CODEWITHSUNDEEP
bash
user247763
user247763

Reputation: 261

Can I access a global variable with the same name as a local variable from inside a function?

Say I have the following code:

# Global variable x
x="Hi"

foo()
{
    # Local variable x
    local x="Hello"
}

Can I access (read and set) the global variable x from inside the function foo()?

Upvotes: 6

Views: 2974

Answers (3)

spawn
spawn

Reputation: 390

Actually, by using a nested function, this can be done.

foo()
{
    local nested_ret
    foo_nested(){
        local x
        x="Hello"
        nested_ret="$x"
    }
    foo_nested "$@"
    x="$nested_ret"
}

x="Hi"
foo
echo "x is $x"

Output: x is Hello

Upvotes: 0

dimo414
dimo414

Reputation: 48884

Once you have shadowed the global variable with a local of the same name there is no way to access the global anymore. Search for "shadow" in the manual.

As mentioned you can access the global variable before declaring the local, but if that isn't what you need (e.g. if you want to write back to the global variable after declaring the local) the only option is to use a different name. It's fairly common to use a _ or __ prefix to denote special or private variables that should not be set at the global scope. That way all (well-behaved) callers will not collide with your variable name.

Upvotes: 2

Maroun
Maroun

Reputation: 96016

Try the following:

  1 x="Hi"
  2
  3 foo()
  4 {
  5   echo $x
  6   x="Hello"
  7   echo $x
  8   local x="Bye"
  9   echo $x
 10 }
 11
 12 foo
 13 echo $x

This will print:

Hi
Hello
Bye
Hello
  • First echo $x prints the value of the global x
  • Next we modify it, and its value is changed as you can see in line 13
  • Inside foo, local x value wasn't changed

Upvotes: 0

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