Why does {. . . .0} evaluate to {}?

I just found {....0} in friend's code. Evaluating it in console returns {} (empty object).

Why is that? What is the meaning of 4 dots in JavaScript?

edited Dec 25 '18 at 15:57

asked Dec 25 '18 at 11:37

Mist

49747

10

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.

– Jeremy Harris
Dec 25 '18 at 18:30

This is more of a "how this expression is parsed" question. Type this in JS console and you'll notice that the 4th dot is colored differently... same color as zero.

– Salman A
Dec 26 '18 at 9:38

1

Always relevant

– MikeTheLiar
Dec 26 '18 at 14:36

@JeremyHarris the magic of HNQ

– Pac0
Dec 26 '18 at 21:31

add a comment |

I just found {....0} in friend's code. Evaluating it in console returns {} (empty object).

Why is that? What is the meaning of 4 dots in JavaScript?

edited Dec 25 '18 at 15:57

asked Dec 25 '18 at 11:37

Mist

49747

10

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.

– Jeremy Harris
Dec 25 '18 at 18:30

This is more of a "how this expression is parsed" question. Type this in JS console and you'll notice that the 4th dot is colored differently... same color as zero.

– Salman A
Dec 26 '18 at 9:38

1

Always relevant

– MikeTheLiar
Dec 26 '18 at 14:36

@JeremyHarris the magic of HNQ

– Pac0
Dec 26 '18 at 21:31

add a comment |

I just found {....0} in friend's code. Evaluating it in console returns {} (empty object).

Why is that? What is the meaning of 4 dots in JavaScript?

edited Dec 25 '18 at 15:57

asked Dec 25 '18 at 11:37

Mist

49747

I just found {....0} in friend's code. Evaluating it in console returns {} (empty object).

Why is that? What is the meaning of 4 dots in JavaScript?

javascript spread-syntax number-literal

edited Dec 25 '18 at 15:57

asked Dec 25 '18 at 11:37

Mist

49747

edited Dec 25 '18 at 15:57

asked Dec 25 '18 at 11:37

Mist

49747

edited Dec 25 '18 at 15:57

asked Dec 25 '18 at 11:37

Mist

49747

asked Dec 25 '18 at 11:37

Mist

49747

asked Dec 25 '18 at 11:37

Mist

49747

10

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.

– Jeremy Harris
Dec 25 '18 at 18:30

This is more of a "how this expression is parsed" question. Type this in JS console and you'll notice that the 4th dot is colored differently... same color as zero.

– Salman A
Dec 26 '18 at 9:38

1

Always relevant

– MikeTheLiar
Dec 26 '18 at 14:36

@JeremyHarris the magic of HNQ

– Pac0
Dec 26 '18 at 21:31

add a comment |

10

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.

– Jeremy Harris
Dec 25 '18 at 18:30

This is more of a "how this expression is parsed" question. Type this in JS console and you'll notice that the 4th dot is colored differently... same color as zero.

– Salman A
Dec 26 '18 at 9:38

1

Always relevant

– MikeTheLiar
Dec 26 '18 at 14:36

@JeremyHarris the magic of HNQ

– Pac0
Dec 26 '18 at 21:31

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.

– Jeremy Harris
Dec 25 '18 at 18:30

This is more of a "how this expression is parsed" question. Type this in JS console and you'll notice that the 4th dot is colored differently... same color as zero.

– Salman A
Dec 26 '18 at 9:38

Always relevant

– MikeTheLiar
Dec 26 '18 at 14:36

@JeremyHarris the magic of HNQ

– Pac0
Dec 26 '18 at 21:31

add a comment |

4 Answers
4

active

oldest

votes

Four dots actually have no meaning. ... is the spread operator, and .0 is short for 0.0.

Spreading 0 (or any number) into an object yields an empty object, therefore {}.

edited Dec 26 '18 at 1:07

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

11

Spreading any number yields an empty object.

– Kresimir
Dec 25 '18 at 11:43

9

Spreading 0 (or any number) yields an empty object not necessarily if you spread a number at any other places apart from an object, it will throw an error eg [...0] throws an error.

