how to pass a function with parameters as a parameter?-CodePudding

There are two shell functions like belows

function call_function {
    func=$1
    desc=$2
    log_file=$3

    $func >> ${log_file} 2>&1
    ...
}

function echo_str {
   str=$1
   echo "$str"
}

How can I pass a shell function with parameters as a parameter?

I tried this:

call_function $( echo_str "test" ) "Echoing something" /var/logs/my_log.log

but only got

command not found

I googled it but nothing helps. Thanks a lot in advance!

CodePudding user response：

If you want to do this right, reorder your arguments to put the function to call last, and use shift to pop off the other arguments -- which will leave the function to call and its arguments in the "$@" array.

call_function() {
    local log_file desc
    log_file=$1; shift || return
    desc=$1; shift || return
    "$@" >>"$log_file" 2>&1
}

echo_str() {
    local str
    str=$1
    echo "$str"
}

#             log_file   desc                           func     args
call_function myfile.log "a function to echo something" echo_str "string to echo"

CodePudding user response：

call_function $( echo_str "test" ) "Echoing something" /var/logs/my_log.log

This $( echo_str "test" ) call will execute echo_str "test" which will result in test, so call_function will execute:

test >> /var/logs/my_log.log 2>&1

So, you either create a dedicated function to log messages easily to a log file:

log_msg() {
    current_date=$(date -u)
    echo "[$current_date] $1" >> $2
}

or change call_function as suggested by @Darkman