Why a local variable is not accessible in a function that is declared after the function definition and before it's call?

user_12946447 0
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I'm new to Lua. I noticed the following pattern in Lua. But I'm not understanding why it's happening.

Here, a is a global variable and it's accessible inside the foo function:

function foo()
  print(a)
end

a = 10

foo() -- 10

But if we declare a as local it's not any more accessible inside the function foo:

function foo()
  print(a)
end

local a = 10

foo() -- nil

Why the global variable is accessible inside the function while the local is not?

lua function
1 Answers

That's because global variables work differently: Local variables abide by lexical scoping. This means that variable visibility is intuitive: The variable is visible only after it has been declared; it is also visible in nested blocks (such as the function block) or other control flow statements (ifs or loops).

In

function foo()
  print(a)
end

local a = 10

foo() prints nil because the local variable a isn't visible yet; the function accesses the global variable _G.a at run time instead, which evaluates to nil. Setting the the global variable a = 10 is equivalent to _G.a = 10 or _G["a"] = 10. When foo is run, it indexes the global table using _G["a"] to find the value for the variable a which yields 10 as expected.

TL;DR: Visibility of global variables depends on order of execution since global variables are just syntactic sugar for accessing a table under the hood; visiblity of local variables abides by lexical scoping - variables must be declared before use

(actually foo indexes the upvalue _ENV which is _G by default if it was not assigned to, but for the sake of simplicity I'm using _G above)

To use a local variable, which is often recommended both for performance & readability reasons, simply declare it before the function:

local a
function foo() print(a) end
a = 10
foo()

local variables inherited by functions are called upvalues in Lua.

user_71853180
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