– Hitesh Kumar
Dec 25 '18 at 16:37

2

@HiteshKumar Spreading non-iterable objects inside an array will indeed throw an error, but that has nothing to do with this question. I am referring to the object-spread mentioned. :)

– NikxDa
Dec 25 '18 at 16:39

2

NikxDa I think that @HiteshKumar made an important point. It's better to be more explicit about cases where your statements holds true. Spreading 0 (or any number) in object literal yields an empty object Contains more useful information..

– Mist
Dec 26 '18 at 0:29

1

@Mist I've updated the answer. I don't think it's needed, but it might be good for clarification. Thanks for the update!

– NikxDa
Dec 26 '18 at 1:08

|
show 2 more comments

Three dots in an object literal are a spread property, e.g.:

  const a = { b: 1, c: 1 };

  const d = { ...a, e: 1 }; // { b: 1, c: 1, e: 1 }

The last dot with a 0 is a number literal .0 is the same as 0.0. Therefore this:

 { ...(0.0) }

spreads all properties of the number object into the object, however as numbers don't have any (own) properties you get back an empty object.

edited Dec 26 '18 at 19:59

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

You lead me to thinking - I can spread any variable, and own keys will be spread into the new object? It works for Function (function x() {}), (x.k = 'v'), ({...x})// {k: 'v'} but doesn't work for Number (x = 10), (x.k = 'v'), ({...x}) // {}

– Mist
Dec 25 '18 at 11:48

3

@mist because numbers (and other primitives) get "boxed" into objects when you work with them as objects, and "unboxed" directly afterwards. Therefore x.k will get lost.

– Jonas Wilms
Dec 25 '18 at 11:50

What does 'boxed' means exactly? E.g. when I used the dot operator (property) I worked with the number as object. If I am correct, that's just one case. Are there other cases when 'boxing' is happening? Does it apply only to numbers? Is there a perf reason or something? I guess this is for other question, and I should study it further. Could you point me to some book or something?

– Mist
Dec 25 '18 at 11:56

1

Thanks! I see why my key on number couldn't work. Yayy boxing!

– Mist
Dec 25 '18 at 12:06

3

Numbers don't have any own enumerable properties. But they do have properties.

– Patrick Roberts
Dec 26 '18 at 3:39

|
show 4 more comments

In a simple terms {...} spread operator in javascript extends one object/array with another.

So, when babelifier tries extending one with another, it has to identify whether it is trying to extend an array or an object.

In the case of array, it iterates over elements.

In the case of object, it iterates over keys.

In this scenario, the babelyfier is trying to extract keys for number by checking the Object's own property call which is missing for number so it returns empty Object.

edited Dec 26 '18 at 18:02

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

11

new word! i.stack.imgur.com/VyYqA.png

– uhoh
Dec 27 '18 at 0:45

add a comment |

Spread operator {...} allows iterables to expand. It means that those data types that can be defined in form of key-value pairs can be expanded. In terms of Object we call key-value pair as Object property and it's value whereas in terms of arrays we can think index as key and element in array as it's value.

let obj = { a: 4, b: 1};

let obj2 = { ...obj, c: 2, d: 4}; // {a: 4, b: 1, c: 2, d: 4}



let arr1 = ['1', '2'];

let obj3 = { ...arr1, ...['3']}; // {0: "3", 1: "2"}

In terms of array, as it takes index as key so here it replaces element '1' of arr1 with '3' because both of them have same index in different array.

With strings too spread operator returns non-empty object. As string is an array of character so it treats string as an array.

let obj4 = {...'hi',...'hello'}   // {0: "h", 1: "e", 2: "l", 3: "l", 4: "o"}

let obj5 = {...'y',...'x'}   // {0: "x" }

But with other primitive data types it return empty object

with Numbers

let obj6 = { ...0.0, ...55} // {}

with Boolean

let obj7 = { ...true, ...false} // {}

In conclusion those data types that can be treated in form of key-value pairs when used with spread operator {...} returns non-empty object otherwise it returns empty object {}

answered Jan 5 at 5:25

JS Engine

666313

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Four dots actually have no meaning. ... is the spread operator, and .0 is short for 0.0.

