syntax explanation needed

There have been many threads that required a re-arrangement of the input
record.  The first proposal was '$1 = $1'

Followups pointed out the the line won't print if $1 = "" or $1 == 0, so
a fix to that was '{$1 = $1}1'

Now I can explain that, I said to myself; since } ends a line, the above
is equivalent to

'   {$1 = $1}
1'

The first action line rearranges the input line, the second expression
is always true, so now we print the rearranged line.

Kenny countered the solution of '{$1 = $1}1' with '($1 = $1)1'.  Now
that I'm not sure I understand.

What does 'expr expr expr' mean?
When is 'expr expr expr' the same as 'expr || expr || expr' ?

what does '($1 = $1)1' do? and why is it better than '{$1 = $1}1' ?

--
Regards,

---Robert

In article <88izd.4994$n05.4427@news.cpqcorp.net>,
Robert Katz <katz@hp.com> wrote:

> There have been many threads that required a re-arrangement of the input
> record.  The first proposal was '$1 = $1'
>
> Followups pointed out the the line won't print if $1 = "" or $1 == 0, so
> a fix to that was '{$1 = $1}1'
>
> Now I can explain that, I said to myself; since } ends a line, the above
> is equivalent to
>
>      '   {$1 = $1}
>       1'
>
> The first action line rearranges the input line, the second expression
> is always true, so now we print the rearranged line.
>
> Kenny countered the solution of '{$1 = $1}1' with '($1 = $1)1'.  Now
> that I'm not sure I understand.
>
> What does 'expr expr expr' mean?
> When is 'expr expr expr' the same as 'expr || expr || expr' ?
>
> what does '($1 = $1)1' do? and why is it better than '{$1 = $1}1' ?

This is a guess, but I think the (...)1 is just awk concatenation of
(...) and 1 so that the result of (...) is always non-zero.  For example

echo 0 | awk '($1 = $1)'

will not print anything, but

echo 0 | awk '($1 = $1)1'

will print

0

Bob Harris

Bob Harris wrote:
> In article <88izd.4994$n05.4427@news.cpqcorp.net>,
>  Robert Katz <katz@hp.com> wrote:
>
> 
>
>
> This is a guess, but I think the (...)1 is just awk concatenation of
> (...) and 1 so that the result of (...) is always non-zero.  For example
>
>         echo 0 | awk '($1 = $1)'
>
> will not print anything, but
>
>         echo 0 | awk '($1 = $1)1'
>
> will print
>
>         0
>
>                                         Bob Harris
Yup Bob,

According to

http://www.opengroup.org/onlinepubs...lities/awk.html

the result of

expr expr is string concatenation, but I don't quite understand how the
resulting string affects the input line.  Are the two expr applied
sequentially from left to right?


--
Regards,

---Robert

In article <o2szd.4997$Pf5.1016@news.cpqcorp.net>,
Robert Katz <katz@hp.com> wrote:

> Bob Harris wrote: 
> Yup Bob,
>
> According to
>
> http://www.opengroup.org/onlinepubs...lities/awk.html
>
> the result of
>
> expr expr is string concatenation, but I don't quite understand how the
> resulting string affects the input line.  Are the two expr applied
> sequentially from left to right?

Starting with

pattern { action }

where  the default action is "print $0" if the pattern is true
(non-zero, or non-null string).

So for most things

($1 = $1)

will be non-zero and/or a non-null string, so the default action of
"print $0" will occur.

However if the input is '0' or a null string, then nothing would be
printed.

But by concatenating 1 to the result of ($1 = $1), you are assured to
always have a non-zero or non-null result, so you would always print $0

Bob Harris