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Convert a String to a Number

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# 1  
Old 03-17-2017
Convert a String to a Number
I read in two numbers from a user but the number is a string.
Code:
#!/bin/bash
read -p "Enter first number: " num1
read -p "Enter second number: " num2

I know you can use the the "expr" or "bc" command to automatically convert the string to a number then add them together. But I don't want to add these numbers, I just want to convert the string to a number. That's it. How would I go about doing this? thank you
# 2  
Old 03-17-2017
Hi Loc...

In shell scripting consider a number as a string of characters in _integer_ form only:-
Code:
Last login: Fri Mar 17 15:30:19 on ttys000
AMIGA:amiga~> number1="101"
AMIGA:amiga~> number2=57
AMIGA:amiga~> number3='8'
AMIGA:amiga~> echo "$(( number1 + number2 - number3 ))"
150
AMIGA:amiga~> _

If however you want to change the 'type' you will have to use something like python:-
Code:
Last login: Fri Mar 17 15:32:35 on ttys000
AMIGA:amiga~> python
Python 2.7.10 (default, Jul 30 2016, 19:40:32) 
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.34)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> x="123"
>>> type(x)
<type 'str'>
>>> y=int(str(x))
>>> type(y)
<type 'int'>
>>> exit()
AMIGA:amiga~> _

# 3  
Old 03-17-2017
In bash use the typeset builtin (declare builtin does the same ). You declare the variable, THEN assign value to it:
Code:
typeset -i var=19

Once a variable is declared as a number it no longer will interpreted as a string, and allows some arithmetic operations.
From Typing variables: declare or typeset

Quote:
declare -i number
# The script will treat subsequent occurrences of "number" as an integer.

number=3
echo "Number = $number" # Number = 3

number=three
echo "Number = $number" # Number = 0
# Tries to evaluate the string "three" as an integer.

Certain arithmetic operations are permitted for declared integer variables without the need for expr or let.

n=6/3
echo "n = $n" # n = 6/3

declare -i n
n=6/3
echo "n = $n" # n = 2
# 4  
Old 03-17-2017
You need to be careful that you validate the input, else you can end up with this:-
Code:
$ typeset -i wally
$ wally=abc
$ echo $wally
0
$ wally=123abc456
bash: 123abc456: value too great for base (error token is "123abc456")
$

Perhaps you would be better to do something more like this after you read in num1 & num2:-
Code:
vnum1=$(tr -cd "[0-9]" < <(echo $num1))
if [ "$vnum1" != $"num1" ]
then
   echo "Invalid input!"
   exit
fi

It's a bit messy, but keeping it all as strings means that the shell doesn't care what the contents are. You can always for it to an integer afterwards should you wish.



I hope that this helps,
Robin
# 5  
Old 03-17-2017
Another way, a bit more shell independent (as long as it is Posix):
Code:
$ num=0001
$ echo "$num"
0001
$ echo "$((num+=0))"
1
$ echo "$num"
1

Last edited by Scrutinizer; 03-17-2017 at 07:09 PM..
# 6  
Old 03-17-2017
FWIW: a slightly better approach than
Code:
typeset -i

is simply to validate input
Example, pure bash validation of a number using bash regex where the variable is named n:
Code:
regex="^([0-9]$"
if [[ $n =~ $regex ]] ; then
  echo "$n is numeric"
else
  echo "$n is not numeric"
fi

# 7  
Old 03-17-2017
@jim, what if num=+1 or -001
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