javascript - 难以理解 javascript 函数级范围

Question

我无法理解 JavaScript 函数级作用域，作为一个 C# 程序员，它对我来说似乎很连贯，我将尝试通过代码来解释它:

代码#1

//Problem
//if same named variable (as in global scope) is used inside function scope,
//then variable defined inside function will be used,global one will be shadowed
var a = 123;
function func() {
    alert(a); //returns undefined,why not just return 123 ?
    //how come js knew that there is variable 'a' will be defined and used in 
    //this function scope ,js is interpreter based ?
    var a = 1; //a is defined inside function
    alert(a); //returns 1 
}
func();

代码#2

//when a variable(inside function) not named as same as the global,
//then it can be used inside function,and global variable is not shadowed
var k = 123;
function func2() {
    alert(k); //returns 123 ,why not 'undefined'
    var c = 1;
    alert(c); //returns 1
}
func2();

所以我的问题是

1. 在 CODE#1 为什么第一次 a 是 undefined，为什么不直接返回 123?如何come js 知道有变量 'a' 将被定义和使用这个函数作用域，js 是基于解释器的吗？

2. CODE#2 为什么 k 不是“未定义”？

http://jsfiddle.net/Nu2Vu/

Answer 1

代码 #1

Hoisting导致所有变量声明都被带到范围的顶部，但它将赋值保留在原处。当存在对变量的引用时，JavaScript 将首先在当前作用域中查找，如果没有找到该变量，它将继续查找作用域链，直到找到该变量。此代码解释如下:

var a = 123; // global var named a declared, assigned a value
function func() {
  var a; // the declaration of the local var a gets
         // hoisted to the top of the scope, but the
         // assignment is left below, so at the point
         // it is initialized with a value of `undefined`
    alert(a); // looks for a local var named a, finds it but
              // it currently has a value of `undefined`

    a = 1; // now the value is assigned to the local a
    alert(a); // returns 1 
}
func();

代码 #2

由于 closure，此代码的行为方式如此.闭包的一个基本定义是，JavaScript 函数不仅可以访问定义在其自身作用域中的变量，还可以访问其父作用域可用的变量。

var k = 123; // declares and assigns a value to global k
function func2() {
    alert(k); // looks for a local var named k, doesn't find it,
              // looks in its parent scope (the global scope in
              // this case) finds k, returns its value of 123
    var c = 1;
    alert(c); //returns 1
}
func2();