Spreading 0 (or any number) into an object yields an empty object, therefore {}.

edited Dec 26 '18 at 1:07

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

11

Spreading any number yields an empty object.

– Kresimir
Dec 25 '18 at 11:43

9

Spreading 0 (or any number) yields an empty object not necessarily if you spread a number at any other places apart from an object, it will throw an error eg [...0] throws an error.

– Hitesh Kumar
Dec 25 '18 at 16:37

2

@HiteshKumar Spreading non-iterable objects inside an array will indeed throw an error, but that has nothing to do with this question. I am referring to the object-spread mentioned. :)

– NikxDa
Dec 25 '18 at 16:39

2

NikxDa I think that @HiteshKumar made an important point. It's better to be more explicit about cases where your statements holds true. Spreading 0 (or any number) in object literal yields an empty object Contains more useful information..

– Mist
Dec 26 '18 at 0:29

1

@Mist I've updated the answer. I don't think it's needed, but it might be good for clarification. Thanks for the update!

– NikxDa
Dec 26 '18 at 1:08

|
show 2 more comments

Four dots actually have no meaning. ... is the spread operator, and .0 is short for 0.0.

Spreading 0 (or any number) into an object yields an empty object, therefore {}.

edited Dec 26 '18 at 1:07

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

11

Spreading any number yields an empty object.

– Kresimir
Dec 25 '18 at 11:43

9

Spreading 0 (or any number) yields an empty object not necessarily if you spread a number at any other places apart from an object, it will throw an error eg [...0] throws an error.

– Hitesh Kumar
Dec 25 '18 at 16:37

2

@HiteshKumar Spreading non-iterable objects inside an array will indeed throw an error, but that has nothing to do with this question. I am referring to the object-spread mentioned. :)

– NikxDa
Dec 25 '18 at 16:39

2

NikxDa I think that @HiteshKumar made an important point. It's better to be more explicit about cases where your statements holds true. Spreading 0 (or any number) in object literal yields an empty object Contains more useful information..

– Mist
Dec 26 '18 at 0:29

1

@Mist I've updated the answer. I don't think it's needed, but it might be good for clarification. Thanks for the update!

– NikxDa
Dec 26 '18 at 1:08

|
show 2 more comments

Four dots actually have no meaning. ... is the spread operator, and .0 is short for 0.0.

Spreading 0 (or any number) into an object yields an empty object, therefore {}.

edited Dec 26 '18 at 1:07

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

Four dots actually have no meaning. ... is the spread operator, and .0 is short for 0.0.

Spreading 0 (or any number) into an object yields an empty object, therefore {}.

edited Dec 26 '18 at 1:07

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

edited Dec 26 '18 at 1:07

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

answered Dec 25 '18 at 11:40

NikxDa

3,12111737

11

Spreading any number yields an empty object.

– Kresimir
Dec 25 '18 at 11:43

9

Spreading 0 (or any number) yields an empty object not necessarily if you spread a number at any other places apart from an object, it will throw an error eg [...0] throws an error.

– Hitesh Kumar
Dec 25 '18 at 16:37

2

@HiteshKumar Spreading non-iterable objects inside an array will indeed throw an error, but that has nothing to do with this question. I am referring to the object-spread mentioned. :)

– NikxDa
Dec 25 '18 at 16:39

2

NikxDa I think that @HiteshKumar made an important point. It's better to be more explicit about cases where your statements holds true. Spreading 0 (or any number) in object literal yields an empty object Contains more useful information..

– Mist
Dec 26 '18 at 0:29

1

@Mist I've updated the answer. I don't think it's needed, but it might be good for clarification. Thanks for the update!

– NikxDa
Dec 26 '18 at 1:08

|
show 2 more comments

11

Spreading any number yields an empty object.

