Using local variables from “child” functions

Question

Consider following code:

function child()
{
    echo $var
}

function parent()
{
    local var=5
    child
}

I\'ve tested it on my machine

Answer 1

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.

Answer 2

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"