The code will output `{a: 1, b: 2}` to the console.

Here's why:

1. An object `obj` is declared with properties `a` and `b` set to `1` and `2` respectively.
2. An empty array `arr` is declared.
3. The line `arr[0] = { ...obj }` is using the spread operator (`...`) to copy all properties from `obj` into a new object. This new object is then assigned to the first position (index 0) of the array `arr`.
4. `console.log(arr[0])` will then output the object at the first position of the array, which is the object `{a: 1, b: 2}`.

Question

The code will output `{a: 1, b: 2}` to the console.

Here's why:

1. An object `obj` is declared with properties `a` and `b` set to `1` and `2` respectively.
2. An empty array `arr` is declared.
3. The line `arr[0] = { ...obj }` is using the spread operator (`...`) to copy all properties from `obj` into a new object. This new object is then assigned to the first position (index 0) of the array `arr`.
4. `console.log(arr[0])` will then output the object at the first position of the array, which is the object `{a: 1, b: 2}`.

Knowee AI · Accepted Answer

The code will output `{a: 1, b: 2}` to the console.

Here's why:

1. An object `obj` is declared with properties `a` and `b` set to `1` and `2` respectively.
2. An empty array `arr` is declared.
3. The line `arr[0] = { ...obj }` is using the spread operator (`...`) to copy all properties from `obj` into a new object. This new object is then assigned to the first position (index 0) of the array `arr`.
4. `console.log(arr[0])` will then output the object at the first position of the array, which is the object `{a: 1, b: 2}`.

6What will the code below output to the console and why?const obj = {a: 1,b: 2};const arr = [];console.log(arr[0] = { ...obj });Review Later1,2{…obj}{a: 1, b: 2}Does not compile

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