Quoting in zsh -x output

Quoting in zsh -x output

[Hrvoje Niksic]

Thanks for looking into this.  Dan, you are correct in your conclusion
that I didn't ask for the original quoting to be preserved, but only
for quoting to be consistent.  The examples I gave in fact confirmed
that.  I hope the patch will go in; thanks for writing it.

Bart, I agree that this patch will not allow pasting of more complex
syntactic constructs.  But that's not the case in Bash either, and I'm
not sure if it's even generally possible with -x.  Being able to
correctly copy one command is already (IMHO) a great thing.

For example, I have a fairly complex script called `reencode' that
accepts simple arguments and uses them to construct the complex
command line with which to call the actual encode utility, `mencoder'.
For debugging and educational purposes, it's useful to see what the
script is doing.  The way I did it is:

    set -x
    mencoder $foo_opts $bar_opts_maybe -o "$dest" "$source" ...
    set +x

With this I get a `make'-like feature of seeing the commands that are
being run.  Full quoting of output brings this to the next level: it
allows me to rerun the command manually, possibly after making a small
adjustment to the options.  `make' doesn't support this, but it's
really neat and very easy to get used to.

Again, thanks for implementing this.

Re: Quoting in zsh -x output

[Bart Schaefer]

On Jul 31, 10:42am, Peter Stephenson wrote:
} Subject: Re: Quoting in zsh -x output
}
} Dan Nelson wrote:
} > I think all bash does is check each argument for spaces, single-quotes,
} > or backslashes, and then quotes and escapes that argument.
} 
} Ah, that's a different kettle of fish entirely.  I don't see any problem
} including the patch.  I doubt if anyone is attached to the current
} ambiguous behaviour of xtrace, are they?

I don't have any particular problem with the patch -- though I'm not sure
it covers all possible places where trace output is produced; I haven't
checked -- but I would point out that there are a whole lot of things
beyond mere argument quoting that prevent zsh's xtrace output from being
cut'n'paste-able.  Individual commands and their arguments might be, but  
for almost any more complex syntactic structure you're out of luck.

-- 
Bart Schaefer                                 Brass Lantern Enterprises
http://www.well.com/user/barts              http://www.brasslantern.com

Zsh: http://www.zsh.org | PHPerl Project: http://phperl.sourceforge.net   

Re: Quoting in zsh -x output

[Peter Stephenson]

Dan Nelson wrote:
> I think all bash does is check each argument for spaces, single-quotes,
> or backslashes, and then quotes and escapes that argument.  It doesn't
> care what quoting the user had on the commandline:
> 
> $ echo one\ two 'one two' "one two"
> + echo 'one two' 'one two' 'one two'

Ah, that's a different kettle of fish entirely.  I don't see any problem
including the patch.  I doubt if anyone is attached to the current
ambiguous behaviour of xtrace, are they?

-- 
Peter Stephenson <pws@csr.com>                  Software Engineer
CSR Ltd., Science Park, Milton Road,
Cambridge, CB4 0WH, UK                          Tel: +44 (0)1223 392070


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Re: Quoting in zsh -x output

[Dan Nelson]

[-- Attachment #1: Type: text/plain, Size: 1500 bytes --]

In the last episode (Jul 30), Peter Stephenson said:
> Hrvoje Niksic wrote:
> > I suggest modifying the zsh `-x' output to quote its arguments the
> > way Bash does.  What do the others think?
> 
> This would be nice, but it's not a trivial change.  The xtrace output
> is done at the last minute after quoting information has been
> stripped.  This is deliberate, in order to output exactly what is
> being evaluated after all substitutions, which is the information you
> can't get any other way.  This is at odds with showing all the
> quotes.  It's possible there's some compromise position which would
> show a lot of everything --- the quotes get stripped at quite a late
> stage.  However, there's no way of doing it perfectly without a great
> deal of work --- consider expansions like "$@".

I think all bash does is check each argument for spaces, single-quotes,
or backslashes, and then quotes and escapes that argument.  It doesn't
care what quoting the user had on the commandline:

$ echo one\ two 'one two' "one two"
+ echo 'one two' 'one two' 'one two'

I did a quick hack that replaced a bunch of fprintf/zputs with
quotedzputs, and it seems to work fine.  It ends up quoting "one'two"
as 'one'\''two' for some reason, single-quoting "one",
backslash-escaping "'", and single-quoting "two".  Weird, but valid.
This is my first zsh hack, and I'm probably doing everything wrong, but
it shows that it's really not difficult.  Even "$@" works.

-- 
	Dan Nelson
	dnelson@allantgroup.com

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--- Src/builtin.c~	Tue Oct 16 04:49:17 2001
+++ Src/builtin.c	Tue Jul 30 15:01:53 2002
@@ -356,10 +356,14 @@ execbuiltin(LinkList args, Builtin bn)
 	if (xtr) {
 	    printprompt4();
 	    fprintf(xtrerr, "%s", name);
-	    if (xarg)
-		fprintf(xtrerr, " %s", xarg);
-	    while (*oargv)
-		fprintf(xtrerr, " %s", *oargv++);
+	    if (xarg) {
+	        fputc(' ', xtrerr);
+	        quotedzputs(xarg, xtrerr);
+	    }
+	    while (*oargv) {
+	        fputc(' ', xtrerr);
+	        quotedzputs(*oargv++, xtrerr);
+	    }
 	    fputc('\n', xtrerr);
 	    fflush(xtrerr);
 	}
--- Src/cond.c~	Wed Oct 24 11:03:02 2001
+++ Src/cond.c	Tue Jul 30 15:29:48 2002
@@ -147,9 +147,14 @@ evalcond(Estate state)
 		singsub(&rt);
 		untokenize(rt);
 	    }
-	    fprintf(xtrerr, " %s %s %s", left, condstr[ctype], rt);
-	} else
-	    fprintf(xtrerr, " -%c %s", ctype, left);
+	    fputc(' ',xtrerr);
+	    quotedzputs(left, xtrerr);
+	    fprintf(xtrerr, " %s ", condstr[ctype]);
+	    quotedzputs(rt, xtrerr);
+	} else {
+	    fprintf(xtrerr, " -%c ", ctype);
+	    quotedzputs(left, xtrerr);
+	}
     }
 
