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












20














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







share|improve this question









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.
















  • 4




    in friends code ... but why ... ?
    – Jonas Wilms
    14 hours ago






  • 12




    Yeah, asking for a friend... ;)
    – Kresimir
    14 hours ago






  • 2




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






  • 1




    @jeremy views on SO are exponential, at some point they are reblogged, posted on twitter and hacker news ... usually that happens to the "good to know, but useless in reality" questions ...
    – Jonas Wilms
    7 hours ago
















20














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







share|improve this question









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.
















  • 4




    in friends code ... but why ... ?
    – Jonas Wilms
    14 hours ago






  • 12




    Yeah, asking for a friend... ;)
    – Kresimir
    14 hours ago






  • 2




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






  • 1




    @jeremy views on SO are exponential, at some point they are reblogged, posted on twitter and hacker news ... usually that happens to the "good to know, but useless in reality" questions ...
    – Jonas Wilms
    7 hours ago














20












20








20


3





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







share|improve this question









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.











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




javascript spread-syntax number-literal






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 10 hours ago





















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.









asked 14 hours ago









Mist

10816




10816




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.





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.






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








  • 4




    in friends code ... but why ... ?
    – Jonas Wilms
    14 hours ago






  • 12




    Yeah, asking for a friend... ;)
    – Kresimir
    14 hours ago






  • 2




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






  • 1




    @jeremy views on SO are exponential, at some point they are reblogged, posted on twitter and hacker news ... usually that happens to the "good to know, but useless in reality" questions ...
    – Jonas Wilms
    7 hours ago














  • 4




    in friends code ... but why ... ?
    – Jonas Wilms
    14 hours ago






  • 12




    Yeah, asking for a friend... ;)
    – Kresimir
    14 hours ago






  • 2




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






  • 1




    @jeremy views on SO are exponential, at some point they are reblogged, posted on twitter and hacker news ... usually that happens to the "good to know, but useless in reality" questions ...
    – Jonas Wilms
    7 hours ago








4




4




in friends code ... but why ... ?
– Jonas Wilms
14 hours ago




in friends code ... but why ... ?
– Jonas Wilms
14 hours ago




12




12




Yeah, asking for a friend... ;)
– Kresimir
14 hours ago




Yeah, asking for a friend... ;)
– Kresimir
14 hours ago




2




2




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




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




1




1




@jeremy views on SO are exponential, at some point they are reblogged, posted on twitter and hacker news ... usually that happens to the "good to know, but useless in reality" questions ...
– Jonas Wilms
7 hours ago




@jeremy views on SO are exponential, at some point they are reblogged, posted on twitter and hacker news ... usually that happens to the "good to know, but useless in reality" questions ...
– Jonas Wilms
7 hours ago












2 Answers
2






active

oldest

votes


















35














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 {}.






share|improve this answer



















  • 1




    Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
    – Mist
    14 hours ago








  • 6




    Spreading any number yields an empty object.
    – Kresimir
    14 hours ago






  • 4




    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
    9 hours 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
    9 hours ago












  • Gotcha. Thanks for the clarification.
    – Hitesh Kumar
    9 hours ago



















21














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 properties you get back an empty object.






share|improve this answer





















  • 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
    14 hours ago






  • 1




    @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
    14 hours 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
    14 hours ago










  • @mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
    – Jonas Wilms
    14 hours ago






  • 1




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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









35














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 {}.






share|improve this answer



















  • 1




    Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
    – Mist
    14 hours ago








  • 6




    Spreading any number yields an empty object.
    – Kresimir
    14 hours ago






  • 4




    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
    9 hours 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
    9 hours ago












  • Gotcha. Thanks for the clarification.
    – Hitesh Kumar
    9 hours ago
















35














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 {}.






share|improve this answer



















  • 1




    Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
    – Mist
    14 hours ago








  • 6




    Spreading any number yields an empty object.
    – Kresimir
    14 hours ago






  • 4




    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
    9 hours 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
    9 hours ago












  • Gotcha. Thanks for the clarification.
    – Hitesh Kumar
    9 hours ago














35












35








35






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 {}.






share|improve this answer














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 {}.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 14 hours ago









NikxDa

2,53811429




2,53811429








  • 1




    Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
    – Mist
    14 hours ago








  • 6




    Spreading any number yields an empty object.
    – Kresimir
    14 hours ago






  • 4




    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
    9 hours 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
    9 hours ago












