Shell Programming and Scripting

Hi,

I have 3 doubts which I posted it here.

Doubt 1:
I have a file containing data:

22 -73 89 10
99 21 15 -77
23 63 -80 91
-22 65 28 97

I am trying to print the fields in the reverse order and replace every field by its absolute (positive) value
(ie.) I am looking for output:

10 89 73 22
77 15 21 99
91 80 63 23
97 28 65 22

when I use the following command

Code:

awk '{ for ( i = NF; i > 0; i-- ) if ($i < 0) $i = -$i; print $i }' <file>

it gives o/p

22 73 89 10
99 21 15 77
23 63 80 91
22 65 28 97

How do I get the output which I need?


Doubt 2:
what is the difference between the following 2 sed commands? And, also please explain the logic behind these commands, it's more obscure to understand.
It does remove all linefeed chars and make multiple lines into a single line.

Code:

sed ':a;$!{N;s/\n//;ba;}'
sed ':a;$!{N;ba;};$s/\n/ /g'

Doubt 3:
what is the difference between the following 2 shell commands?
I know they both sets the positional parameters to the values of $var

Code:

set $var  
set -- $var

The print $i is not inside the for loop so it simply evaluates to print $0 which of course prints the whole line.

The sed scripts mainly differ in how newlines are neutralized. The first will read a line, remove its newline, and loop. The second reads in lines as long as there are any, and then finally replaces all newlines with a space each, as a global replacement.

The double dashes are just a security device; if $var were to contain "-x" then the command would be set -x which does not set the positional parameters of the shell. The double dash says that even if the following argument were to start with a dash, it's not an option. It's good form to always put the double dash when you want to set the positional parameters.

It might be better to post separate threads for separate questions.

1.) awk '{ gsub(/-/,"",$0); for ( i = NF; i > 0; i-- ) printf("%s ",$i); printf("\n");}' <file>

the another way to do the 1st :

Code:

awk '{for (i=NF; i>=1; i--) printf ("%d ",($i>0) ? $i : -1*$i); printf "\n"}'  /path/of/file

Code:

sed ':a;$!{N;s/\n//;ba;}'

maybe i can explain some details of this script:
:a : the label of the branch (u will see ba in the end of script), u can use any alphabet as u wish, so that's a loop of sed
$!: except the last line will do the following actions :
N: read the next line into the pattern space, so, there are two lines in the pattern space
s/\n/ /: replace the \n with the " ", so that will combine the two lines to one
ba: return to the top of the script until the all lines have been read, print it!

i hope that will helpful :-)

Great explanations. Thanks to era, sudhamacs, gnsxhj

Now, I can explain this script's logic by my own:

Code:

sed ':a;$!{N;ba;}; $s/\n/ /g'

:a -- label of the branch that acts as a loop of sed until ba, any alphabet can be used
$! -- except the last line do the following actions:
N -- read next line into the pattern space (adding up multiple lines separated by '\n')
ba -- return to the top of the script until all the lines are read and then print it
s/\n/ /g -- replace '\n' with "<space>" globally in the pattern space, that combines all the lines in to one

Code:

nawk '{
for (i=NF;i>=1;i--)
printf("%s ",($i>0?$i:(0-$i)))
print ""
}' filename