为什么console.log（）不能获取传递的变量的快照？ [英] Why doesn't console.log() take a snapshot of the passed variables?

问题描述

今天我遇到了一些很奇怪的行为与javascript。我想我现在已经弄清楚了，但我想知道我的想法是怎么回事真的发生，或者如果有一些其他的魔法参与。所以这是我的代码：

I've ran into some really weird behavior with javascript today. I think I got it somehow figured out now, but I'd like to know if what I think is going on is really happening or if there is some other magic involved. So this is my code:

    var SomeObject = {};

    SomeObject.foo = function(a, b) {
       var baz = this.bar(a, b);
       console.log(baz);
       console.log(baz.left);
       SomeObject.magicalStuff(baz);
    };

    SomeObject.bar = function(a, b) {
        return {left: a-b, top: b-a};
    };

    SomeObject.magicalStuff = function(position) {
        position.left = 0;
    };

    SomeObject.foo(100, 50);

jsFiddle

这个输出类似（取决于浏览器）：

The output of this is something like (depending on the browser):

> Object
50

如果展开对象（在Chrome，Safari或Firefox Firebug）你得到的是：

If you expand the "Object" (in Chrome, Safari or Firefox (Firebug) what you get is:

> Object
    left: 0
    top: -50

我会期望：

> Object
    left: 50
    top: -50

$ b b

我想的是，console.log（）真的只是发布一个对控制台的引用，一旦你点击展开符号读取，但不是那样的失败console.log（）作为调试工具的目的？
我总是期望console.log（）来传递我传递给它的东西，看到一个在实际控制台之后的语句真令人惊讶。 log（）改变console.log（）调用的输出。

What I think is going on is that console.log() really just "posts" a reference to the console, which gets read once you click on the "expand" symbol. But doesn't that kind of defeat the purpose of console.log() as a debugging instrument? I always expected console.log() to "snapshot" the stuff I pass to it. It is really surprising to see a statement which comes after the actual console.log() change the output of that very console.log() call.

或者还有其他事情发生吗？

Or is there something else going on?

编辑：我也想知道浏览器开发者是否有正确的理由来实现这样的console.log（我想有一个，否则它不会在主要浏览器）。

I'm also wondering if there is a sound reason for browser developers to implement console.log like this (I guess there is one, otherwise it wouldn't be consistent across major browsers).

推荐答案

有 console.dir（）

一般来说，每个级别的嵌套属性都不能打印，因为对象可以包含循环引用，例如 var a = {}; var b = {a：a}; ab = b;

实现完美的克隆它将不得不基本上只是转储整个内存，并且日志将需要非常长。考虑 console.log（window） ...

Implementing a perfect clone method is very hard - I guess it would have to basically just dump the whole memory, and logging would take awfully long. Think about console.log(window)...

这篇关于为什么console.log（）不能获取传递的变量的快照？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！