– Kresimir
Dec 25 '18 at 11:43

9

Spreading 0 (or any number) yields an empty object not necessarily if you spread a number at any other places apart from an object, it will throw an error eg [...0] throws an error.

– Hitesh Kumar
Dec 25 '18 at 16:37

2

@HiteshKumar Spreading non-iterable objects inside an array will indeed throw an error, but that has nothing to do with this question. I am referring to the object-spread mentioned. :)

– NikxDa
Dec 25 '18 at 16:39

2

NikxDa I think that @HiteshKumar made an important point. It's better to be more explicit about cases where your statements holds true. Spreading 0 (or any number) in object literal yields an empty object Contains more useful information..

– Mist
Dec 26 '18 at 0:29

1

@Mist I've updated the answer. I don't think it's needed, but it might be good for clarification. Thanks for the update!

– NikxDa
Dec 26 '18 at 1:08

Spreading any number yields an empty object.

– Kresimir
Dec 25 '18 at 11:43

Spreading 0 (or any number) yields an empty object not necessarily if you spread a number at any other places apart from an object, it will throw an error eg [...0] throws an error.

– Hitesh Kumar
Dec 25 '18 at 16:37

@HiteshKumar Spreading non-iterable objects inside an array will indeed throw an error, but that has nothing to do with this question. I am referring to the object-spread mentioned. :)

– NikxDa
Dec 25 '18 at 16:39

NikxDa I think that @HiteshKumar made an important point. It's better to be more explicit about cases where your statements holds true. Spreading 0 (or any number) in object literal yields an empty object Contains more useful information..

– Mist
Dec 26 '18 at 0:29

@Mist I've updated the answer. I don't think it's needed, but it might be good for clarification. Thanks for the update!

– NikxDa
Dec 26 '18 at 1:08

|
show 2 more comments

Three dots in an object literal are a spread property, e.g.:

  const a = { b: 1, c: 1 };

  const d = { ...a, e: 1 }; // { b: 1, c: 1, e: 1 }

The last dot with a 0 is a number literal .0 is the same as 0.0. Therefore this:

 { ...(0.0) }

spreads all properties of the number object into the object, however as numbers don't have any (own) properties you get back an empty object.

edited Dec 26 '18 at 19:59

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

You lead me to thinking - I can spread any variable, and own keys will be spread into the new object? It works for Function (function x() {}), (x.k = 'v'), ({...x})// {k: 'v'} but doesn't work for Number (x = 10), (x.k = 'v'), ({...x}) // {}

– Mist
Dec 25 '18 at 11:48

3

@mist because numbers (and other primitives) get "boxed" into objects when you work with them as objects, and "unboxed" directly afterwards. Therefore x.k will get lost.

– Jonas Wilms
Dec 25 '18 at 11:50

What does 'boxed' means exactly? E.g. when I used the dot operator (property) I worked with the number as object. If I am correct, that's just one case. Are there other cases when 'boxing' is happening? Does it apply only to numbers? Is there a perf reason or something? I guess this is for other question, and I should study it further. Could you point me to some book or something?

– Mist
Dec 25 '18 at 11:56

1

Thanks! I see why my key on number couldn't work. Yayy boxing!

– Mist
Dec 25 '18 at 12:06

3

Numbers don't have any own enumerable properties. But they do have properties.

– Patrick Roberts
Dec 26 '18 at 3:39

|
show 4 more comments

Three dots in an object literal are a spread property, e.g.:

  const a = { b: 1, c: 1 };

  const d = { ...a, e: 1 }; // { b: 1, c: 1, e: 1 }

The last dot with a 0 is a number literal .0 is the same as 0.0. Therefore this:

 { ...(0.0) }

spreads all properties of the number object into the object, however as numbers don't have any (own) properties you get back an empty object.

edited Dec 26 '18 at 19:59

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

You lead me to thinking - I can spread any variable, and own keys will be spread into the new object? It works for Function (function x() {}), (x.k = 'v'), ({...x})// {k: 'v'} but doesn't work for Number (x = 10), (x.k = 'v'), ({...x}) // {}

– Mist
Dec 25 '18 at 11:48

3

@mist because numbers (and other primitives) get "boxed" into objects when you work with them as objects, and "unboxed" directly afterwards. Therefore x.k will get lost.

