why does printf output include a trailing 'D'

I am a newbie C coder. I am wondering why the following program's printf output show a trailing 'D' after the decimal integer. Note the trailing D after the 8 in the output snippet below.

compiler: gcc version 2.95.4 20020320 [FreeBSD]


here is the output:

---- output ----

% ./count.exe
123
123
8D
^
this is the 'D' I am talking about.

---- snip ----

Code:

---


#include <stdio.h>

/* count characters in input; 1st version */
main()
{
long nc;

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

---

Because you have a 'd' in the line. So your output is actually "8d", not "8D" as you have in your example. IE: That isn't your actual output, it's your illustration of it, which has a 'D' that really is a 'd' in the above code.


Quzah

Actually the %ld tells printf that hte correcsponding argument is a long integer.

let me get out a different Program that does not use the %ld specification. it is still strange because the output count displays as '2D'

here is the output:

--- output ---

hello
there
2D

--- output ---

Code:

---


#include <stdio.h>

/* count lines in input */
main()
{
  int c, nl;
 
  nl=0;
  while ((c = getchar()) != EOF)
    if (c == '\n')
      ++nl;
  printf("%d\n", nl);
}

---

Yeah, don't mind me. I have no idea what I was thinking.

They say just waking up is like being drunk. This isn't entirely true, because when you're just waking up, you want to be asleep. Where as when you're drunk, you're usually just fine being the way you are. ;)


Quzah.

You are terminating the program by hitting control-D, right? So, the terminal is echo'ing part of that as a "D". It would probably echo ^D if not for the fact that you overwrite the ^ with a digit. You'll see the D disappears too if you get nl to 10 or above (2 digits).

The D doesn't appear if you tell the same program to read stdin from a file, ie foo < somefile.

okay CWR - thats it. thank you!