CODEWITHSUNDEEP
bashpipe
Andrew H. Hunter
Andrew H. Hunter

Reputation: 561

terminate a shell script without waiting for early parts of pipeline

Consider the following script:

#!/bin/bash
function long_running {
    for i in $(seq 1 10); do
        echo foo
        sleep 100
    done
}

long_running | head -n 1

This produces the expected output (one line "foo") but sleeps (for the specified 100 seconds) before terminating. I would like the script to terminate immediately when head does. How can I force bash to actually quit immediately? Even changing the last line to

long_running | (head -n 1; exit)

or similar doesn't work; I can't get set -e, another common suggestion, to work even if I force a failure with, say, (head -n 1; false) or the like.

(This is a simplified version of my real code (obviously) which doesn't sleep; just creates a fairly complex set of nested pipelines searching for various solutions to a constraint problem; as I only need one and don't care which I get, I'd like to be able to make the script terminate by adding head -n 1 to the invocation...)

Upvotes: 6

Views: 477

Answers (1)

jaypal singh
jaypal singh

Reputation: 77095

How about sending the function to head like this -

#!/bin/bash
function long_running {
    for i in $(seq 1 10); do
        echo foo
        sleep 100
    done
}

head -n 1 <(long_running)

Obviously if you will increase the -n to a greater number, the sleep would kick in but would exit once head is completed.

Upvotes: 3

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