar() != EOF)
        ++nc;
    printf("%ld\n", nc);
}

I'm on Mac OS X, so I compile it, run it, type in "12345", press enter for newline (I believe newline is the sixth character?) and then press ctrl-D to send an EOF.

It prints out "6D".

Why is the "D" there?

How do I write a program to just count the 5 chars in "12345" and not the newline?

Should I just subtract one at the end?

How开发者_运维技巧 do I get it to stop printing the "D"?

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What happens is that the terminal actually echoes your control-D (and prints it out as ^D) when you type it, but then your program overwrites that line with a number and a line-feed. So your one-digit number overwrites the ^, but the D stays there.

If you enter more than 10 characters, or if change the code by adding a space at the end of your format string ("%ld \n"), then your program would overwrite the ^D (though it would still have been echoed by your terminal)

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The program is working correctly: it's printing 6.

You see 6D because the terminal window printed ^D then went back to the beginning of the line, and your program prints 6 over the ^ leaving the following D. Try redirecting the output to a file, or giving it enough input that the answer has more than one digit, and you won't see the D.

The answer is 6 instead of 5 because of the newline, yes. If you don't want to count the newline at the end, subtract 1. If you don't want to count any newlines, subtract 1 whenever you see a newline.