From 1891a5bc86553f816c62ef7517b757bc5ffb7366 Mon Sep 17 00:00:00 2001
From: CH-DanReif <35710107+CH-DanReif@users.noreply.github.com>
Date: Thu, 19 Apr 2018 14:48:44 -0700
Subject: [PATCH] Created SC2106 (markdown)

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+## SC2106: This only exits the subshell caused by the pipeline.
+
+### Problematic code:
+
+```sh
+for i in a b c; do
+  echo hi | grep -q bye | break
+done
+```
+
+### Correct code:
+
+```sh
+for i in a b c; do
+  echo hi | grep -q bye || break
+done
+```
+### Rationale:
+
+The most common cause of this issue is probably using a single `|` when `||` was intended.  The reason this message appears, though, is that a construction like this, intended to surface a failure inside of a loop:
+
+```sh
+for i in a b c; do false | break; done; echo ${PIPESTATUS[@]}
+```
+
+may appear to work:
+```
+$ for i in a b c; do false | break; done; echo ${PIPESTATUS[@]}
+1 0
+```
+
+What's actually happening, though, becomes clear if we add some `echo`s; the entire loop completes, and the `break` has no effect.
+
+```sh
+$ for i in a b c; do echo $i; false | break; done; echo ${PIPESTATUS[@]}
+a
+b
+c
+1 0
+$ for i in a b c; do false | break; echo $i; done; echo ${PIPESTATUS[@]}
+a
+b
+c
+0
+```
+
+Because bash processes pipelines by creating subshells, control statements like `break` only take effect in the subshell.
+
+### Related resources:
+
+* Contrast with the related, but different, problem in [this link](https://unix.stackexchange.com/questions/166546/bash-cannot-break-out-of-piped-while-read-loop-process-substitution-works).
+* [Bash Reference Manual: Pipelines](https://www.gnu.org/software/bash/manual/bash.html#Pipelines), esp.:
+     > Each command in a pipeline is executed in its own subshell.
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