Question

Can I declare a UNIX shell script variable a character or an integer?

Asked by: jjc_mn

Can I declare a UNIX shell script variable a character or an integer? I'm trying to get yesterdays date so I put this in my script. It seems to work but on the 8th I got an error doing the math(unfortunately I didn't write it down). I thought maybe if I declared it an INT things would work better.  Is that possible?
dateday=`date +%d`
((dateday=dateday-1))

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Asked On
2007-12-10 at 12:31:04ID23013982
Tags

shell

,

script

,

unix

,

variable

,

declare

Topic

Unix Systems Programming

Participating Experts
3
Points
500
Comments
6

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Answers

 

by: ozoPosted on 2007-12-10 at 13:47:19ID: 20445531

It shouldn't matter

 

by: evilrixPosted on 2007-12-10 at 14:18:08ID: 20445806

Shell scripts use Duck Typing.

http://en.wikipedia.org/wiki/Duck_typing

Basically, if it looks like a duck, smells like a duck and sounds like a duck it probably is a duck!

So, the type of the variable is inferred based upon how you use it. This is why bash has different predicate operators for testing (for example) equality and inequality on strings and integers. Use a string type operator and bash will assume it's a string, likewise for number operators. Of course, if you try to use a number operator on a string type that cannot possibly be inferred as a number bash will complain!

In your example, ((dateday=dateday-1)), you are treating dateday as a string so bash then complains when you try to perform maths with it. You might be able to convince it otherwise if you use -eq instead.

If, like me, you come from a strongly typed background (C++) it can take a while to get used to, but it's not all bad :)

       STRING1 = STRING2
              the strings are equal

       STRING1 != STRING2
              the strings are not equal

       INTEGER1 -eq INTEGER2
              INTEGER1 is equal to INTEGER2

       INTEGER1 -ne INTEGER2
              INTEGER1 is not equal to INTEGER2

I hope this helps to clarify for you.

-Rx.

 

by: evilrixPosted on 2007-12-10 at 14:20:37ID: 20445823

Very simple example: -

# Ok, bash assumes strings for both tests: -
:~$ test 1 == 1
:~$ test A == 1
 
# Not ok, bash can't figure out how a can be compared with 1 as a number!
:~$ test 1 -eq 1
:~$ test a -eq 1
-bash: test: a: integer expression expected
                                              
1:
2:
3:
4:
5:
6:
7:
8:

Select allOpen in new window

 

by: omarfaridPosted on 2007-12-10 at 17:13:42ID: 20446586

Hi,

As ozo said it doesn't matter.

e.g
dateday=`date +%d`
dateday=`expr $dateday - 1`
echo $dateday

 

by: jjc_mnPosted on 2007-12-11 at 08:40:09ID: 31414008

>In your example, ((dateday=dateday-1)), you are treating dateday as a string so bash then complains when you try to perform maths with it.
evilrix, it does work most (but not on the 8th for some reason) time so it usually thinks of it as as a number (I use the bash shell).  

If I get the error again I'll try omarfarid way to do the math and see if that make a difference and let you all know via a new question. I'll copy the screen output next time too.


 

by: jjc_mnPosted on 2008-01-08 at 09:03:54ID: 20610378

replacing ((dateday=dateday-1)) with dateday=`expr $dateday - 1` fixed the problem that always comes up on the 8th For whatever reason

20120131-EE-VQP-002

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