Whenever I pass explicit parameters to runhaskell command using a variable, in order to make a specific package visible, for example:
#!/bin/bash
args="'-package pandoc'"
runhaskell --ghc-arg="$args"<<EOF
import Text.Pandoc
main :: IO ()
main = print $ "Hello World"
EOF
I get the error:
: error: Can't find '-package pandoc' *** Exception: ExitFailure 1
However, if I enter the contents of that variable explicitly:
#!/bin/bash
runhaskell --ghc-arg='-package pandoc'<<EOF
import Text.Pandoc
main :: IO ()
main = print $ "Hello World"
EOF
The script works as expected. Does someone know why is this the case ?
You are passing this escaped a second time. You should work with:
#!/bin/bash
args="-package pandoc" # no quotations in the string
runhaskell --ghc-arg="$args"<<EOF
import Text.Pandoc
main :: IO ()
main = print $ "Hello World"
EOF
By using "$args" it is already clear that you use this as a single parameter, but if you use single quotes inside the string, these will be passed as content of the parameter. In other words you will call ghc-arg with '-package pandoc' as content of the first argument, where the single quotes are thus not interpreted by the shell, but given to the runhaskell executable. Since this does not start with a hyphen (-) likely the interpreter assumes it is the name of the file to run and not a flag, and thus looks for a file with that name in the working directory.
You can also pass multiple parameters, without quotes for $args:
#!/bin/bash
args="-package=pandoc -package=pandoc-types" # no quotations in the string
runhaskell -- $args <<EOF
import Text.Pandoc
main :: IO ()
main = print $ "Hello World"
EOF
The echo command can for example show the content of the parameter:
$ echo 'foo'
foo
$ echo '"foo"'
"foo"
$ echo "foo"
foo
$ echo "'foo'"
'foo'
The "outer" quotes (whether ' or ") are thus interpreted by the shell, but the inner quotes (if present) are passed as parameter value.