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# 1  
Old 03-05-2018
While read
Hello from Paris,

Here is a script that i wrote. I nammed this script sum.sh :

Code:
#! /bin/bash

sum=0
cat $1 | while read line
do
#set $line
sum=`expr $sum  +  1`
done
echo $sum

When I execute this script with a 4 lines file as argument, it returns 0 and not 4 as i expected :

Quote:
ghassen@ghassen-Lenovo-V110-15ISK:~/Bash$ cat fic
line 1
line 2
line 3
line 4
ghassen@ghassen-Lenovo-V110-15ISK:~/Bash$ ./sum.sh fic
0
ghassen@ghassen-Lenovo-V110-15ISK:
Somebody could explain to me why and how i can change this script to get 4 as return ?

Thank you !
# 2  
Old 03-05-2018
such as it is (although there's a program for that, called wc Smilie):
Code:
sum=0
while read line; do
  sum=$((sum + 1))
done < $1
echo $sum

Code:
$ cat -n myFile
     1	this
     2	is
     3	a
     4	file
     5	here

$ ./myScript
5

This User Gave Thanks to Scott For This Post:
Ghassen
# 3  
Old 03-05-2018
The reason is when you pipe the file to a while loop, it actually gets executed in a sub-shell:-
Code:
cat $1 | while read line

So the scope of the variable sum value is within the while loop. This is the reason why it prints 0 outside the while loop

You can resolve this by removing the pipe:-
Code:
sum=0
while read line
do
#set $line
sum=`expr $sum  +  1`
done < "$1"
echo $sum

These 2 Users Gave Thanks to Yoda For This Post:
Ghassen Scott
# 4  
Old 03-05-2018
Quote:
Originally Posted by Yoda
The reason is when you pipe the file to a while loop, it actually gets executed in a sub-shell:-
Code:
cat $1 | while read line

So the scope of the variable sum value is within the while loop. This is the reason why it prints 0 outside the while loop

You can resolve this by removing the pipe:-
Code:
sum=0
while read line
do
#set $line
sum=`expr $sum  +  1`
done < "$1"
echo $sum

Being a Korn shell user, myself, I often overlook the pipe / sub-shell thing in Bash.. it's good that you pointed that out to the OP Smilie
# 5  
Old 03-05-2018
Thank you both for your answers !
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