Javascript:为什么我可以访问在全局范围内的函数内声明的内部名称？

Question

在chrome开发控制台中,我创建了一个带有两个嵌入功能的函数f

> var a = 'ga';
  var b = 'gb';
  var c = 'gc';
  var f = function(){
      var a = 'fa';
      var b = 'fb';
      ff = function(){
          var a = 'ffa';
          fff = function(){
              console.log("a,b,c is: " + a + "," + b + "," + c);
          };
          fff();
      };
      ff();
  };
< undefined

然后,我输入ff到控制台,发现我仍然可以访问它,而它是在内部范围内定义的f

> ff     // why can I still access the name ff ?
< function (){
         var a = 'ffa';
         fff = function(){
             console.log("a,b,c is: " + a + "," + b + "," + c);
         };
         fff();
     }

这个名字也是如此 fff

> fff   // why can I still access the name fff ?
< function (){
             console.log("a,b,c is: " + a + "," + b + "," + c);
         }

我是一名C/C++开发人员,目前正忙于javascript.

这种现象对我来说似乎很难理解.
因为在Cpp中,访问内部范围内的名称是错误的.
例如:

#include <iostream>

using namespace std;

int main(int argc, char *argv[]){
    auto f = [](){
        std::cout << "in f() now" << std::endl;
        auto ff = [](){
            std::cout << "in ff() now" << std::endl;
            auto fff = [](){
                std::cout << "in fff() now" << std::endl;
            };
            fff();
        };
        ff();
    };

    f(); //it's okay
    ff(); // not okay, error: use of undeclared identifier 'ff'
    fff(); // not okay too, error: use of undeclared identifier 'fff'

    return 0;
}

即使在python中,我们也不能这样做:

def f():
    print("in f() now")
    def ff():
        print("in ff() now")
        def fff():
            print("in fff() now")
        fff()
    ff()

f()   # okay
ff()  # NameError: name 'ff' is not defined
fff() # NameError: name 'fff' is not defined

所以,我很奇怪为什么即使我不在内部也可以访问内部范围内的名称？

提前致谢!

Answer 1

var在全局上下文中生成没有变量的变量.

在执行赋值时,为未声明的变量赋值会隐式地将其创建为全局变量(它将成为全局对象的属性).