Scope of Local Variables in Shell Functions

After reading [24.2. Local Variables](http://tldp.org/LDP/abs/html/localvar.html) , I thought that declaring a variable var with the keyword local meant that var's value was only accessible within the block of code delimited by the curly braces of a function. However, after running the following example, I found out that var can also be accessed, read and written from the functions invoked by that block of code -- i.e. even though var is declared local to outerFunc, innerFunc is still able to read it and alter its value. [**Run It Online**](http://www.tutorialspoint.com/execute_bash_online.php?PID=0Bw_CjBb95KQMd1ptTmpoOWUyS3M) #!/usr/bin/env bash function innerFunc() { var='new value' echo "innerFunc: [var:${var}]" } function outerFunc() { local var='initial value' echo "outerFunc: before innerFunc: [var:${var}]" innerFunc echo "outerFunc: after innerFunc: [var:${var}]" } echo "global: before outerFunc: [var:${var}]" outerFunc echo "global: after outerFunc: [var:${var}]" Output: global: before outerFunc: [var:] # as expected, var is not accessible outside of outerFunc outerFunc: before innerFunc: [var:initial value] innerFunc: [var:new value] # innerFunc has access to var ?? outerFunc: after innerFunc: [var:new value] # the modification of var by innerFunc is visible to outerFunc ?? global: after outerFunc: [var:] **Q: Is that a bug in my shell (bash 4.3.42, Ubuntu 16.04, 64bit) or is it the expected behavior ?** **EDIT:** Solved. As noted by @MarkPlotnick, this is indeed the expected behavior.