How to add arithmetic variables in a script

Question

I want to accumulate the line size of a number of files contained in a folder. I have written the following script:

let a=0
let num=0
for i in folder/*
do
        num=`cat $i | wc -l`
        a=$a+$num
done
echo $a

What i am geting at the end of the script is 123+234+432+... and not the result of the arithmetic operation of addition.

score 77 · Accepted Answer · edited Nov 07 '12 at 13:54

77

Your arithmetic evaluation syntax is wrong. Use any of the following (the first is extremely portable but slow, the second is POSIX and portable except to the Bourne shell and earlier versions of the Almquist shell, the last three require ksh, bash or zsh):

a=`expr "$a" + "$num"`

a=$(($a+$num))

((a=a+num))

let a=a+num

((a+=num))

Or you can just skip the entire for loop and just do:

wc -l folder/*

Or, if you only want the total:

cat folder/* | wc -l

Or with zsh and its mult_ios option:

wc -l < folder/*

edited Nov 07 '12 at 13:54

Stéphane Chazelas

544,893

answered Nov 07 '12 at 10:36

manatwork

31,277

2

No need for the $ inside ((...)). a=$((a+num)) is already fine. – Weijun Zhou Feb 07 '18 at 08:08
if a variable is not set or having null value, would this work, if no any other way to handle that. – Bharat Sep 14 '18 at 13:10
$((a+num)) is slightly faster than $(($a+$num)), tested in bash. – lxvs Jul 11 '22 at 07:10

score 7 · Answer 2 · edited Jan 22 '16 at 20:02

7

you can also use this code

    a=`expr $a + $num`
    echo $a

and MAKE SURE THAT THERE IS A SPACE ON BOTH SIDES OF + IN "$a + $num"

edited Jan 22 '16 at 20:02

Jeff Schaller

67,283
35
116
255

answered Nov 07 '12 at 10:42

user1678213

594

always give space on both sides of operator when using expr command for calculation. – user1678213 Nov 07 '12 at 10:45

score 4 · Answer 3 · answered Dec 05 '18 at 14:54

The answer needs to specify in which shell the code is valid. For instance in the bourne Shell (sh) only the following instructions are valid:

a=$((a+num))
a=$(($a+$num))

while the other possibilities listed by @manatwork may be valid in bourne again shell (bash)

score 2 · Answer 4 · answered Oct 19 '15 at 04:56

2

You could declare the type of variable first:

    declare -i a=0
    declare -i num=0

answered Oct 19 '15 at 04:56

Code42

121

Leo · Answer 5 · 2018-01-30T22:22:35.870

1

Sorry, previous edit was for a different post. Here, just a small modification to the original script:

let a=0
let num=0
for i in folder/*
do
        num=`cat $i | wc -l`
        a=$(echo $a+$num|bc)
done
echo $a

edited Jan 30 '18 at 22:22

answered Jan 30 '18 at 22:03

Leo

11

1

Voted down, as the answer doesn't appear to address the question. – Time4Tea Jan 30 '18 at 22:07
Sorry guys, now corrected – Leo Jan 30 '18 at 22:22

score -1 · Answer 6 · answered Aug 05 '18 at 14:28

-1

i make it like this

MY_VARIABLE=$((num1+num2))

answered Aug 05 '18 at 14:28

Hatem Badawi

197

4

This adds nothing to what has already been said. – Kusalananda Aug 05 '18 at 14:33
1

it is the brief answer – Hatem Badawi Aug 05 '18 at 16:46

How to add arithmetic variables in a script

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