     if (ctype >= COND_EQ && ctype <= COND_GE) {
--- Src/exec.c.orig	Wed Oct 24 06:16:32 2001
+++ Src/exec.c	Tue Jul 30 14:56:50 2002
@@ -1275,7 +1275,7 @@ makecline(LinkList list)
 
 	for (node = firstnode(list); node; incnode(node)) {
 	    *ptr++ = (char *)getdata(node);
-	    zputs(getdata(node), xtrerr);
+	    quotedzputs(getdata(node), xtrerr);
 	    if (nextnode(node))
 		fputc(' ', xtrerr);
 	}
@@ -1544,8 +1544,10 @@ addvars(Estate state, Wordcode pc, int e
 		untokenize(peekfirst(vl));
 		val = ztrdup(ugetnode(vl));
 	    }
-	    if (xtr)
-		fprintf(xtrerr, "%s ", val);
+	    if (xtr) {
+		quotedzputs(val, xtrerr);
+		fputc(' ', xtrerr);
+	    }
 	    if (export && !strchr(name, '[')) {
 		if (export < 0 && isset(RESTRICTED) &&
 		    (pm = (Param) paramtab->removenode(paramtab, name)) &&
@@ -1583,8 +1585,10 @@ addvars(Estate state, Wordcode pc, int e
 	*ptr = NULL;
 	if (xtr) {
 	    fprintf(xtrerr, "( ");
-	    for (ptr = arr; *ptr; ptr++)
-		fprintf(xtrerr, "%s ", *ptr);
+	    for (ptr = arr; *ptr; ptr++) {
+		quotedzputs(*ptr, xtrerr);
+		fputc(' ', xtrerr);
+	    }
 	    fprintf(xtrerr, ") ");
 	}
 	setaparam(name, arr);
@@ -3187,7 +3191,7 @@ execshfunc(Shfunc shf, LinkList args)
 	    for (lptr = firstnode(args); lptr; incnode(lptr)) {
 		if (lptr != firstnode(args))
 		    fputc(' ', xtrerr);
-		fprintf(xtrerr, "%s", (char *)getdata(lptr));
+		quotedzputs((char *)getdata(lptr), xtrerr);
 	    }
 	fputc('\n', xtrerr);
 	fflush(xtrerr);

Re: Quoting in zsh -x output

[Peter Stephenson]

Hrvoje Niksic wrote:
> I suggest modifying the zsh `-x' output to quote its arguments the way
> Bash does.  What do the others think?

This would be nice, but it's not a trivial change.  The xtrace output is
done at the last minute after quoting information has been stripped.
This is deliberate, in order to output exactly what is being evaluated
after all substitutions, which is the information you can't get any
other way.  This is at odds with showing all the quotes.  It's possible
there's some compromise position which would show a lot of everything
--- the quotes get stripped at quite a late stage.  However, there's no
way of doing it perfectly without a great deal of work --- consider
expansions like "$@".

You could try the `verbose' option, which works at the input end of the
command chain.

(In case anybody assumes otherwise, I have zero time to work on this
myself for the foreseeable future.)

-- 
Peter Stephenson <pws@csr.com>                  Software Engineer
CSR Ltd., Science Park, Milton Road,
Cambridge, CB4 0WH, UK                          Tel: +44 (0)1223 392070


**********************************************************************
The information transmitted is intended only for the person or
entity to which it is addressed and may contain confidential 
and/or privileged material. 
Any review, retransmission, dissemination or other use of, or
taking of any action in reliance upon, this information by 
persons or entities other than the intended recipient is 
prohibited.  
If you received this in error, please contact the sender and 
delete the material from any computer.
**********************************************************************

Quoting in zsh -x output

[Hrvoje Niksic]

Bash has a nice feature of its `-x' output: the arguments that contain
whitespace or shell metacharacters are quoted, which allows one to
copy the output and paste it into another shell.  I find this very
useful.

Here is an example:

bash$ set -x
bash$ echo "foo bar"
+ echo 'foo bar'
foo bar
bash$ echo "foo*"
+ echo 'foo*'
foo*
bash$ echo "foo'bar"
+ echo 'foo'\''bar'
foo'bar
bash$ echo "foo"
+ echo foo
foo

The last example shows that the quotes are printed only when needed.
Arguments containing whitespace and metacharacters are quoted, and
embedded single quotes are handled correctly.  This makes sure that
pasting the output produces the same results.

Zsh, on the other hand, prints this:

zsh$ set -x
zsh$ echo "foo bar"
+zsh:2> echo foo bar
foo bar
zsh$ echo "foo*"
+zsh:3> echo foo*
foo*
zsh$ echo "foo'bar"
+zsh:4> echo foo'bar
foo'bar
zsh$ echo "foo"
+zsh:5> echo foo
foo

I suggest modifying the zsh `-x' output to quote its arguments the way
Bash does.  What do the others think?