Javascript scoping of variables

Question

The output is 15 (in f, x is taken to be 10 and y is 7) with the following:

var x = 5;
function f(y) { return (x + y) - 2};
function g(h) { var x = 7; return h(x         


        

Answer 1

vars are not blocked scoped, so the last line is equivalent to

x = 10; z = g(f); console.log(z)

It should be clearer now that the value of x was changed to 10 before f was executed.
It is also important to note that free variables are evaluated when a function is called, not when it was defined. And of course the value of a variable can change between the definition of a function and the call of the function, just like in your example.

In the third line, x is local to g and therefore is completely independent from the "outer" variable x.

See also What is the scope of variables in JavaScript?

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A much simpler example that demonstrates this behavior would be:

var x = 5;
function foo() {
  console.log(x);
}
x = 10;
foo(); // outputs 10