JS：进行闭包时，如果我不存储为变量，内部函数如何访问外部函数参数？

Question

I don't know how these closures 2 accomplish the same thing. I understand how example 1 below would work since the variable foo is storing the value of the argument but what I can not understand is that taking away the variable in example 2 and just accessing the outer argument directly from the inner func, how does it have any reference to that?

 //ex1: function firstex1(x) { let foo = x; console.log(foo) function second() { return foo + 100 } return second } console.log("Example 1: "+ firstex1(1)()); //ex:2 function firstex2(x) { console.log(x) function second() { return x + 100 } return second } console.log("Example 2: "+ firstex2(1)()); 

Answer 1

In example 2 a argument x is passed to the outer function. Inside the outer function every entity has access to x as it's scope is covered in the full code block of that function. When an inner function tries to access that x it can do it easily as it has access to that scope because it itself is present in that scope. It's just like when you declare a global variable and try to access it inside a function. The function has access to all the global variables.

  var count=0; function a() { console.log(count++) } a(); 

The above will return count+1 and so will

  function first(x) { console.log(x) function second() { return x + 100 } return second } console.log(first(1)());