Scopes at functions?

Question

function x() {
  var a = 2;
  y();
}

function y() {
  a += 3;
}
x();
//returns a is not defined

If i declare a function (y) outside the function (x) where it's been called, why it isn't inherit the outer function's variables and values? There's a way to solve the problem described above?

Mamun · Accepted Answer · 2018-03-17T13:11:08.970

1

Variable a is declared as part of the function x(). That's why a is available only inside x(). To solve the problem you have to pass variable a as the function parameter:

function x() {
  var a = 2;
  y(a);
}

function y(a) {
  a += 3;
  console.log(a);
}
x();

edited Mar 17 '18 at 13:11

answered Mar 17 '18 at 12:53

Mamun

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Even with closure it isn't possible to inherit the variables and values of `x()` without passing the variables of `x()` as the `y()`'s parameter? – Gergő Horváth Mar 17 '18 at 13:19
@GergőHorváth, in your code closure is irrelevant. – Mamun Mar 17 '18 at 13:25

Scopes at functions?

1 Answers1