Process a line in bash as is appears

Question

Let's say I have 2 bash commands:

$ A
apple
pear
lemon

A is writing this really fast but waiting 1 minute after "lemon" to terminate successfully.

I want every line being processed as input for B separately and instantly. but I do not want to wait for A to terminate but rather call a new B for every fast appearing line. B would then add something to the output of A let's say:

$ A <for every line appearing> B
1 apple
1 pear
1 banana

How can I do this with bash?

update

here are the original commands:

$sudo alive6 -l eth1 -W 0.2 | sed -e 's/Alive: \(.*\) \[ICMP echo-reply\]/\1%eth0/' -e'/Scanned.*/d' -e'//d'

responds instantly, while

$sudo alive6 -l eth1 -W 0.2 | sed -e 's/Alive: \(.*\) \[ICMP echo-reply\]/\1%eth0/' -e'/Scanned.*/d' -e'//d' | while read l; do echo $l; done

seems to wait for alive6 to finish

Answer 1

A | while read -r l; do B &; done

B is run in the background to address the following requirement: line being processed as input for B separately and instantly.

Answer 2

Use unbuffer to disable A's line buffering.

$ unbuffer A <for every line appearing> B

Answer 3

In Bash:

while read -r -u 9 line || [ -n "$line" ]
do
    B "$line"
done 9< <(A)

• Works even if A prints backslashes.
• Works even if the output of A does not end in a newline.
• Works even if B reads standard input, but not if it for some reason reads file descriptor 9. In that case you'll have to create a pipe with mkfifo.
• Runs B synchronously, processing the output of A asynchrously.