Bash function always ignores the else block

Question

When I run the below code with exact two arguments, the else block doesn't get executed.

If I take out the if else block out of the function, everything works fine.

#!/bin/bash
usage() {

if [[ $# -gt 2 || $# -lt 2 ]]; then
        echo "insufficient args"
else
        if [[ $# -eq 2 ]]; then
                echo "continuing with the script"
        fi
fi

}

usage

The problem is that usage is receiving 0 arguments. Call usage with: usage $@ — buhao
– buhao, Commented May 8, 2020 at 3:21
No problem! If you could accept my answer, I'd appreciate it. — buhao
– buhao, Commented May 8, 2020 at 3:27
Use usage "$@", not usage $@. See mywiki.wooledge.org/Quotes#When_Should_You_Quote.3F — dimo414
– dimo414, Commented May 8, 2020 at 4:12

buhao · Accepted Answer · 2020-05-08 06:17:22Z

1

In this situation, the function usage is receiving 0 arguments from the call.

Change the call to usage $@, which will pass the command line arguments to the usage function.

#!/bin/bash
usage() {

if [[ $# -gt 2 || $# -lt 2 ]]; then
        echo "insufficient args"
else
        if [[ $# -eq 2 ]]; then
                echo "continuing with the script"
        fi
fi

}

usage "$@"

edited May 8, 2020 at 6:17

answered May 8, 2020 at 3:24

buhao

1612 gold badges3 silver badges13 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

dimo414 Over a year ago

You should always quote arrays, including $@. So this should be usage "$@", not usage $@.

Collectives™ on Stack Overflow

Bash function always ignores the else block

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related