bash script within EOF unable to pass arguments to function [duplicate]

Question

I have the following bash script, it is running in a file named some_file.sh, the contents look like

sudo -i -u $USER bash << EOF


func(){
    echo $1

}

func 'Test message for channel'

EOF

This returns nothing in the argument for the function even though the function is invoked with an argument, what am I doing wrong?

To invoke it I do

bash some_file.sh

@Inder : The HERE-document is basically like a double-quoted string and undergoes parameter expansion. Hence, $1 is already expanded before sudo is invoked. If you would call it as bash some_file.sh FOO, you would see the effect of the expansion better. — user1934428
– user1934428, Commented Feb 22, 2021 at 13:16
@user1934428 valuable insight, thanks for that clarification it was really helpful — Inder
– Inder, Commented Feb 22, 2021 at 13:42

Jonathan Leffler · Accepted Answer · 2021-02-22 11:35:39Z

3

Quote the << 'EOF' to prevent the $ in the function from being expanded before the function is defined.

sudo -i -u $USER bash << 'EOF'

func(){
    echo $1
}

func 'Test message for channel'

EOF

See the Bash manual on Here Documents. This behaviour applies to all Bourne shell derivatives — Bash, POSIX shells, ksh, etc.

answered Feb 22, 2021 at 11:35

Jonathan Leffler

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