I'm having a hard time understanding how bash propagates `-e' to a subshell. Considering the example script below: - shouldn't C, D and E give no output? I know `-e' isn't in effect if the test is part of a `&&', `||' or `!', but shouldn't `-e' be *in* effect because the `&&', `||' or `!' are defined outside of the subshell? - shouldn't the lefthand and righthand subshells of the `&&' behave the same in FGH, IJK and PQRS? Especially since LMNO works like expected?
I'm running GNU bash, version 3.2.39(1)-release (i486-pc-linux-gnu) ---8<--------------------------------------------------- #!/bin/bash -e # Investigating `-e' subshell behaviour echo $(false; echo A) # output: A; no exit within `$()' (false; echo B) # output: none; exits within `()' (false; echo C) || true # output: C; no exit within `()' (false; echo D) && true # output: D; no exit within `()' !(false; echo E) # output: E; no exit within `()' (false; echo F) && (echo G; false; echo H) # output: FG; exits within right `()' only !(false; echo I) || (echo J; false; echo K) # output: IJ; exits within right `()' only echo $(set -e; echo L; false; echo M) && echo $(set -e; echo N; false; echo O) # output: LN; exits within `$()' if `-e' set explicitly (set -e; echo P; false; echo Q) && (set -e; echo R; false; echo S) # output: PQR; exits within right `()' only ---8<--------------------------------------------------- Regards, Freddy Vulto
