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First of all, I have little experience using Bash and I apologize for my bad English. Maybe it's obvious.

I am trying to understand why Bash drop value of the variable in this oneliner.

echo "Alpha;Beta;Gamma" | IFS=";" read First Second Third; echo $First $Second $Third

no output

But

echo "Alpha;Beta;Gamma" | (IFS=";" read First Second Third; echo $First $Second $Third)

has the right output

Alpha Beta Gamma.

I guess the read command opens a subshell and when it is close the variables lose their values.

If I am right, how to prevent it?

The goal was to separate a CSV like structure into variables.

Thanks in advice!

BR,

b

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  • Thank you for the hint! Partially yes, because there is no loop at all. And I expect that the last pipe will be executed in the parent shell, therefore the variables would be defined there and keep their value. Commented Jan 28, 2024 at 18:12
  • In the meantime i dug up some info from this toppic link. This type of redirection is working while read var ; do x=55 ; done < <(echo fred) echo "$x". But I don't understand why. Commented Jan 28, 2024 at 18:24
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    The last command in a pipeline is not executed in the parent shell (unless you set the lastpipe option -- but it's not set by default). Whether that last command is a while loop or a simple command like read does not matter. As for why the done < <(echo fred) version works, it's because that's not a pipe, so the while loop isn't forced into a subshell. Commented Jan 28, 2024 at 19:02
  • "I guess the read command opens a subshell and when it is close the variables lose their values.", it's not read, but the pipeline. True, read is one to naturally appear in cases like this, but see Gilles' example of a=0; a=1 | a=2; echo $a in their answer to the while read question. Commented Jan 28, 2024 at 20:03
  • @ilkkachu That is my problem exactly. The last part (right side of the pipe) lose the value. In the example it keeps it. That is what confuse me. Commented Jan 28, 2024 at 21:09

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