– Jonas Wilms
Dec 25 '18 at 11:50

What does 'boxed' means exactly? E.g. when I used the dot operator (property) I worked with the number as object. If I am correct, that's just one case. Are there other cases when 'boxing' is happening? Does it apply only to numbers? Is there a perf reason or something? I guess this is for other question, and I should study it further. Could you point me to some book or something?

– Mist
Dec 25 '18 at 11:56

1

Thanks! I see why my key on number couldn't work. Yayy boxing!

– Mist
Dec 25 '18 at 12:06

3

Numbers don't have any own enumerable properties. But they do have properties.

– Patrick Roberts
Dec 26 '18 at 3:39

|
show 4 more comments

Three dots in an object literal are a spread property, e.g.:

  const a = { b: 1, c: 1 };

  const d = { ...a, e: 1 }; // { b: 1, c: 1, e: 1 }

The last dot with a 0 is a number literal .0 is the same as 0.0. Therefore this:

 { ...(0.0) }

spreads all properties of the number object into the object, however as numbers don't have any (own) properties you get back an empty object.

edited Dec 26 '18 at 19:59

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

Three dots in an object literal are a spread property, e.g.:

  const a = { b: 1, c: 1 };

  const d = { ...a, e: 1 }; // { b: 1, c: 1, e: 1 }

The last dot with a 0 is a number literal .0 is the same as 0.0. Therefore this:

 { ...(0.0) }

spreads all properties of the number object into the object, however as numbers don't have any (own) properties you get back an empty object.

edited Dec 26 '18 at 19:59

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

edited Dec 26 '18 at 19:59

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

answered Dec 25 '18 at 11:42

Jonas Wilms

57.6k43051

You lead me to thinking - I can spread any variable, and own keys will be spread into the new object? It works for Function (function x() {}), (x.k = 'v'), ({...x})// {k: 'v'} but doesn't work for Number (x = 10), (x.k = 'v'), ({...x}) // {}

– Mist
Dec 25 '18 at 11:48

3

@mist because numbers (and other primitives) get "boxed" into objects when you work with them as objects, and "unboxed" directly afterwards. Therefore x.k will get lost.

– Jonas Wilms
Dec 25 '18 at 11:50

What does 'boxed' means exactly? E.g. when I used the dot operator (property) I worked with the number as object. If I am correct, that's just one case. Are there other cases when 'boxing' is happening? Does it apply only to numbers? Is there a perf reason or something? I guess this is for other question, and I should study it further. Could you point me to some book or something?

– Mist
Dec 25 '18 at 11:56

1

Thanks! I see why my key on number couldn't work. Yayy boxing!

– Mist
Dec 25 '18 at 12:06

3

Numbers don't have any own enumerable properties. But they do have properties.

– Patrick Roberts
Dec 26 '18 at 3:39

|
show 4 more comments

You lead me to thinking - I can spread any variable, and own keys will be spread into the new object? It works for Function (function x() {}), (x.k = 'v'), ({...x})// {k: 'v'} but doesn't work for Number (x = 10), (x.k = 'v'), ({...x}) // {}

– Mist
Dec 25 '18 at 11:48

3

@mist because numbers (and other primitives) get "boxed" into objects when you work with them as objects, and "unboxed" directly afterwards. Therefore x.k will get lost.

– Jonas Wilms
Dec 25 '18 at 11:50

What does 'boxed' means exactly? E.g. when I used the dot operator (property) I worked with the number as object. If I am correct, that's just one case. Are there other cases when 'boxing' is happening? Does it apply only to numbers? Is there a perf reason or something? I guess this is for other question, and I should study it further. Could you point me to some book or something?

– Mist
Dec 25 '18 at 11:56

1

Thanks! I see why my key on number couldn't work. Yayy boxing!

