Question

假设我有这段代码：

// somewhere else
var outFn = (() => {
  var local = 5;
  var fn = () => { console.log(local); };
  return fn;
})();

// I have a reference to outFn
outFn(); // prints 5

如何引用outFn的本地范围，即如何使用我对local的引用来获取对outFn变量的引用？

Answer 1

不要写下最后一个括号：

var outFn = (() => {
  var local = 5;
  var fn = () => { console.log(local); };
  return fn;
}) //(); <-- No need those parentheses

然后：

  var localFn= outFn(); 
  localFn() ; // prints 5

如果您还想同时访问local变量，请将其附加到fn，如下所示：

var outFn = (() => {
  var fn = () => { console.log(fn.local); };
  fn.local = 5;
  return fn;
});

现在您可以阅读local：

  var localFn= outFn(); 
  localFn.local ; // ===5

从函数中获取对局部作用域的引用

1 个答案: