How can I pass a block as an argument to a bash function?

Question

I'd like to create a bash function that can get a bash code block like time does:

time {
  echo 123
  echo "Something else"
  sleep 1
}

Basically I'd like to be able to wrap the block with my bash code.

Edit: Updated my question after the first answer of @DavidPostill♦ and the comments: For example, I'd like to wrap a code block with 2>&1 > /dev/null and also the time it with time. Should I write a program outside bash to do that?

function myfunction() {
  time { $1 } 2>&1 /dev/null
}
myfunction { sleep 1 }

Edit 2: Upon further reading it seems like it isn't possible as time is a special case for bash.

Note: time can do this because its a keyword in Bash. It's recognized by the parser. If you use /usr/bin/time instead (or command time or some function or a builtin), then it will be a syntax error. What exactly do you want your function to do? Maybe we can do this using a different syntax. — Kamil Maciorowski, Oct 04 '20 at 12:56
Can't do this in bash. The closest you get is to define a function containing the code you want as a block, and then pass the function name to your function that wants a block. — glenn jackman, Oct 04 '20 at 14:14
I'd like to wrap a code block with 2>&1 > /dev/null and also time it with time. Should I write a program outside bash to do that? — funerr, Oct 04 '20 at 18:24
Somewhat similar: Wrapping time (and similar keywords) in call from another script. — Kamil Maciorowski, Oct 04 '20 at 18:31
So, it seems it isn't possible with the syntax I had in mind, right? what are my options, per the edit? — funerr, Oct 04 '20 at 22:58

score 1 · Answer 1 · answered Apr 30 '21 at 14:20

You can't pass blocks around directly as if they were objects, but you can pass the file descriptor (FD) of a redirect block (<()) and time the output of that (or do whatever to it). The inner {} 2>&1 is needed to capture the stderr on the same FD:

function myfunction() {
  time cat $1 >/dev/null
}
myfunction <({ 
  echo 123
  echo "Something else"
  sleep 1 
} 2>&1)

score 0 · Answer 2 · answered Oct 04 '20 at 12:57

I'd like to be able to wrap the function with my bash code and pass arguments

Functions with parameters sample
#!/bin/bash 
function quit {
   exit
}  
function e {
    echo $1 
}  
e Hello
e World
quit
echo foo 
The function e prints the first argument it receives.

Arguments, within functions, are treated in the same manner as arguments given to the script.

Source BASH Programming - Introduction HOW-TO: Functions

How can I pass a block as an argument to a bash function?

2 Answers2

I'd like to be able to wrap the function with my bash code and pass arguments

Further reading