  • Gotcha. Thanks for the clarification.
    – Hitesh Kumar
    9 hours ago














  • 1




    Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
    – Mist
    14 hours ago








  • 6




    Spreading any number yields an empty object.
    – Kresimir
    14 hours ago






  • 4




    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
    9 hours 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
    9 hours ago












  • Gotcha. Thanks for the clarification.
    – Hitesh Kumar
    9 hours ago








1




1




Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
– Mist
14 hours ago






Oh, I just learned that you can spread a number. Wouldn't expect that. Thanks!
– Mist
14 hours ago






6




6




Spreading any number yields an empty object.
– Kresimir
14 hours ago




Spreading any number yields an empty object.
– Kresimir
14 hours ago




4




4




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
9 hours 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
9 hours 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
9 hours 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
9 hours ago














Gotcha. Thanks for the clarification.
– Hitesh Kumar
9 hours ago




Gotcha. Thanks for the clarification.
– Hitesh Kumar
9 hours ago













21














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 properties you get back an empty object.






share|improve this answer





















  • 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
    14 hours ago






  • 1




    @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
    14 hours 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
    14 hours ago










  • @mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
    – Jonas Wilms
    14 hours ago






  • 1




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
















21














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 properties you get back an empty object.






share|improve this answer





















  • 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
    14 hours ago






  • 1




    @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
    14 hours 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
    14 hours ago










  • @mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
    – Jonas Wilms
    14 hours ago






  • 1




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














21












21








21






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 properties you get back an empty object.






share|improve this answer












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 properties you get back an empty object.







share|improve this answer












share|improve this answer



share|improve this answer










answered 14 hours ago









Jonas Wilms

54.4k42649




54.4k42649












  • 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
    14 hours ago






  • 1




    @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
    14 hours 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
    14 hours ago










  • @mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
    – Jonas Wilms
    14 hours ago






  • 1




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






  • 1




    @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
    14 hours 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
    14 hours ago










  • @mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
    – Jonas Wilms
    14 hours ago






  • 1




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




1




1




@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
14 hours 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
14 hours 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
14 hours 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
14 hours ago












@mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
– Jonas Wilms
14 hours ago




@mist It applies to all primitives (number, boolean, string), and it means that x.k is the same as (new Number(x)).x, and thats because JavaScript's design philosophy is everything is an object and thats how they implemented it (not the most elegant solution though in my eyes)
– Jonas Wilms
14 hours ago




1




1




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




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










Mist is a new contributor. Be nice, and check out our Code of Conduct.










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Mist is a new contributor. Be nice, and check out our Code of Conduct.












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Cypress Hill

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Dieser Artikel behandelt die Musikgruppe. Zum Musikalbum siehe Cypress Hill (Album). Cypress Hill Sen Dog, Bobo und B-Real Allgemeine Informationen Genre(s) Hip-Hop, Crossover, Reggae Gründung 1988 Website www.cypresshill.com Aktuelle Besetzung Rap B-Real Rap Sen Dog DJ DJ Muggs Percussion Eric Bobo (seit 1994) Cypress Hill ist eine US-amerikanische Hip-Hop-Gruppe aus Los Angeles. Sie verkauften über 18 Millionen Alben weltweit. [1] Die Gruppe besteht aus DJ Muggs , den Rappern B-Real und Sen Dog sowie seit 1994 Percussionist Bobo , der vorher im Umkreis der Beastie Boys tätig gewesen war. Sie wurden international vor allem auch wegen ihres Engagements für die Legalisierung des Cannabiskonsums bekannt. Der Name „Cypress Hill“ (Deutsch: Zypressenhügel) bezieht sich auf den Wohnort der Gründungsmitglieder, da fast alle in der Cypress Avenue in South Gate aufwuchsen. Bis auf DJ Muggs sind alle Mitglieder Hispanics. ...
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up vote 6 down vote favorite 2 I am in trouble: my boss, a multi-millionaire, has just come up with a scheme, but it has problems and he wants me to solve them. The whole thing started when he saw this picture: This gave him an idea: invent, develop, and sell "superpowers". These "powers" would not be biological (he's rich, but not that rich), but technological. Basically, he wants to have Small Hero Five-and-a-half (tm) working for him, constantly cranking out new tech. After he has his team invent the tech (with possible help from Anderson Robotics), he would start to mass-produce it (think: bat-mobile factory), and sell the gadgets (why sell the rights, when there is still more money to be made?) to would-be super-heroes. This is where the problems start: the reason no-one has done this yet, is ...
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