Bash parallel pipelined a for loop

Question

I would like to know if there is a way to execute the body of the loop as soon as my generator has emited IFS. The default behaviour is for bash to slurp the loop and then execute it.

Take this code:

time for i in $(echo 3; sleep 2; echo 1); do sleep ${i}; done

It runs in 6s. Because bash first execute echo 6; sleep 2; echo 1 fully and slurp the input. And then once 2s are spent here and the echo 1 finished it follows it with the body of the loop. Executing sleep 3 and then sleep 1.

What I would expect is for it to run in 4s because:

• first echo 3 is executed, triggering sleep 3.
• while sleep 3 execute the sleep 2; echo 1 code has time to run.
• when sleep 3 finishes, the body of the loop is free again and sleep 1 executes.

Note I don't want & done, I want a single body of my loop executed at one time, just pipeline the generator + loop execution.

You might think that & do does it (as it would be unambiguous vs & done), but not it doesn't, it fails with a syntax error.

Note I know I can do this with xargs -P or gnu-parallel but I would like to know if this is possible to do with loops.

Answer 1

Well I've just found it, still posting. To have this behaviour, you must use the while loop with read. The equivalent to my earlier for loop is this:

time (echo 3; sleep 2; echo 1) | while read i; do sleep ${i}; done

Correctly execute in 4s.