Re: Characters saved mismatch?

[Kaushal Modi <kaushal.modi@gmail.com>]

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On Sun, May 7, 2017, 5:16 AM Angelo Graziosi <angelo.graziosi@alice.it>
wrote:

>
>
> Someone should explain the meaning of "Wrote ‘c:/msys64/tmp/foo.text’ (8
> characters)".
>

As others stated, that's because each newline is counted as 1 character
too.

If it refers to the number of characters, my example contains 6
> characters: f-o-o-b-a-r and not 8.
>

No, it contains 8 characters:

1. f
2. o
3. o
4. Newline (Just 1 character, does not matter if it is 1 byte on unix or 2
bytes on Windows. This is character count, not byte count.)
5. b
6. a
7. r
8. Newline

As I wrote, in Windows Emacs uses DOS style, more precisely 'utf-8-dos'.
> That should mean 1 byte/ch and CR+LF for end line (RET). This mean that
>
> foo RET
> bar RET
>
> should contain (3+2) * 2 = 10 bytes as, 'ls' shows..
>

As written about emacs sees the newline as just 1 character. Emacs is
printing character count, while ls is printing byte count, and thus the
difference.

Visualize that newline character as just 1 character as the "\n" used in
regexps to match newlines.

Then, where does "Wrote ‘c:/msys64/tmp/foo.text’ (8 characters)" came
> from, on Windows?
>

As explained above.

> --

Kaushal Modi

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