Bash output in function not as expected

Question

I have the following simple example of which I don't understand how it comes to the output:

#!/bin/bash

function wtf() {
    echo -e "test1" >&1
    sleep 2
    echo -e "test2" >&1
}

a=$(wtf)
echo $a

The output on the terminal is AFTER 2 seconds test1 test2

When I change the last two lines just to wtf then the output is

 test1
 test2 #after 2 seconds

Why is in the first version the test1 test2 in the same line?
Why does the output of the line test1 test2 need 2 seconds to show up since only the second test2 should show up after 2 seconds?

janos · Accepted Answer · 2016-05-06 20:34:50Z

2

The $(...) captures the stdout of the commands inside the subshell. The echo -e statements are not printed to the stdout of the calling shell, but get buffered in the subshell. If you remove the echo $a command then nothing will be printed at all.

The subshell in this example doesn't exit until wtf completes. That's why you have to wait 2 seconds, so that the calling shell gets back control, and does echo $a, printing all the output that was captured by the subshell.

answered May 6, 2016 at 20:34

janos

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2 Comments

wasp256 Over a year ago

Thx for the clarification, is there a way to print to the stdout anyways in the function?

janos Over a year ago

stdout will be captured by the subshell, so no, you cannot. You could print to stderr though, with echo test3 >&2

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