Error in string Concatenation in Shell Scripting

Question

I am beginner to Shell scripting.

I have used a variable to store value A="MyScript". I tried to concatenate the string in subsequent steps $A_new. To my surprise it didn't work and $A.new worked.

Could you please help me in understanding these details?

Thanks

Does the info in [this question](http://stackoverflow.com/questions/4181703/how-can-i-concatenate-string-variables-in-bash) help? — Michael Scarn, May 07 '15 at 01:31
@MichaelScarn though it was not I was looking for but thanks for sharing such nice information. — Prradep, May 07 '15 at 18:46

score 3 · Answer 1 · edited Jun 20 '20 at 09:12

Shell variable names are composed of alphabetic characters, numbers and underscores.

3.231 Name

In the shell command language, a word consisting solely of underscores, digits, and alphabetics from the portable character set. The first character of a name is not a digit.

So when you wrote $A_new the shell interpreted the underscore (and new) as part of the variable name and expanded the variable A_new.

A period is not valid in a variable name so when the shell parsed $A.new for a variable to expand it stopped at the period and expanded the A variable.

The ${A} syntax is designed to allow this to work as intended here.

You can use any of the following to have this work correctly (in rough order of preferability):

echo "${A}_new"
echo "$A"_new
echo $A\_new

The last is least desirable because you can't quote the whole string (or the \ doesn't get removed. So since you should basically always quote your variable expansions you would end up probably doing echo "$A"\_new but that's no different then point 2 ultimately so why bother.

This is exact explanation for your query. Also you can specify explicitly by any of the mentioned ways : 1. ${A}_new 2. $A\\_new — Amit Kumar, May 07 '15 at 06:56
Thank you @Etan for your detailed answer. It clears my doubt and concepts. — Prradep, May 07 '15 at 18:19

Rajesh · Answer 2 · 2015-05-07T14:58:18.273

1

This happens because the underscore is the valid character in variable names. Try this way: ${A}_new or "$A"_new

edited May 07 '15 at 14:58

answered May 07 '15 at 01:39

Rajesh

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Ya. `$A_new`, in fact, is `${A_new}`. – Robin Hsu May 07 '15 at 03:49
You mean `${A}_new` not `{$A}_new`. – Etan Reisner May 07 '15 at 13:38
Thanks @Rajesh. I didn't get your statement Robin. – Prradep May 07 '15 at 18:21

score 0 · Accepted Answer · answered May 09 '15 at 19:42

The name of a variable can contain letters ( a to z or A to Z), numbers ( 0 to 9) or the underscore character ( _).

Shell does not require any variable declaration as in programming languages as C , C++ or java. So when you write $A_new shell consider A_new as a variable, which you have not assigned any value therefore it comes to be null.

To achieve what you mentioned use as : ${A}_new

Its always a good practice to enclose variable names in braces after $ sign to avoid such situation.

Error in string Concatenation in Shell Scripting

3 Answers3

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