concatenating regular expressions

Can someone explain this output?

$ print 'a b c' | awk '{ print gsub(/x/, "") }'
0

$ print 'a b c' | awk '{ print gsub(/x/, "") gsub(/y/, "") }'
00

$ print "a b c" | awk '0'
$ print "a b c" | awk '00'

$ print "a b c" | awk 'gsub(/x/, "")'
$ print "a b c" | awk 'gsub(/y/, "")'
$ print "a b c" | awk 'gsub(/x/, "") gsub(/y/, "")'
a b c

--
Regards,

---Robert

On Sun, 17 Apr 2005 at 22:06 GMT, Robert Katz wrote:
> Can someone explain this output?
>
> $ print 'a b c' | awk '{ print gsub(/x/, "") }'
> 0
>
> $ print 'a b c' | awk '{ print gsub(/x/, "") gsub(/y/, "") }'
> 00
>
> $ print "a b c" | awk '0'
> $ print "a b c" | awk '00'
>
> $ print "a b c" | awk 'gsub(/x/, "")'
> $ print "a b c" | awk 'gsub(/y/, "")'
> $ print "a b c" | awk 'gsub(/x/, "") gsub(/y/, "")'
> a b c

See my response to your other post ("0 0" is a string).


--
Chris F.A. Johnson                  http://cfaj.freeshell.org/shell
 ========================================
===========================
My code (if any) in this post is copyright 2005, Chris F.A. Johnson
and may be copied under the terms of the GNU General Public License

In article <u5B8e.4031$z9.355@news.cpqcorp.net>,
Robert Katz <katz@hp.com> wrote:

> Can someone explain this output?
>
> $ print 'a b c' | awk '{ print gsub(/x/, "") }'
> 0

Default Pattern, print return value from gsub, aka 0, no match found

> $ print 'a b c' | awk '{ print gsub(/x/, "") gsub(/y/, "") }'
> 00

Default Pattern, concatenate results from 2 gsub calls (00 because there
was no result found for either substitution), print concatenated result

> $ print "a b c" | awk '0'
> $ print "a b c" | awk '00'

Default Action, do nothing because pattern is 0, aka FALSE

> $ print "a b c" | awk 'gsub(/x/, "")'
> $ print "a b c" | awk 'gsub(/y/, "")'

Default Action, do nothing because gsub returned 0 (no match) and that
is the Pattern

> $ print "a b c" | awk 'gsub(/x/, "") gsub(/y/, "")'
> a b c

This one is tricky.  Both gsub's return 0, because they do not find
anything, _HOWEVER_, you then concatenate the result with is a string
operation, you do not have the number 00, you have the string "00" and
it is not a null string, so you non-null strings are treated a TRUE,
thus, the Default Action is taken, aka print $0

The basic rule of Grep is:

Pattern { Action }

Action performed if Pattern is true

Default Action:  print $0
Example:
/a/
$1 == "a"
$0 == "a b c"
NF != 0
Default Pattern: true
Example:
{ gsub(/a/,""); print }
{ gsub(/x/,""); print }
{ x = 2 + 2; print x }

I suggest you read the following web page:
<http://h30097.www3.hp.com/docs/base...B_HTML/ARH9WBTE
/WKXXXXXX.HTM>

I think it will help a lot in getting a starting understanding of Awk.
After that, if you really want to learn more "The Awk Programming
Language" book, by Aho,,Kernighan Weinberger is very good, but a bit
expensive.

Bob Harris

Bob Harris wrote:
> In article <u5B8e.4031$z9.355@news.cpqcorp.net>,
>  Robert Katz <katz@hp.com> wrote:
>
> 
>
>
> Default Action, do nothing because pattern is 0, aka FALSE
>
> 
>
>
> Default Action, do nothing because gsub returned 0 (no match) and that
> is the Pattern
>
> 
>
Here is really the essence of my question.

Is this what happens?  Since in the line
$ print "a b c" | awk 'gsub(/x/, "")'

no substitutions are made, the line becomes equivalent to
$ print "a b c" | awk '0'

which will not print anything, but two consecutive returns get concatenated 
to a string, so that

$ print "a b c" | awk 'gsub(/x/, "") gsub(/y/, "")'
becomes
$ print "a b c" | awk '"00"'
a b c

and not
$ print "a b c" | awk '00'

>
> This one is tricky.  Both gsub's return 0, because they do not find
> anything, _HOWEVER_, you then concatenate the result with is a string
> operation, you do not have the number 00, you have the string "00" and
> it is not a null string, so you non-null strings are treated a TRUE,
> thus, the Default Action is taken, aka print $0

Does this imply that
'expr expr ... '

will always print regardless of the return values of any of the expressions?

--
Regards,

---Robert

On Mon, 18 Apr 2005 at 02:21 GMT, Robert Katz wrote:
> Bob Harris wrote: 
>
> Does this imply that
>   'expr expr ... '
>
> will always print regardless of the return values of any of the expressions?[/colo
r]

If the string concatenation is not empty, yes; but not for:

awk 'substr("a",1,0) substr("x",1,0)'

--
Chris F.A. Johnson                  http://cfaj.freeshell.org/shell
 ========================================
===========================
My code (if any) in this post is copyright 2005, Chris F.A. Johnson
and may be copied under the terms of the GNU General Public License

Robert Katz wrote:
<snip>
> Does this imply that
>  'expr expr ... '
>
> will always print regardless of the return values of any of the
> expressions?
>

No. Not if the result of expr expr is itself the null string, e.g. if
each expr results in a null string it's the same as:

PS1> print "a b c" | gawk '""'
PS1> print "a b c" | gawk '""""'

which doesn't print. See my answer in your previous thread for details.

Ed.

Robert Katz wrote:
<snip>
> Does this imply that
>  'expr expr ... '
>
> will always print regardless of the return values of any of the
> expressions?
>

No. Not if the result of expr expr is itself the null string, e.g. if
each expr results in a null string it's the same as:

PS1> print "a b c" | gawk '""'
PS1> print "a b c" | gawk '""""'

which doesn't print. See my answer in your previous thread for details.

Ed.