Concatenate variables using expansion in bash

Question

I'm trying to simplify a compilation line for gcc using exapansion variables properties of bash scripting.

Let's say I've defined a variable SRC=src/ containing a folder of source files, and an additional variable C_SOURCES=(source1.c source2.c ...) countaining the source files themselves.

If I write something like "${SRC}${C_SOURCES[@]}", the result is that only the first value in the C_SOURCES variable gets the value in SRC, so: src/source1.c source2.c .... Which is not what I expected.

How should I rewrite the line with variable expansion so all values in C_SOURCES variable get the value in SRC variable?

Don't know if it is a typo, but source 2.c will be two different elements in the array unless you quote it — 123
– 123, Commented Apr 28, 2016 at 9:53

123 · Accepted Answer · 2016-04-28 09:56:39Z

3

You can use printf

C_SOURCES=('source1.c' 'source 2.c' 'source3.c' )

SRC='src/'

printf "'$SRC%s' " "${C_SOURCES[@]}"

outputs

'src/source1.c' 'src/source 2.c' 'src/source3.c'

answered Apr 28, 2016 at 9:56

123

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2 Comments

elcortegano Over a year ago

Actually, it works for me if i remove the single quotation marks around the $SRC%s term. Thank you very much!

123 Over a year ago

@EuGENE Didn't know what you want to use the end product for so yeah just tweak it to fit your purposes :)

John Zwinck · Accepted Answer · 2016-04-28 09:43:33Z

1

Try this:

PATHS=($(for x in ${C_SOURCES[@]}; do echo "$SRC/$x"; done))

I chose to include the / just to be sure. This won't handle filenames or paths with spaces in them properly.

answered Apr 28, 2016 at 9:43

John Zwinck

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1 Comment

elcortegano Over a year ago

It does not seem to work in the gcc instruction, as using echo in a for loop introduces newline characters at the end of "$SRC/$x". But thank you for your advice.

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