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

[Antonio Diaz Diaz]

John Cowan wrote:
> The Posix standard says "Within the text of that shell command line, the
> unescaped character '%' shall be replaced with the remembered pathname".

Which is what GNU ed does.

> 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.

What "current GNU behavior" are you talking about?

> 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.

And is also what GNU ed has done since 1994.

> 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.

Agreed.

> So I argue that the sequence '\%' in a '!' command should send '%' to the
> shell without a backslash, and that the man page be fixed.

Agreed.

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

Not so agreed. Now you can write !printf '\n' to get the output '\n'.

$ printf '%s\n' 'r !printf "% modified on \n \%d\n" 23' ',n' Q | ed -s fil1
printf "fil1 modified on \n \%d\n" 23
1       fil1 modified on
2        \23

Best regards,
Antonio.