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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"))'
Per Lundberg
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2 Answers2

1

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.

Socowi
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Zig Razor
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  • Thanks. I already documented this (at the same time as writing the question) here: https://stackoverflow.com/a/59388263/227779 :-) – Per Lundberg Dec 18 '19 at 11:16
  • 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. – Per Lundberg Dec 18 '19 at 12:01
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    Oh sorry, i don't see your answer from my mobile. Regards – Zig Razor Dec 18 '19 at 12:03
  • Fair enough. I will upvote your answer now in good will. :) Merry Christmas! – Per Lundberg Dec 18 '19 at 14:24
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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")
Per Lundberg
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