传递console.log作为函数参数抛出“Uncaught TypeError：非法调用”

当测试二叉树生成和遍历代码的JS示例时，我用console.log（）替换了print（）以创建控制台输出。 “未被捕获......非法调用”在'case this'引发：：func（this.value）;'。为什么这个函数不能像其他函数那样作为函数参数传递？从注释代码中可以看到，将“tree”对象传递给console.log只是输出函数代码，而不是.value。任何人都可以提供有关此异常情况的指导或更好的记录测试输出的方法吗？传递console.log作为函数参数抛出“Uncaught TypeError：非法调用”

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
    <title></title> 
    <script type="text/javascript"> 
     //20140122-this example code originally had "print" statements in place of console.log, however, console.log cannot be passed as a function parameter. 

     function BinaryTree(value, left, right) { 
      this.value = value; 
      this.left = left; 
      this.right = right; 
     } 

     BinaryTree.prototype.preorder = function(f) { this.walk(f, ['this', 'left', 'right']); }; 
     BinaryTree.prototype.inorder = function(f) { this.walk(f, ['left', 'this', 'right']); }; 
     BinaryTree.prototype.postorder = function(f) { this.walk(f, ['left', 'right', 'this']); }; 
     BinaryTree.prototype.walk = function(func, order) { 
      for (var i in order) 
       switch (order[i]) { 
       case "this": 
        func(this.value); 
        break; 
       case "left": 
        if (this.left) this.left.walk(func, order); 
        break; 
       case "right": 
        if (this.right) this.right.walk(func, order); 
        break; 
       } 
     }; 
     BinaryTree.prototype.levelorder = function(func) { 
      var queue = [this]; 
      while (queue.length != 0) { 
       var node = queue.shift(); 
       func(node.value); 
       if (node.left) queue.push(node.left); 
       if (node.right) queue.push(node.right); 
      } 
     }; 

     // convenience function for creating a binary tree 
     function createBinaryTreeFromArray(ary) { 
      var left = null, right = null; 
      if (ary[1]) left = createBinaryTreeFromArray(ary[1]); 
      if (ary[2]) right = createBinaryTreeFromArray(ary[2]); 
      return new BinaryTree(ary[0], left, right); 
     } 

     var tree = createBinaryTreeFromArray([1, [2, [4, [7]], [5]], [3, [6, [8], [9]]]]); 

     console.log("*** preorder ***"); tree.preorder(console.log); 
     console.log("*** inorder ***"); tree.inorder(console.log); 
     console.log("*** postorder ***"); tree.postorder(console.log); 
     console.log("*** levelorder ***"); tree.levelorder(console.log); 
//  console.log("*** preorder ***"); console.log(tree.preorder); 
//  console.log("*** inorder ***"); console.log(tree.inorder); 
//  console.log("*** postorder ***"); console.log(tree.postorder); 
//  console.log("*** levelorder ***"); console.log(tree.levelorder); 
// </script> 
</head> 
<body> 
</body> 
</html>

来源

2014-01-23 robWDC

http://stackoverflow.com/questions/8904782/uncaught-typeerror-illegal-invocation-in-javascript – Andreas

当您将对象的方法作为参数传递给某个函数时，会丢失上下文。为了保持它，只需将绑定方法到上下文：

tree.levelorder(console.log.bind(console));

例：

[1,2,3].forEach(console.log) 
> TypeError: Illegal invocation 

[1,2,3].forEach(console.log.bind(console)) 
> 1 0 [1, 2, 3] 
> 2 1 [1, 2, 3] 
> 3 2 [1, 2, 3]

来源

2014-01-23 14:06:32 Tibos

传递console.log作为函数参数抛出“Uncaught TypeError：非法调用”

回答

相关问题