Typescript3.1.3 + Generics, error in type assignment
Consider this situation (dummy example):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this);
// ...
class Animal<F extends Farmer<Animal<F>>>
public farmer: F;
constructor(f: F)
this.farmer = f;
Anyone can explain why in the Farmer
constructor (where I pass this
argument to new Animal
) the above code raises this error?
TS2345: Argument of type 'this' is not assignable to parameter of type 'Farmer<Animal<this>>'.
Type 'Farmer<A>' is not assignable to type 'Farmer<Animal<this>>'.
Type 'A' is not assignable to type 'Animal<this>'.
Type 'Animal<Farmer<A>>' is not assignable to type 'Animal<this>'.
Type 'Farmer<A>' is not assignable to type 'this'.
The following solve the issue:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this as Farmer<A>); // works fine
Or as alternative:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal<Farmer<A>>(this); // works fine
The more strange to me is the following:
class Farmer<A extends Animal<Farmer<A>>>
// public animal: A; // removed this line
constructor()
let a = new Animal(this); // now this one works fine too
While first two solutions are quite clear, the latter is the one that leaves me more doubtful.
Can anyone explain what's happen?
Edit
The following is fine (consider optional the param in Animal
constructor):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal();
a.farmer = this; // works fine
And here I can't see any difference in types, compared with the first case.
typescript generics typescript3.0
add a comment |
Consider this situation (dummy example):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this);
// ...
class Animal<F extends Farmer<Animal<F>>>
public farmer: F;
constructor(f: F)
this.farmer = f;
Anyone can explain why in the Farmer
constructor (where I pass this
argument to new Animal
) the above code raises this error?
TS2345: Argument of type 'this' is not assignable to parameter of type 'Farmer<Animal<this>>'.
Type 'Farmer<A>' is not assignable to type 'Farmer<Animal<this>>'.
Type 'A' is not assignable to type 'Animal<this>'.
Type 'Animal<Farmer<A>>' is not assignable to type 'Animal<this>'.
Type 'Farmer<A>' is not assignable to type 'this'.
The following solve the issue:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this as Farmer<A>); // works fine
Or as alternative:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal<Farmer<A>>(this); // works fine
The more strange to me is the following:
class Farmer<A extends Animal<Farmer<A>>>
// public animal: A; // removed this line
constructor()
let a = new Animal(this); // now this one works fine too
While first two solutions are quite clear, the latter is the one that leaves me more doubtful.
Can anyone explain what's happen?
Edit
The following is fine (consider optional the param in Animal
constructor):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal();
a.farmer = this; // works fine
And here I can't see any difference in types, compared with the first case.
typescript generics typescript3.0
Wow, usually in TypeScript you can sidestep F-bounded polymorphism with polymorphicthis
, but here you have mutually recursive type parameters so you can't. The compiler can't verify thatA
is equivalent toAnimal<Farmer<A>>
(it only knows thatA
is a subtype of that) so it balks atnew Animal(this)
without those type assertions.
– jcalz
Nov 11 '18 at 20:27
The reason why the last solution "works" is becauseFarmer<A>
no longer depends onA
structurally, which is all TypeScript really cares about. That is,Farmer<A>
andFarmer<B>
are equivalent types, and there is no more mutual recursion to care about.
– jcalz
Nov 11 '18 at 20:28
I'm not sure if someone has a cleaner solution that doesn't involve type assertions.. if not I'll turn these comments into an answer at some point. Cheers
– jcalz
Nov 11 '18 at 20:29
@jcalz, useful points. thanks. I edited the question adding another case.
– Andrea
Nov 12 '18 at 7:29
add a comment |
Consider this situation (dummy example):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this);
// ...
class Animal<F extends Farmer<Animal<F>>>
public farmer: F;
constructor(f: F)
this.farmer = f;
Anyone can explain why in the Farmer
constructor (where I pass this
argument to new Animal
) the above code raises this error?
TS2345: Argument of type 'this' is not assignable to parameter of type 'Farmer<Animal<this>>'.
Type 'Farmer<A>' is not assignable to type 'Farmer<Animal<this>>'.
Type 'A' is not assignable to type 'Animal<this>'.
Type 'Animal<Farmer<A>>' is not assignable to type 'Animal<this>'.
