Function parameters of the function inside a /bin/bash command

Question

I had to use some bash code called from /bin/bash as a command followed -c option. I got something that is difficult to understand. This passed as expected:

root@punk:/home/s# function a () { echo $1 "world";}&&a hello
hello world
root@punk:/home/s#

But this missed the parameter of the function

root@punk:/home/s# /bin/bash -c "function a () { echo $1 "world";}&&a hello"
world
root@punk:/home/s#

How to fix the second case?

pynexj · Accepted Answer · 2019-02-25 10:00:00Z

1

Since you're using double quotes, with

/bin/bash -c "function a () { echo $1 "world";} && a hello"

, the current value of $1 will be replaced which must be empty in your environment so it'll become

/bin/bash -c "function a () { echo "world";} && a hello"

which is actually

/bin/bash -c "function a () { echo world;} && a hello"

.

You can use single quotes for the outer quoting:

[STEP 101] # /bin/bash -c 'function a () { echo $1 "world";} && a hello'
hello world
[STEP 102] #

edited Feb 25, 2019 at 10:00

answered Feb 25, 2019 at 9:54

pynexj

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