problem with expr command

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1

:slight_smile:
hi Unix gurus,
Pls consider the following piece of code
str='hello'
length=echo $str|wc -c
echo $length
y= ` expr \( 80 - $length \) `
echo $y
:confused:
The last echo stmt is displaying 0 as the result.
If i put direct value like 6 instead of $length in i 3rd stmt it is giving the correct result.
I came to know that $length is not getting accessed inside of expr command.

any help pls.

cheers
Ravi Raj Kumar

2

You are not assigning value to variable length correctly, try using backquotes there as well

length=`echo $str|wc -c`
3

Another way of doing it (bash)

str='hello'
length=${#str}
y=$(( 80 - $length ))
4

:slight_smile:
hi Ripat,
Thank u for ur reply.The method u suggested is working fine.
But i want it to be worked in the following way.

str='hello'
length=`echo $str|wc -c`
echo $length
y= ` expr \( 80 - $length \) `
echo $y

cheers
Ravi Raj Kumar

5

str='hello'
length=`echo $str|wc -c`
echo $length
y=`expr 80 - $length`
echo $y

6

try this

y=`expr 23 - $length`
7

:slight_smile:

Fine the above code is working.
But i wonder that ,why its not giving result,if we put parantheses.
we must to put parantheses in arithmetc expressions na.
suppose we have a long expression like 10 + 20 -30 + 40 *2 ,then will the above technique works :confused:

any help pls

cheers
Ravi Raj Kumar :slight_smile:

8 
expr 10 + 20 - 30 + \( 40 \* 2 \)

This works if you give a space between operator and operand

y=`expr \( 23 - $len \)`
9

you can also use let

let " y=(80-$x)*10/2+$c " etc. between " " just aritmetical operations no ekstra things.