Consider following code:
function child()
{
echo $var
}
function parent()
{
local var=5
child
}
I've tested it on my machine and it seems to work but I wasn't able to find anything definitive describing such usage of local variables. Namely, when I declare a local variable in one function and from that function I call some other function, can I use the variable in the latter (and even nest it deeper)? Is it legal in bash and is it standard for all versions?
bash
uses dynamic scoping. The value of var
in child
is not determined by where child
is defined, but by where it is called. If there is no local definition in the body of child
, the next place the shell looks is in the body of the function from which child
is called, and so forth. The local
modifier creates a variable in a function that is local to that call, so it does not affect the value of the variable from any enclosing scopes. It is, though, visible to any enclosed scope.
a () { echo "$var"; }
b () { local var="local value"; a; }
var="global value"
a # outputs "global value"
b # outputs "local value"
Just complimenting what user 123 already mentioned in the comments
From bash man pages
Variables local to the function may be declared with the local builtin command. Ordinarily, variables and their values are shared between
the function and its caller.
bash3.2.25->cat function.ss
#!/bin/bash
function child()
{
echo "child function called"
echo $var
}
function parent_global()
{
echo "parent_global called, child invoked outside parent"
var=5
}
function parent()
{
echo "parent called, child invoked inside parent"
local var=5
child
}
parent
parent_global
child
bash3.2.25->./function.ss
parent called, child invoked inside parent
child function called
5
parent_global called, child invoked outside parent
child function called
5
So unless specified otherwise (the local builtin), variables are visible outside the function scope, global if you will.