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This is actually from ECMAScript 2018's spread syntax and ECMAScript 2015's object destructuring .
{ ...state, animals } creates a shallow copy of the state object, with a new property called animals (with the value of animals object within it).
Since you are a Vuex user, this conforms to the rule of immutable update pattern s, which is to prevent the 'original' state object from being altered or mutated. You should read up on the ways of handling common operations such as add/update/delete, using the immutable pattern.
What does { …obj1, obj2 } do exactly [duplicate], This question already has an answer here: What do these three dots in React do ? 19 answers Let's say we have two objects: const state = { fishes: { /* some obj� What does { …obj1, obj2 } do exactly [duplicate] Ask Question Asked 8 months ago. Active 8 months ago. Viewed 246 times 1. 1. This question already has answers here: What do these three dots in React do? (23 answers) Closed 9 months ago. Let's say we h
It means Object.assign({}, state, { animals: animals} }.
Object to primitive conversion, What happens when objects are added obj1 + obj2 , subtracted obj1 - obj2 or We can fine-tune string and numeric conversion, using special object methods. There is no control whether toString returns exactly a string,� The variables obj1 and obj2 contain a reference. Each to a separate memory location. The important thing to understand here is that the variables, obj1 and obj2 (which could be an Object, Array or
What is does is it spreads all the properties of state into the new object - making a shallow copy. All the properties in the original object (state) will be copied into the new object (the one you're passing into replaceState). Here's a simple demonstration:
let obj1 = { obj: "1" };
let obj2 = { ...obj1, obj2: true };
console.log(obj1);
console.log(obj2);
Java Integer equals() method with Examples, Java Integer equals() Method. The equals() method is a method of Integer class under java.lang package. false obj1 and obj2 are equal. True or false? = true� /execute does not require the targets to be OP'd and is very necessary to use in order to do what you're trying to do. /scoreboard operations requires one of the two selectors to resolve to a single target, so you cannot use @a with @a. Using /execute: /execute @a ~ ~ ~ scoreboard players operation @a[c=1] internalCash = @a[c=1] cash
C++ Midterm Flashcards, Assume a class C exists with objects obj1, obj2, and obj3. For the statement obj3 = obj1 Of the following C++ operators, which can be overloaded. >= Operator� function hasSameData if both objects have the same number of keys (properties) if every key of obj1 matches with the corresponding key of obj2 and values of the same keys of both objects match. return true return false
How to Check if Two Objects Have Equal Values, So I'm going to say, "const obj1" and then this object is going to have a name property, and then So you would think that if you did something like this where you said, "obj1 === obj2," you would think that So what exactly am I doing here ? and in my program, I do: public static Main() { Class1 obj1 = new Class1(); Class1 obj2 = new Class1(); obj2.Parent = obj1; } does obj2.Parent contain a pointer to obj1 or is it a copy? If I wanted a pointer, would I use obj2.Parent = ref obj1; J.
How to Use C++ Operator Overloading with an Example Program, But, what if you want to do something like 'obj1 = obj2 + obj3' (where all This is what exactly can be done through operator overloading. Is using obj2.prototype = Object.create(obj1.prototype) better practice then obj2.prototype = obj1.prototype or does it make no difference? 7 comments. share. save hide report. 100% Upvoted. This thread is archived. New comments cannot be posted and vote
Comments { ... } is a object spread operatorI think you're looking for this : stackoverflow.com/questions/31048953/… Just fyi, array spread is not same as Object spread. Object spread is a part of ES7 and while array spread is of ES6 @Rajesh Thanks for pointing that out! It does indeed get confusing at times since spread applies for both objects and arrays after the proposals. "with a new property called animals" - but if the state object already contains a property animals, will it get overwritten? Also, doesn't the destructuring result in multiple parameters passed to the replaceState() function in my example? @NikolayDyankov yes, I believe it does. It in facts create a shallow copy from the 'original' state object. You can try testing it out by mutating the copied object and run console.log() on both of them to see the differences! However, I have to bring your attention to the fact that, if animals is a nested object, you will have to shallow copy the various levels of it. This link has an excellent example redux.js.org/recipes/structuring-reducers/… . I hope I am not boring you too much with all these URLs, but it is essential for you to understand this important concept. This is more of a comment. It would help readers if you add explanation as to what is happening, What's its purpose etc As far as I know Object.assign() makes a shallow merge of two objects, but why are you destructuring the second one? Actually, I am not destructuring the second one, but using the old, verbose ES5 object notation. Ohhh right. I started to see destructuring everywhere lol
