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awk syntax doubt - trailing 1 value

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# 1  
Old 10-26-2008
awk syntax doubt - trailing 1 value
In the below awk syntax, what does the value '1' signify?

awk '{....}1' file

some eg:
Code:
awk '{gsub(/[^[:cntrl:]]/,"")}1'
awk 'BEGIN{ORS=""}1'
awk '{ORS=(!(NR%5)?"":"\n")}1'

# 2  
Old 10-26-2008
awk evaluates the 1 as true and the prints the entire line by default, including a newline.
Code:
echo 'foo' | awk '/foo/{print}1'

Take a look here: http://sunsite.ualberta.ca/Documenta...l#SEC_Contents
# 3  
Old 10-28-2008
Quote:
Originally Posted by danmero
awk evaluates the 1 as true and the prints the entire line by default, including a newline.
Code:
echo 'foo' | awk '/foo/{print}1'

Take a look here: The GNU Awk User's Guide: Table of Contents
But, what does 1 serve here or what is the role/purpose of 1?
Anyway, the default action of awk is to print the record right, even if the 'print' statement is missing?
Last edited by royalibrahim; 10-28-2008 at 06:34 AM..
# 4  
Old 10-28-2008
Quote:
Originally Posted by royalibrahim
But, what does 1 serve here or what is the role/purpose of 1?
Anyway, the default action of awk is to print the record right, even if the 'print' statement is missing?
Don't take everything for granted.
Code:
$ echo 'foo' | awk '/foo/'            #default action
foo

$ echo 'foo' | awk '/foo/{print}'     #defined action
foo
$ echo 'foo' | awk '/foo/{$1="bar"}'  #defined action

$ echo 'foo' | awk '/foo/{print}1'    #defined action and default action
foo
foo

$ echo 'foo' | awk '/foo/{$1="bar"}1'
bar

# 5  
Old 10-29-2008
Thanks danmero, I got the essence !! hat's off
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