Re: Escaped percent sign in shell escape (bang) command (ed 1.16)

[John Cowan]

Oops, sent prematurely:

I further argue that '\\' should send a single backslash to the shell, so
that you can write !echo '\\n" to get the output '\n'.

On Tue, Jul 13, 2021 at 11:49 AM John Cowan <cowan@ccil.org> wrote:

> The Posix standard says "Within the text of that shell command line, the
> unescaped character '%' shall be replaced with the remembered pathname".
> Now if "unescaped" means anything at all here, it means "not preceded by a
> backslash", just like everywhere else in Posix.  So the current GNU
> behavior is unequivocally wrong.
>
> Both the GNU and BSD man pages say that escape sequences are not treated
> specially in the "!" command, and this also seems to be wrong or at best
> misleading.  The BSD behavior of "echo '\%' is to output "\%", which is
> certainly not illegal.
>
> But we must ask if it is the most useful behavior.  I contend that it is
> not.  If you want to pass a '%' character to the shell, the obvious way to
> do so is by writing '\%', just as the way to get '*' interpreted literally
> in a regex is to write '\*'.  The BSD behavior is not only surprising, it
> makes it impossible to send a simple '%' to the shell at all.  This is the
> problem which escape sequences were designed to solve.
>
> So I argue that the sequence '\%' in a '!' command should send '%' to the
> shell without a backslash, and that the man page be fixed.  I further argue
> that '\\' s
>