How can I assign the output of a function with parameters to a variable using bash?

Question

Similar to How can I assign the output of a function to a variable using bash?, but slightly different.

If I have a function like this:

function scan {
  echo "output"
}

...I can easily assign this to a variable like this:

VAR=$(scan)

Now, what if my function takes one or more parameters, how can I pass them to the function using the "shell expansion" syntax? E.g. this:

function greet {
  echo "Hello, $1"
}

# Does not work
VAR=$(greet("John Doe"))

The above produces an error like this with my bash (version 5.0.3(1)-release):

$ ./foo.sh 
./foo.sh: command substitution: line 8: syntax error near unexpected token `"John Doe"'
./foo.sh: command substitution: line 8: `greet("John Doe"))'

Socowi · Accepted Answer · 2019-12-18 09:19:53Z

2

in this code :

function greet {
  echo "Hello, $1"
}

# Does not work
VAR=$(greet("John Doe"))

the error is the parenthesis passing parameters. try to do this:

function greet {
  echo "Hello, $1"
}

# works
VAR=$(greet "John Doe")

it should work.

Explanation: when you use the $ an the parenthesis you have to write inside the parenthesis command as in a shell so the parameters are passed without parenthesis.

edited Dec 18, 2019 at 9:19

Socowi

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answered Dec 18, 2019 at 8:49

Zig Razor

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4 Comments

Per Lundberg Over a year ago

Thanks. I already documented this (at the same time as writing the question) here: stackoverflow.com/a/59388263/227779 :-)

Per Lundberg Over a year ago

Why would I upvote it when it mostly just duplicates my answer which was written before? I discovered the answer to my question while writing it, but decided to write down the answer Q&A-style. Thanks for writing a reply, but your answer doesn't add much more than what my answer already stated.

Zig Razor Over a year ago

Oh sorry, i don't see your answer from my mobile. Regards

Per Lundberg Over a year ago

Fair enough. I will upvote your answer now in good will. :) Merry Christmas!

Per Lundberg · Accepted Answer · 2019-12-18 08:39:12Z

0

I have obviously been writing to much Java code lately. Drop the parentheses when calling the function and everything works flawlessly:

function greet {
  echo "Hello, $1"
}

VAR=$(greet "John Doe")

answered Dec 18, 2019 at 8:39

Per Lundberg

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