Question

我一直在熟悉javascript闭包并浏览了这篇文章 http://blog.morrisjohns.com/javascript_closures_for_dummies.html

由于关闭，示例5不能按预期工作。如何修改

result.push( function() {alert(item + ' ' + list[i])} );

使代码有效？

function buildList(list) { 
  var result = [];
  for (var i = 0; i < list.length; i++) {
    var item = 'item' + list[i];
    result.push( function() {alert(item + ' ' + list[i])} );
  }
  return result;
}

function testList() {
  var fnlist = buildList([1,2,3]);
  // using j only to help prevent confusion - could use i
  for (var j = 0; j < fnlist.length; j++) {
    fnlist[j]();
  }
}

testList();

谢谢！

Answer 1

function buildList(list) { 
  var result = [];
  for (var i = 0; i < list.length; i++) {
    (function(i) {
      var item = 'item' + list[i];
      result.push(function() {
        alert(item + ' ' + list[i]);
      });
    })(i);
  }
  return result;
}

Answer 2

您需要来自不同函数的i和item作为不同范围内的单独变量（因此它们可以具有不同的值而不是共享）。由于Javascript只有函数作用域，因此需要创建一个包含这些变量的包装函数

function buildList(list) { 
  var result = [];

  function make_f(item, i){
    //item and i are now private variables of make_f
    //and wont be shared by the returned closure
    // the other variable in scope (list) is not shadowed so
    // it is still shared but that is not a problem since we
    // never change its value.
    return function() { alert(item + ' ' + list[i]) };
  }

  for (var i = 0; i < list.length; i++) {
    var item = 'item' + list[i];
    result.push( make_f(item, i) );
  }
  return result;
}

您也可以使用立即调用的匿名函数（(function(){}())模式）执行相同的操作。

function buildList(list) { 
  var result = [];

  for (var i = 0; i < list.length; i++) {
    var item = 'item' + list[i];
    result.push( (function (item, i){
        return function(){function() {alert(item + ' ' + list[i])};
    })(item, i) );
  }

  return result;
}

Answer 3

把它放进......另一个封闭！

function buildList(list) { 
  var result = [];
  for (var i = 0; i < list.length; i++) {
    result.push((function(item, listItem){
      return function() {
        alert(item + ' ' + listItem);
      };
    })('item' + list[i], list[i]));
  }
  return result;
}

function testList() {
  var fnlist = buildList([1,2,3]);
  // using j only to help prevent confusion - could use i
  for (var j = 0; j < fnlist.length; j++) {
    fnlist[j]();
  }
}

testList();

Answer 4

var具有函数作用域，但如果使用let，则将获得块作用域。

所以我会做： for（让我= 0;我

修复此Javascript闭包的正确方法是什么？

4 个答案: