Why does "wc -c" print a value of one more with echo?

Question

When running

echo abcd | wc -c

it returns 5.
But the word abcd is only 4 characters long.

Is echo printing some special character after the word abcd?

And can I prevent echo from printing that?

Answer 1

echo print newline (\n) at end of line

echo abcd | xxd
0000000: 6162 6364 0a          abcd.

With some echo implementations, you can use -n :

-n do not output the trailing newline

and test:

echo -n abcd | wc -c
4

With some others, you need the \c escape sequence:

\c: Suppress the <newline> that otherwise follows the final argument in the output. All characters following the '\c' in the arguments shall be ignored.

echo -e 'abcd\c' | wc -c
4

Portably, use printf:

printf %s abcd | wc -c
4

(note that wc -c counts bytes, not characters (though in the case of abcd they are generally equivalent). Use wc -m to count characters).

Answer 2

If you run echo without the -n option, it writes a newline character after the arguments.  Otherwise, if you typed echo foo, your next shell prompt would appear to the right of the foo.  So wc is counting the newline.

How can I prevent echo from printing that?

echo -n abcd | wc -c

Answer 3

By default, echo will print a newline character (\n) after the string (which is why you see the shell prompt on the next line instead of the same line where abcd is printed.)

You can try adding a -n parameter to echo to disable the trailing newline character.

echo -n abcd | wc -c
4