bash add integer to string

Question

I've been smashing my head against the wall for 2 hours already but can't get it work right.
Here is what I have:

String, say 133, with "\r" at the end, so it is "133\r".
Integer, which I want to add to the number from the string above.

I tried sed to remove "\r" with sed from that string and just (( )) with hope that math expression will truncate "\r" but still no success. Please, point me to my mistakes!

Thanks in advance!
Below is the file I read my string from and let's pretend that I'm forced to read this all the file (Screenshot from Notepad++).

Does the string end with an ASCII carriage return, or the two literal characters '\' and 'r'? — chepner
– chepner, Commented Jun 6, 2013 at 12:45
LF corresponds to \n, not \r !!! LF is for unix systems and \r is for dos, and corresponds to CR (see here:en.wikipedia.org/wiki/Newline) — bendaizer
– bendaizer, Commented Jun 6, 2013 at 13:02
@bendaizer, I'm totally confused then, cause when I try to sum string and int I'm getting error )2yntax error: invalid arithmetic operator (error token is "\r" — mr.nothing
– mr.nothing, Commented Jun 6, 2013 at 13:07

blue · Accepted Answer · 2013-06-06 12:32:54Z

4

You could use awk:

awk 'BEGIN { var="133\r"; print var+3; }'

outputs 136

or you could use bash parameter expansion:

VAR="133\r"
VAR2=${VAR%\\r}
echo $VAR

outputs: 133, so now that VAR2 contains the number stripped of the \r you can add an integer to it.

answered Jun 6, 2013 at 12:32

blue

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glenn jackman · Accepted Answer · 2013-06-06 13:00:52Z

3

$ a=$'133\r'
$ printf "%s" "$a" | od -c
0000000   1   3   3  \r
0000004
$ echo $((a+3))
")syntax error: invalid arithmetic operator (error token is "
$ echo $(( ${a%[[:space:]]} + 3 ))
136

answered Jun 6, 2013 at 13:00

glenn jackman

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2 Comments

mr.nothing Over a year ago

This is the state of the art! Thanks you a lot for your help!

glenn jackman Over a year ago

of course, the key os to avoid getting the CR into the string in the first place, not dealing with it after the fact. This is a quick dos2unix clone: sed 's/\r$//' file | ...

chepner · Accepted Answer · 2013-06-06 12:48:09Z

3

No need for sed just to remove a substring. bash can do it for you.

#!/usr/bin/env bash

mystr="133\r"
myint=2

newstring=${mystr%\\r}               # Remove the substring "\r" from $mystr
myanswer=$(expr $newstring + $myint) # Add the two numbers
echo $myanswer                       # 135

edited Jun 6, 2013 at 12:48

chepner

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answered Jun 6, 2013 at 12:43

James

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5 Comments

mr.nothing Over a year ago

unfortunately it spits out the following: expr: non-integer argument

chepner Over a year ago

See my comment in the question regarding what is actually at the end of your string.

chepner Over a year ago

Use the standard $(( newstring + myint )) in place of the expr command.

mr.nothing Over a year ago

@chepner When I change $(expr $newstring + $myint) to $(( newstring + myint )) it says ")2yntax error: invalid arithmetic operator (error token is "\r

chepner Over a year ago

You have a literal carriage return, not a two-character sequence, at the end of the string. Use ${mystr%?} as a simple standard way to remove it. If there is a chance that mystr ends with something other than a carriage return, you can use the bash-specific (although slated to be included in the standard, I believe) ${mystr%$'\r'}.

Community · Accepted Answer · 2017-05-23 10:32:47Z

2

Simple solution

a="133\r"
b=${a/\\r/}

or even simpler

b=${a%\\r}

as it was suggested by chepner

edited May 23, 2017 at 10:32

CommunityBot

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answered Jun 6, 2013 at 12:35

Grzegorz Żur

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1 Comment

chepner Over a year ago

Even simpler (and standard): ${a%\\r}.

unxnut · Accepted Answer · 2013-06-06 12:33:34Z

1

You can get the number using sed as

sed 's/\([0-9]*\).*/\1/'

answered Jun 6, 2013 at 12:33

unxnut

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Comments

bendaizer · Accepted Answer · 2013-06-06 13:06:42Z

1

If you want to remove \r with sed, don't forget to escape the \, this works for me :

echo "133\r" | sed 's/\\r//'

and outputs 133

EDIT :

as I said in my comment : "LF corresponds to \n, not \r !!! LF is for unix systems and \r is for dos, and corresponds to CR (see here:en.wikipedia.org/wiki/Newline)"

So you should try this instead :

echo "133\n" | sed 's/\\n//'

edited Jun 6, 2013 at 13:06

answered Jun 6, 2013 at 12:36

bendaizer

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1 Comment

mr.nothing Over a year ago

That outputs absolutely the same what it read from the file (please, find my screenshot in the edited post)

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