– Mist
Dec 25 '18 at 12:06

3

Numbers don't have any own enumerable properties. But they do have properties.

– Patrick Roberts
Dec 26 '18 at 3:39

You lead me to thinking - I can spread any variable, and own keys will be spread into the new object? It works for Function (function x() {}), (x.k = 'v'), ({...x})// {k: 'v'} but doesn't work for Number (x = 10), (x.k = 'v'), ({...x}) // {}

– Mist
Dec 25 '18 at 11:48

@mist because numbers (and other primitives) get "boxed" into objects when you work with them as objects, and "unboxed" directly afterwards. Therefore x.k will get lost.

– Jonas Wilms
Dec 25 '18 at 11:50

What does 'boxed' means exactly? E.g. when I used the dot operator (property) I worked with the number as object. If I am correct, that's just one case. Are there other cases when 'boxing' is happening? Does it apply only to numbers? Is there a perf reason or something? I guess this is for other question, and I should study it further. Could you point me to some book or something?

– Mist
Dec 25 '18 at 11:56

Thanks! I see why my key on number couldn't work. Yayy boxing!

– Mist
Dec 25 '18 at 12:06

Numbers don't have any own enumerable properties. But they do have properties.

– Patrick Roberts
Dec 26 '18 at 3:39

|
show 4 more comments

In a simple terms {...} spread operator in javascript extends one object/array with another.

So, when babelifier tries extending one with another, it has to identify whether it is trying to extend an array or an object.

In the case of array, it iterates over elements.

In the case of object, it iterates over keys.

In this scenario, the babelyfier is trying to extract keys for number by checking the Object's own property call which is missing for number so it returns empty Object.

edited Dec 26 '18 at 18:02

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

11

new word! i.stack.imgur.com/VyYqA.png

– uhoh
Dec 27 '18 at 0:45

add a comment |

In a simple terms {...} spread operator in javascript extends one object/array with another.

So, when babelifier tries extending one with another, it has to identify whether it is trying to extend an array or an object.

In the case of array, it iterates over elements.

In the case of object, it iterates over keys.

In this scenario, the babelyfier is trying to extract keys for number by checking the Object's own property call which is missing for number so it returns empty Object.

edited Dec 26 '18 at 18:02

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

11

new word! i.stack.imgur.com/VyYqA.png

– uhoh
Dec 27 '18 at 0:45

add a comment |

In a simple terms {...} spread operator in javascript extends one object/array with another.

So, when babelifier tries extending one with another, it has to identify whether it is trying to extend an array or an object.

In the case of array, it iterates over elements.

In the case of object, it iterates over keys.

In this scenario, the babelyfier is trying to extract keys for number by checking the Object's own property call which is missing for number so it returns empty Object.

edited Dec 26 '18 at 18:02

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

In a simple terms {...} spread operator in javascript extends one object/array with another.

So, when babelifier tries extending one with another, it has to identify whether it is trying to extend an array or an object.

In the case of array, it iterates over elements.

In the case of object, it iterates over keys.

In this scenario, the babelyfier is trying to extract keys for number by checking the Object's own property call which is missing for number so it returns empty Object.

edited Dec 26 '18 at 18:02

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

edited Dec 26 '18 at 18:02

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

answered Dec 26 '18 at 11:51

Rajendra kumar Vankadari

1,207912

11

new word! i.stack.imgur.com/VyYqA.png

– uhoh
Dec 27 '18 at 0:45

add a comment |

11

new word! i.stack.imgur.com/VyYqA.png

– uhoh
Dec 27 '18 at 0:45

new word! i.stack.imgur.com/VyYqA.png

– uhoh
Dec 27 '18 at 0:45

add a comment |

let obj = { a: 4, b: 1};

let obj2 = { ...obj, c: 2, d: 4}; // {a: 4, b: 1, c: 2, d: 4}



let arr1 = ['1', '2'];

let obj3 = { ...arr1, ...['3']}; // {0: "3", 1: "2"}

In terms of array, as it takes index as key so here it replaces element '1' of arr1 with '3' because both of them have same index in different array.

