关于js函数的嵌套

Question

JavaScript的函数在查找变量时从自身函数定义开始，从“内”向“外”查找。如果内部函数定义了与外部函数重名的变量，则内部函数的变量将“屏蔽”外部函数的变量。请看：

function foo(){

                var x=1;

                function bar(){

                    var x="a";

                    alert("x in bar()="+x);

                }

                alert("x in foo()="+x);

                bar();

            }

怎么理解上面的话呢？如果调用foo，县弹出x in foo()=1，后弹出x in bar()=a  这不是很正常嘛。

Answer 1

弹出的顺序没问题的啊，不知道你想问什么？

Answer 2

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