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schlingel

but yeah, i know there is a hack with "&", but if i'm not mistaken the program's output gets lost that way?

No, it does exactly what you want. The only thing "lost" is the process state. Normally, you could access the exit value via shell variable. That's lost in that case.

But STDOUT is just printed to the terminal as it's written.

Don't be the product, buy the product!

Schweinderl