Is Chrome’s JavaScript console lazy about evaluating objects?
Asked Answered
M

7

177

I’ll start with the code:

var s = ["hi"];
console.log(s);
s[0] = "bye";
console.log(s);

Simple, right? In response to this, the Firefox console says:

[ "hi" ]
[ "bye" ]

Wonderful, but Chrome’s JavaScript console (7.0.517.41 beta) says:

[ "bye" ]
[ "bye" ]

Have I done something wrong, or is Chrome’s JavaScript console being exceptionally lazy about evaluating my array?

Screenshot of the console exhibiting the described behavior.

Midgut answered 30/10, 2010 at 5:51 Comment(14)
I observe the same behavior in Safari -- so it's probably a webkit thing. Pretty surprising. I'd call it a bug.Candlenut
@mplungjan - that's not true. the first line declares a "plain old" array with a single element at index 0. The third line simply assigns a new value to that element. both cases are working with a simple numerically indexed array.Candlenut
if this is a bug, why this bug wasn't found and fixed is beyond my comprehension.Existential
To me it looks like a bug. On Linux Opera and Firefox display the expected result, Chrome and other Webkit-based browsers do not. You might want to report the issue to the Webkit devs: webkit.org/quality/reporting.htmlCollin
DOH - you are of course correct. I was not fully awakeLeesaleese
I found the same issue with Firebug for Firefox. Really disappointing. I suspected a shuffle function was behaving strangely, until I decided to check with jsbin and used .toString(). Here's the jsbin code. In the console counter part, from line 8 onward, the original array looks shuffled too.Mess
Man this was driving me crazy.Axon
See also console.log() async or sync? for a generic explanationGlacial
as of March 2016, this issue is no more.Stoplight
April 2020, having this issue in Chrome. Wasted 2 hours looking for a bug in my code that turned out to be a bug in Chrome.Christoperchristoph
Also worth noting that the blue i icon’s tooltip says “Value below was evaluated just now.”.Subdue
I solved mine by removing the setTimeout methodUnstopped
Related: console.log() shows the changed value of a variable before the value actually changes - it might be a better dupe target.Therefrom
I always lose time on this issue to. But I can only find others having this problem, but no always working solutionEquimolecular
M
97

Thanks for the comment, tec. I was able to find an existing unconfirmed Webkit bug that explains this issue: https://bugs.webkit.org/show_bug.cgi?id=35801 (EDIT: now fixed!)

There appears to be some debate regarding just how much of a bug it is and whether it's fixable. It does seem like bad behavior to me. It was especially troubling to me because, in Chrome at least, it occurs when the code resides in scripts that are executed immediately (before the page is loaded), even when the console is open, whenever the page is refreshed. Calling console.log when the console is not yet active only results in a reference to the object being queued, not the output the console will contain. Therefore, the array (or any object), will not be evaluated until the console is ready. It really is a case of lazy evaluation.

However, there is a simple way to avoid this in your code:

var s = ["hi"];
console.log(s.toString());
s[0] = "bye";
console.log(s.toString());

By calling toString, you create a representation in memory that will not be altered by following statements, which the console will read when it is ready. The console output is slightly different from passing the object directly, but it seems acceptable:

hi
bye
Midgut answered 30/10, 2010 at 18:54 Comment(5)
Actually, with associative arrays or other objects, this could be a real problem, since toString doesn't produce anything of value. Is there an easy work-around for objects in general?Midgut
webkit landed a patch for this a few months agoStoryteller
do this: console.log(JSON.parse(JSON.stringify(s));Eeg
I just wanted to mention that in the current Chrome version the console is delayed and outputting values wrong again (or was it ever right). For instance, I was logging an array and popping the top value after logging it, but it was showing up without the popped value. Your toString() suggestion was really helpful in getting to where I needed to get to see the values.Lheureux
Inserting a breakpoint from the code with debugger; is also a great option. (Or manually adding the breakpoint from the Developers Tools if it’s feasible).Ume
E
32

From Eric's explanation, it is due to console.log() being queued up, and it prints a later value of the array (or object).

There can be 5 solutions:

1. arr.toString()   // not well for [1,[2,3]] as it shows 1,2,3
2. arr.join()       // same as above
3. arr.slice(0)     // a new array is created, but if arr is [1, 2, arr2, 3] 
                    //   and arr2 changes, then later value might be shown
4. arr.concat()     // a new array is created, but same issue as slice(0)
5. JSON.stringify(arr)  // works well as it takes a snapshot of the whole array 
                        //   or object, and the format shows the exact structure
Existential answered 7/10, 2012 at 22:11 Comment(2)
Any solution that copies a list/object will work. My favourite shallow copy for objects is available since ECMAScript 2018: copy = {...orig}Fetation
@Fetation worth mentioning you shallow copy would turn the array into an objectCircumsolar
K
7

You can clone an array with Array#slice:

console.log(s); // ["bye"], i.e. incorrect
console.log(s.slice()); // ["hi"], i.e. correct

