ecmascript-6函数来反转数组中的值

Question

Why doesn't this two codes do the same thing ?

Can someone explain to me where is the trick ?

 function randomize(arr){ for(var i = 0; i < arr.length; i++){ const random = Math.floor(Math.random()*arr.length); let temp = arr[i]; let newR = arr[random]; [temp, newR] = [newR, temp]; } return arr; } console.log(randomize([1,2,3,4,5])); 

 function randomize(arr){ for(var i = 0; i < arr.length; i++){ const random = Math.floor(Math.random()*arr.length); [arr[i], arr[random]] = [arr[random], arr[i]]; } return arr; } console.log(randomize([1,2,3,4,5])); 

I am most grateful for your read !

Answer 1

With the following, you are swapping temp and newR however, the original array items stay intact and are not swapped. Hence, the output remains the same as input.

 let temp = arr[i];   
 let newR = arr[random];
 [temp, newR] = [newR, temp];

With the later you are swapping the array items, hence, the array gets updated.

[arr[i], arr[random]] = [arr[random], arr[i]];

Answer 2

Because in JavaScript you cannot pass primitive values by reference, but only by value. Only compound values (Object, Array) can be assigned by reference.

let arr = ["one", 2, {name: "foo"}];

if you do

let item = arr[0];

then you are just copying it the value. Changing item will not modify the value of arr[i] , because there is no reference to it.

But if you would do following

let obj = arr[2];

you would have a reference to the object at the 3rd position. you could reassign obj with something different, that would not change arr[2] . But if you access the name property and change it, you are changing a property of a referenced object.

obj.name = "bar";
console.log(obj === arr[2]);
console.log(arr[2].name === "bar");

obj = "something else";
console.log(arr[2] !== obj);