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I think the inferred type should be this: { value?: number: undefined } for { ...a, ...b } and this: { value: number } for { ...b, ...a } .
– Vanuan
Mar 10 '20 at 22:41
Found a relevant issue: github.com/microsoft/TypeScript/issues/13195
– Vanuan
Mar 10 '20 at 22:51
You got me here, I thought it is trivial, but yep see the problem now
– Maciej Sikora
Mar 10 '20 at 22:56
Looks to be working for number | undefined : typescriptlang.org/play/…
– Vanuan
Mar 10 '20 at 23:46
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Got it. So the result of the spread can't be optional in this case. Which makes sense if the first spread has a required value.
– Vanuan
Mar 12 '20 at 12:22
But it means that optional fields can't be assigned a value that can be undefined.
– Vanuan
Mar 12 '20 at 12:23
Similarly deleting non-optional fields should be prohibited. Which means that the type of a variable shouldn't ever change.
– Vanuan
Mar 12 '20 at 12:24
With the special spread function above, you can assign undefined to an optional field (say a?: number ) of the rightmost object argument, given some other spread object argument with a as required. The result type correctly outputs a?: number | undefined . In contrast, normal spread would mark a of the result type as a: number (required). I am not sure what you mean by "Similarly deleting non-optional fields should be prohibited."
– ford04
Mar 12 '20 at 12:47
Best thing you can do now is give issue a vote and/or use a workaround (cast, function) in those edge cases.
– ford04
Mar 12 '20 at 14:48
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It's great! Works for first case: cutt.ly/btafI9N But not for the other. Maybe it's ok.
– Vanuan
Mar 11 '20 at 0:10
Please, check my answer and tell whether it sounds plausible.
– Vanuan
Mar 11 '20 at 0:44
Hi @Vanuan what means the other case?
– Maciej Sikora
Mar 11 '20 at 9:16
When merging { value: undefined } with { value: number } . It's guaranteed that the result would be { value: number } . That is createObject(a: MyTypeOptional, b: MyTypeRequired): will return MyTypeOptional while it should return MyTypeRequired . So it works for the first order but not for the second. Have you clicked the link? shouldSucceed variable should not have error
– Vanuan
Mar 12 '20 at 12:07
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Consider you have an object with a non-nullable required field:
And you want to update it with fields of another object, with an optional field:
So you go ahead and create a function:
What would be the inferred return type of this function?
Experimentation suggests it would conform to the MyTypeRequired interface, even though the second spread has an optional field.
If we change the order, the inferred type doesn't change, even though the runtime type would be different.
Why does TypeScript have such behavior and how to workaround this issue?
Object spread types are resolved as follows :
Call and construct signatures are stripped, only non-method properties are preserved, and for properties with the same name , the type of the rightmost property is used .
Object literals with generic spread expressions now produce intersection types , similar to the Object.assign function and JSX literals. For example:
This works well, until you have an optional property for the rightmost argument. {value?: number; } can either mean 1) a missing property or 2) a property with undefined value - TS cannot distinguish those two cases with the optional modifier ? notation. Let's take an example:
makes sense, the property key a can be set or not set. We have no other choice than expressing latter with a undefined type.
a 's type is determined by the rightmost argument. So if a in u1 exists, it would be string , otherwise the spread operator would just take a from the first argument t1 with type number . So string | number also makes kinda sense. Note : There is no undefined here. TS assumes, the property doesn't exist at all, or it is a string . The result would be different, if we give a an explicit property value type of undefined :
The last one is simple: a from t1 overwrites a from u1 , so we get number back.
I wouldn't bet on a fix , so here is a possible workaround with a separate Spread type and function:
Hope, it makes sense! Here is sample for above code.
Workaround can be done by type level function. Consider:
I have added additional fields in order to check the behavior. What I am doing is, when the second object has optional field (possible undefined ) then consider that by adding this to the union, so the result will be T1[K] | undefined . The second part is just merging all other fields which are in T2 but not in T1 .
Warning! This is my interpretation (theory) of what happens. It isn't confirmed.
The issue seems to be an intended behavior of handling optional fields.
TypeScript expects that the value would be of type number | never .
The type of spread object should be either { value: never | number } or { value?: number } .
Unfortunately, due to the ambiguous usage of optional field vs undefined TypeScript infers the spread object as such:
So the optional field type is cast into number | undefined when spreading.
What doesn't make sense is that when we spread a non-optional type it seems to override the optional type:
Looks like a bug? But no, here TypeScript doesn't cast optional field into number | undefined . Here it casts it to number | never .
We can confirm this by changing optional into explicit |undefined :
would be inferred as { value: number | undefined; }
And when we change it to optional or undefined:
it is broken again and inferred as { value: number }
I would say, don't use optional fields until TypeScript team fixes them.
There's an open issue about this limitation of optional fields:
https://github.com/microsoft/TypeScript/issues/13195
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