With strings too spread operator returns non-empty object. As string is an array of character so it treats string as an array.

let obj4 = {...'hi',...'hello'}   // {0: "h", 1: "e", 2: "l", 3: "l", 4: "o"}

let obj5 = {...'y',...'x'}   // {0: "x" }

But with other primitive data types it return empty object

with Numbers

let obj6 = { ...0.0, ...55} // {}

with Boolean

let obj7 = { ...true, ...false} // {}

In conclusion those data types that can be treated in form of key-value pairs when used with spread operator {...} returns non-empty object otherwise it returns empty object {}

answered Jan 5 at 5:25

JS Engine

666313

add a comment |

let obj = { a: 4, b: 1};

let obj2 = { ...obj, c: 2, d: 4}; // {a: 4, b: 1, c: 2, d: 4}



let arr1 = ['1', '2'];

let obj3 = { ...arr1, ...['3']}; // {0: "3", 1: "2"}

In terms of array, as it takes index as key so here it replaces element '1' of arr1 with '3' because both of them have same index in different array.

With strings too spread operator returns non-empty object. As string is an array of character so it treats string as an array.

let obj4 = {...'hi',...'hello'}   // {0: "h", 1: "e", 2: "l", 3: "l", 4: "o"}

let obj5 = {...'y',...'x'}   // {0: "x" }

But with other primitive data types it return empty object

with Numbers

let obj6 = { ...0.0, ...55} // {}

with Boolean

let obj7 = { ...true, ...false} // {}

In conclusion those data types that can be treated in form of key-value pairs when used with spread operator {...} returns non-empty object otherwise it returns empty object {}

answered Jan 5 at 5:25

JS Engine

666313

add a comment |

let obj = { a: 4, b: 1};

let obj2 = { ...obj, c: 2, d: 4}; // {a: 4, b: 1, c: 2, d: 4}



let arr1 = ['1', '2'];

let obj3 = { ...arr1, ...['3']}; // {0: "3", 1: "2"}

In terms of array, as it takes index as key so here it replaces element '1' of arr1 with '3' because both of them have same index in different array.

With strings too spread operator returns non-empty object. As string is an array of character so it treats string as an array.

let obj4 = {...'hi',...'hello'}   // {0: "h", 1: "e", 2: "l", 3: "l", 4: "o"}

let obj5 = {...'y',...'x'}   // {0: "x" }

But with other primitive data types it return empty object

with Numbers

let obj6 = { ...0.0, ...55} // {}

with Boolean

let obj7 = { ...true, ...false} // {}

In conclusion those data types that can be treated in form of key-value pairs when used with spread operator {...} returns non-empty object otherwise it returns empty object {}

answered Jan 5 at 5:25

JS Engine

666313

let obj = { a: 4, b: 1};

let obj2 = { ...obj, c: 2, d: 4}; // {a: 4, b: 1, c: 2, d: 4}



let arr1 = ['1', '2'];

let obj3 = { ...arr1, ...['3']}; // {0: "3", 1: "2"}

In terms of array, as it takes index as key so here it replaces element '1' of arr1 with '3' because both of them have same index in different array.

With strings too spread operator returns non-empty object. As string is an array of character so it treats string as an array.

let obj4 = {...'hi',...'hello'}   // {0: "h", 1: "e", 2: "l", 3: "l", 4: "o"}

let obj5 = {...'y',...'x'}   // {0: "x" }

But with other primitive data types it return empty object

with Numbers

let obj6 = { ...0.0, ...55} // {}

with Boolean

let obj7 = { ...true, ...false} // {}

In conclusion those data types that can be treated in form of key-value pairs when used with spread operator {...} returns non-empty object otherwise it returns empty object {}

answered Jan 5 at 5:25

JS Engine

666313

answered Jan 5 at 5:25

JS Engine

666313

answered Jan 5 at 5:25

JS Engine

666313

answered Jan 5 at 5:25

JS Engine

666313

add a comment |

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