Converting string to integer in BASH
 

lets say i want to convert a string "76" to integer
76. How would I do this in BASH?

I remember in Pascal I could do something like char ('7')+39
or some such, and that would convert between character and a number
corresponding to that character. Can't seem to be able to figure
this out in BASH. What I actually wanted to do was to compare
two numbers one of which I got from top command. I finally
figured out that I could compare two strings as well, so the problem
is solved. But it would be nice to know how to convert between
integers and strings.

thanks a bunch in advance.

I don't think you need to actually convert anything...

teddy@toshiba~$ a="76"
teddy@toshiba~$ echo $((a+3))
79
teddy@toshiba~$ echo $((a-12))
64
teddy@toshiba~$

This is covered in the Advanced Bash-Scripting Guide: 4.3. Bash Variables Are Untyped.

I am gonna spoil your day.

See, no conversions made, this is correct code.

Even if you get $this and $that passed as cmd line parameters. The advantage of the second comparison is that you can find out if one varibale is greater than the other:

Actually string comparisons should be quoted:

if [ "$this" = "$that" ]
then
fi

Not necessarily. Quoting is needed when to avoid errors when strings contain certain characters. So it is better practice. But for strings containing [0-9a-zA-Z] non quoted comparision works fine. For the sake of clearness and to show the untypedness (?) I advertently omitted the quotes here, although I always use them in production code.

BTW, you forgot a '=' in the comparison. ;)

jlinkels

Actually for string comparison = and == are the same in bash ;)

I purposely left out the extra '=' as it seems to be irrelevant to the purpose of the code. Also, I know you can get way with not quoting strings, but it is good practice -especially when using single brackets.
Something like this will usually give errors when using single brackets:
but double brackets may not.

You made made open the Bash scripting guide, and you are right, the '=' and '==' operator are equal. Pun intended.

However, assuming that you really meant "that" and not "$that" (A constant and not a variable) it doesn't give errors in my version of bash which is 3.1.17. I am not sure about other versions, nor about about other shells. Quoting might be the safest way to go in most cases.

jlinkels

a=expr'$a/1'

Not sure of the quotes, but this is what the OP wants.

I tried something like this once for some iteration and noted that adding 0.0 didn't work.

a=expr'$a+0.000' gave someting like 760.0000

End

I'd like to know what magic shell are you running on!?! :confused:

Surely conversion isn't needed?
 

I'm on GNU bash, version 4.0.33(1)-release (i486-pc-linux-gnu)
In my shell script, I need to use the static number of cores available in the system minus one; I'm doing the following:
The output isn't a number, but the string "8-1" :confused: Any feedback please?

Indeed, it is what you're asking to the shell since
just does string concatenation. To perform an arithmetic operation you need an arithmetic operator! Post #2 in this thread provides the solution.

An aside note: please, don't resurrect old threads like this. Better to start a new one. By the way, the Advanced Bash Scripting Guide already have all the answers to your question (see specifically §4.3, chapter 13 and §8.3).

The expression $var-1 is a string concatenation as you found out. To do arithmeticMore tersley you could use
but they are not as transparent to programmers who do not know languages with decrement operators.

EDIT: you beat me to it, colucix :)

EDIT2: I didn't realise we were resurrecting the dead! 2007 and all that!

Data types are not declared for bash script.
http://www.tldp.org/LDP/abs/html/