'console.log() shows the changed value of a variable before the value actually changes
This bit of code I understand. We make a copy of A and call it C. When A is changed C stays the same
var A = 1;
var C = A;
console.log(C); // 1
A++;
console.log(C); // 1
But when A is an array we have a different sitiuation. Not only will C change, but it changes before we even touch A
var A = [2, 1];
var C = A;
console.log(C); // [1, 2]
A.sort();
console.log(C); // [1, 2]
Can someone explain what happened in the second example?
Solution 1:[1]
Console.log() is passed a reference to the object, so the value in the Console changes as the object changes. To avoid that you can:
console.log(JSON.parse(JSON.stringify(c)))
Please be warned that if you log objects in the latest versions of Chrome and Firefox what you get logged on the console is a reference to the object, which is not necessarily the 'value' of the object at the moment in time you call
console.log(), but it is the value of the object at the moment you open the console.
Solution 2:[2]
The latest guidance from Mozilla as of May 2022:
Don't use
console.log(obj), useconsole.log(JSON.parse(JSON.stringify(obj))).This way you are sure you are seeing the value of
objat the moment you log it. Otherwise, many browsers provide a live view that constantly updates as values change. This may not be what you want.
Solution 3:[3]
Arrays are objects. Variables refer to objects. Thus an assignment in the second case copied the reference (an address) to the array from "A" into "C". After that, both variables refer to the same single object (the array).
Primitive values like numbers are completely copied from one variable to another in simple assignments like yours. The "A++;" statement assigns a new value to "A".
To say it another way: the value of a variable may be either a primitive value (a number, a boolean, null, or a string), or it may be a reference to an object. The case of string primitives is a little weird, because they're more like objects than primitive (scalar) values, but they're immutable so it's OK to pretend they're just like numbers.
Solution 4:[4]
EDIT: Keeping this answer just to preserve useful comments below.
@Esailija is actually right - console.log() will not necessarily log the value the variable had at the time you tried to log it. In your case, both calls to console.log() will log the value of C after sorting.
If you try and execute the code in question as 5 separate statements in the console, you will see the result you expected (first, [2, 1], then [1, 2]).
Solution 5:[5]
Though it's not going to work in every situation, I ended up using a "break point" to solve this problem:
mysterious = {property:'started'}
// prints the value set below later ?
console.log(mysterious)
// break, console above prints the first value, as god intended
throw new Error()
// later
mysterious = {property:'changed', extended:'prop'}
Solution 6:[6]
The issue is present in Safari as well. As others have pointed out in this and similar questions, the console is passed a reference to the object, it prints the value of the object at the time the console was opened. If you execute the code in the console directly for example, the values print as expected. Instead of JSON stringifying, I prefer to spread arrays (e.g. in your case console.log([...C]);) and objects: the result is quite the same, but the code looks a bit cleaner. I have two VS code snippets to share.
"Print object value to console": {
"prefix": "clo",
"body": [
"console.log(\"Spread object: \", {...$0});"
],
"description": "Prints object value instead of reference to console, to avoid console.log async update"
},
"Print array value to console": {
"prefix": "cla",
"body": [
"console.log(\"Spread array: \", [...$0]);"
],
"description": "Prints array value instead of reference to console, to avoid console.log async update"
}
In order to get the same output as with console.log( JSON.parse(JSON.stringify(c))), you can leave out the string part if you wish. Incidentally, the spread syntax often saves time and code.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Wyck |
| Solution 2 | |
| Solution 3 | Andrew |
| Solution 4 | |
| Solution 5 | |
| Solution 6 |
