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 2 days ago

asked 2 days ago

Mist

35937

New contributor

9

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.
– Jeremy Harris
2 days ago

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
yesterday

1

Always relevant
– MikeTheLiar
yesterday

@JeremyHarris the magic of HNQ
– Pac0
yesterday

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 2 days ago

asked 2 days ago

Mist

35937

New contributor

9

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.
– Jeremy Harris
2 days ago

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
yesterday

1

Always relevant
– MikeTheLiar
yesterday

@JeremyHarris the magic of HNQ
– Pac0
yesterday

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 2 days ago

asked 2 days ago

Mist

35937

New contributor

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 2 days ago

asked 2 days ago

Mist

35937

New contributor

edited 2 days ago

asked 2 days ago

Mist

35937

New contributor

edited 2 days ago

asked 2 days ago

Mist

35937

New contributor

asked 2 days ago

Mist

35937

asked 2 days ago

Mist

35937

New contributor

Mist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

9

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.
– Jeremy Harris
2 days ago

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
yesterday

1

Always relevant
– MikeTheLiar
yesterday

@JeremyHarris the magic of HNQ
– Pac0
yesterday

add a comment |

9

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.
– Jeremy Harris
2 days ago

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
yesterday

1

Always relevant
– MikeTheLiar
yesterday

@JeremyHarris the magic of HNQ
– Pac0
yesterday

Viewed almost 2500 times in 6 hours? It appears your friend is using the spread operator in a different context.
– Jeremy Harris
2 days ago

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
yesterday

Always relevant
– MikeTheLiar
yesterday

@JeremyHarris the magic of HNQ
– Pac0
yesterday

add a comment |

3 Answers
3

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 2 days ago

answered 2 days ago

NikxDa

2,78311531

9

Spreading any number yields an empty object.
– Kresimir
2 days ago

8

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
2 days ago

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
2 days ago

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
2 days ago

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
2 days ago

|
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 yesterday

answered 2 days ago

Jonas Wilms

54.8k42749

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
2 days ago

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
2 days ago

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
2 days ago

1

Thanks! I see why my key on number couldn't work. Yayy boxing!
– Mist
2 days ago

3

Numbers don't have any own enumerable properties. But they do have properties.
– Patrick Roberts
2 days ago

|
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 yesterday

answered yesterday

Rajendra kumar Vankadari

1,087911

7

new word! i.stack.imgur.com/VyYqA.png
– uhoh
yesterday

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

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 2 days ago

answered 2 days ago

NikxDa

2,78311531

9

Spreading any number yields an empty object.
– Kresimir
2 days ago

8

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
2 days ago

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
2 days ago

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
2 days ago

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
2 days ago

|
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 2 days ago

answered 2 days ago

NikxDa

2,78311531

9

Spreading any number yields an empty object.
– Kresimir
2 days ago

8

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
2 days ago

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
2 days ago

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
2 days ago

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
2 days ago

|
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 2 days ago

answered 2 days ago

NikxDa

2,78311531

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 2 days ago

answered 2 days ago

NikxDa

2,78311531

edited 2 days ago

answered 2 days ago

NikxDa

2,78311531

answered 2 days ago

NikxDa

2,78311531

answered 2 days ago

NikxDa

2,78311531

9

Spreading any number yields an empty object.
– Kresimir
2 days ago

8

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
2 days ago

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
2 days ago

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
2 days ago

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
2 days ago

|
show 2 more comments

9

Spreading any number yields an empty object.
– Kresimir
2 days ago

8

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
2 days ago

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
2 days ago

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
2 days ago

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
2 days ago

Spreading any number yields an empty object.
– Kresimir
2 days ago

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
2 days ago

@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
2 days ago

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
2 days ago

@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
2 days ago

|
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 yesterday

answered 2 days ago

Jonas Wilms

54.8k42749

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
2 days ago

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
2 days ago

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
2 days ago

1

Thanks! I see why my key on number couldn't work. Yayy boxing!
– Mist
2 days ago

3

Numbers don't have any own enumerable properties. But they do have properties.
– Patrick Roberts
2 days ago

|
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 yesterday

answered 2 days ago

Jonas Wilms

54.8k42749

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
2 days ago

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
2 days ago

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
2 days ago

1

Thanks! I see why my key on number couldn't work. Yayy boxing!
– Mist
2 days ago

3

Numbers don't have any own enumerable properties. But they do have properties.
– Patrick Roberts
2 days ago

|
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 yesterday

answered 2 days ago

Jonas Wilms

54.8k42749

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 yesterday

answered 2 days ago

Jonas Wilms

54.8k42749

edited yesterday

answered 2 days ago

Jonas Wilms

54.8k42749

answered 2 days ago

Jonas Wilms

54.8k42749

answered 2 days ago

Jonas Wilms

54.8k42749

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
2 days ago

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
2 days ago

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
2 days ago

1

Thanks! I see why my key on number couldn't work. Yayy boxing!
– Mist
2 days ago

3

Numbers don't have any own enumerable properties. But they do have properties.
– Patrick Roberts
2 days ago

|
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
2 days ago

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
2 days ago

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
2 days ago

1

Thanks! I see why my key on number couldn't work. Yayy boxing!
– Mist
2 days ago

3

Numbers don't have any own enumerable properties. But they do have properties.
– Patrick Roberts
2 days ago

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
2 days ago

@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
2 days ago

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
2 days ago

Thanks! I see why my key on number couldn't work. Yayy boxing!
– Mist
2 days ago

Numbers don't have any own enumerable properties. But they do have properties.
– Patrick Roberts
2 days ago

|
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 yesterday

answered yesterday

Rajendra kumar Vankadari

1,087911

7

new word! i.stack.imgur.com/VyYqA.png
– uhoh
yesterday

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 yesterday

answered yesterday

Rajendra kumar Vankadari

1,087911

7

new word! i.stack.imgur.com/VyYqA.png
– uhoh
yesterday

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 yesterday

answered yesterday

Rajendra kumar Vankadari

1,087911

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 yesterday

answered yesterday

Rajendra kumar Vankadari

1,087911

edited yesterday

answered yesterday

Rajendra kumar Vankadari

1,087911

answered yesterday

Rajendra kumar Vankadari

1,087911

answered yesterday

Rajendra kumar Vankadari

1,087911

7

new word! i.stack.imgur.com/VyYqA.png
– uhoh
yesterday

add a comment |

7

new word! i.stack.imgur.com/VyYqA.png
– uhoh
yesterday

new word! i.stack.imgur.com/VyYqA.png
– uhoh
yesterday

add a comment |

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