javascript - 如何使用javascript中的闭包访问函数内另一个范围内的变量？

Question

我有以下功能 makeStopwatch我正在努力通过以更好地理解 javascript 闭包:

var makeStopwatch = function() {
  var elapsed = 0;
  var stopwatch = function() {
    return elapsed;
  };
  var increase = function() {
    elapsed++;
  };

  setInterval(increase, 1000);
  return stopwatch;
};

var stopwatch1 = makeStopwatch();
var stopwatch2 = makeStopwatch();

console.log(stopwatch1());
console.log(stopwatch2());

当我 console.log调用stopwatch1和 stopwatch2我得到 0每次分别返回。

据我了解 makeStopwatch 的预期功能变量 elapsed将是 0如果由内部函数返回 stopwatch .内部函数increase递增变量 elapsed .那么setInterval电话 increase延迟 1 秒后。最后，stopwatch这次再次返回更新后的值，预计为 1 .

但这行不通，因为在 makeStopwatch 里面, 内部 stopwatch , increase , 和 setInterval功能都在彼此独立的范围内？

我怎样才能修改它以按照我的理解工作，以便 elapsed递增并且该值被关闭并保存，这样当我分配 makeStopwatch 时到变量 stopwatch1并调用stopwatch1返回更新后的值？

Answer 1

var makeStopwatch = function() {
  var elapsed = 0;

  // THIS stopwatch function is referenced later
  var stopwatch = function() {
    return elapsed;
  };

  var increase = function() {
    elapsed++;
  };
  // This setInterval will continue running and calling the increase function.
  // we do not maintain access to it.
  setInterval(increase, 1000);

  // THIS returns the stopwatch function reference earlier.  The only way
  // we can interact with the closure variables are through this function.
  return stopwatch;
};

var stopwatch1 = makeStopwatch();
// This runs the makeStopwatch function.  That function *RETURNS* the
// inner stopwatch function that I emphasized above.

console.log(stopwatch1());
// stopwatch1 is a reference to the inner stopwatch function.  We no longer
// have access to the elapsed variable or the function increase.  However
// the `setInterval` that is calling `increase` is still running.  So every
// 1000ms (1 second) we increment elapsed by 1.

因此，如果我们将上述所有代码放入控制台，然后偶尔调用 console.log(stopwatch1())，它将 console.log self 们创建以来的秒数秒表。