A function that you can use instead of console.log that doesn't have this problem is as follows:

console.logShallowCopy = function () {
    function slicedIfArray(arg) {
        return Array.isArray(arg) ? arg.slice() : arg;
    }

    var argsSnapshot = Array.prototype.map.call(arguments, slicedIfArray);
    return console.log.apply(console, argsSnapshot);
};

For the case of objects, unfortunately, the best method appears to be to debug first with a non-WebKit browser, or to write a complicated function to clone. If you are only working with simple objects, where order of keys doesn't matter and there are no functions, you could always do:

console.logSanitizedCopy = function () {
    var args = Array.prototype.slice.call(arguments);
    var sanitizedArgs = JSON.parse(JSON.stringify(args));

    return console.log.apply(console, sanitizedArgs);
};

All of these methods are obviously very slow, so even more so than with normal console.logs, you have to strip them off after you're done debugging.

Kristopher answered 27/1, 2011 at 17:42 Comment(0)
D
5

This has been patched in Webkit, however when using the React framework this happens for me in some circumstances, if you have such problems just use as others suggest:

console.log(JSON.stringify(the_array));
Discourteous answered 30/4, 2015 at 0:35 Comment(1)
Can confirm. This is literally the worst when trying to log out ReactSyntheticEvents. Even a JSON.parse(JSON.stringify(event)) doesn't get the right depth/accuracy. Debugger statements are the only real solution I've found to get the correct insight.Deweydewhirst
S
2

Looks like Chrome is replacing in its "pre compile" phase any instance of "s" with pointer to the actual array.

One way around is by cloning the array, logging fresh copy instead:

var s = ["hi"];
console.log(CloneArray(s));
s[0] = "bye";
console.log(CloneArray(s));

function CloneArray(array)
{
    var clone = new Array();
    for (var i = 0; i < array.length; i++)
        clone[clone.length] = array[i];
    return clone;
}
Spacious answered 31/10, 2010 at 14:20 Comment(9)
That's good, but because it's a shallow copy, there is still the possibility of a more subtle problem. And what about objects that aren't arrays? (Those are the real problem now.) I don't think that what you're saying about "pre compile" is accurate. Also, there is an error in the code: clone[clone.length] should be clone[i].Midgut
No error, I've executed it and it was OK. clone[clone.length] is exactly like clone[i], as the array start with length of 0, and so does the loop iterator "i". Anyway, not sure how it will behave with complex objects but IMO it's worth a try. Like I said, that's not a solution, it's a way around the problem..Spacious
@Shadow Wizard: Good point: clone.length will always be equal to i. It won't work for objects. Perhaps there is a solution with "for each".Midgut
Objects you mean this? var s = { param1: "hi", param2: "how are you?" }; if so I just tested and when you have s["param1"] = "bye"; it's working fine as expected. Can you please post example of "it won't work for objects"? I'll see and try to climb that one as well.Spacious
@Shadow Wizard: Obviously, your function will fail to clone properties and will not work on any objects without a length property. The webkit bug affects all objects, not just arrays.Midgut
@Domenic Because I wasn't familiar with slice back then.Spacious
@Shadow Wizard Fair enough :). I took over Anonymous's answer with a slice-based solution.Donough
@Dom why not new Answer then? :)Spacious
Well, it was already slice, but it needed some ESL and clarity fixes. The main content remains Mr. Mysterious Anonymous's so I thought I'd do an edit instead.Donough
S
2

the shortest solution so far is to use array or object spread syntax to get a clone of values to be preserved as in time of logging, ie:

console.log({...myObject});
console.log([...myArray]);

however be warned as it does a shallow copy, so any deep nested non-primitive values will not be cloned and thus shown in their modified state in the console

Seringapatam answered 12/4, 2020 at 7:52 Comment(1)
Nice idea, but it works only for one level deep data, it fails für each deeper nested objects/arraysEquimolecular
C
1

This is already answered, but I'll drop my answer anyway. I implemented a simple console wrapper which doesn't suffer from this issue. Requires jQuery.

It implements only log, warn and error methods, you will have to add some more in order for it to be interchangeable with a regular console.

var fixedConsole;
(function($) {
    var _freezeOne = function(arg) {
        if (typeof arg === 'object') {
            return $.extend(true, {}, arg);
        } else {
            return arg;
        }
    };
    var _freezeAll = function(args) {
        var frozen = [];
        for (var i=0; i<args.length; i++) {
            frozen.push(_freezeOne(args[i]));
        }
        return frozen;
    };
    fixedConsole = {
        log: function() { console.log.apply(console, _freezeAll(arguments)); },
        warn: function() { console.warn.apply(console, _freezeAll(arguments)); },
        error: function() { console.error.apply(console, _freezeAll(arguments)); }
    };
})(jQuery);
Cordovan answered 28/2, 2013 at 15:31 Comment(0)

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