Type 'Farmer<A>' is not assignable to type 'this'.
The following solve the issue:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this as Farmer<A>); // works fine
Or as alternative:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal<Farmer<A>>(this); // works fine
The more strange to me is the following:
class Farmer<A extends Animal<Farmer<A>>>
// public animal: A; // removed this line
constructor()
let a = new Animal(this); // now this one works fine too
While first two solutions are quite clear, the latter is the one that leaves me more doubtful.
Can anyone explain what's happen?
Edit
The following is fine (consider optional the param in Animal
constructor):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal();
a.farmer = this; // works fine
And here I can't see any difference in types, compared with the first case.
typescript generics typescript3.0
Consider this situation (dummy example):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this);
// ...
class Animal<F extends Farmer<Animal<F>>>
public farmer: F;
constructor(f: F)
this.farmer = f;
Anyone can explain why in the Farmer
constructor (where I pass this
argument to new Animal
) the above code raises this error?
TS2345: Argument of type 'this' is not assignable to parameter of type 'Farmer<Animal<this>>'.
Type 'Farmer<A>' is not assignable to type 'Farmer<Animal<this>>'.
Type 'A' is not assignable to type 'Animal<this>'.
Type 'Animal<Farmer<A>>' is not assignable to type 'Animal<this>'.
Type 'Farmer<A>' is not assignable to type 'this'.
The following solve the issue:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal(this as Farmer<A>); // works fine
Or as alternative:
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal<Farmer<A>>(this); // works fine
The more strange to me is the following:
class Farmer<A extends Animal<Farmer<A>>>
// public animal: A; // removed this line
constructor()
let a = new Animal(this); // now this one works fine too
While first two solutions are quite clear, the latter is the one that leaves me more doubtful.
Can anyone explain what's happen?
Edit
The following is fine (consider optional the param in Animal
constructor):
class Farmer<A extends Animal<Farmer<A>>>
public animal: A;
constructor()
let a = new Animal();
a.farmer = this; // works fine
And here I can't see any difference in types, compared with the first case.
typescript generics typescript3.0
typescript generics typescript3.0
edited Nov 12 '18 at 7:28
Andrea
asked Nov 11 '18 at 17:22
AndreaAndrea
404414
404414
Wow, usually in TypeScript you can sidestep F-bounded polymorphism with polymorphicthis
, but here you have mutually recursive type parameters so you can't. The compiler can't verify thatA
is equivalent toAnimal<Farmer<A>>
(it only knows thatA
is a subtype of that) so it balks atnew Animal(this)
without those type assertions.
– jcalz
Nov 11 '18 at 20:27
The reason why the last solution "works" is becauseFarmer<A>
no longer depends onA
structurally, which is all TypeScript really cares about. That is,Farmer<A>
andFarmer<B>
are equivalent types, and there is no more mutual recursion to care about.
– jcalz
Nov 11 '18 at 20:28
I'm not sure if someone has a cleaner solution that doesn't involve type assertions.. if not I'll turn these comments into an answer at some point. Cheers
– jcalz
Nov 11 '18 at 20:29
@jcalz, useful points. thanks. I edited the question adding another case.
– Andrea
Nov 12 '18 at 7:29
add a comment |
Wow, usually in TypeScript you can sidestep F-bounded polymorphism with polymorphicthis
, but here you have mutually recursive type parameters so you can't. The compiler can't verify thatA
is equivalent toAnimal<Farmer<A>>
(it only knows thatA
is a subtype of that) so it balks atnew Animal(this)
without those type assertions.
– jcalz
Nov 11 '18 at 20:27
The reason why the last solution "works" is becauseFarmer<A>
no longer depends onA
structurally, which is all TypeScript really cares about. That is,Farmer<A>
andFarmer<B>
are equivalent types, and there is no more mutual recursion to care about.
– jcalz
Nov 11 '18 at 20:28
I'm not sure if someone has a cleaner solution that doesn't involve type assertions.. if not I'll turn these comments into an answer at some point. Cheers
– jcalz
Nov 11 '18 at 20:29
@jcalz, useful points. thanks. I edited the question adding another case.
– Andrea
Nov 12 '18 at 7:29
Wow, usually in TypeScript you can sidestep F-bounded polymorphism with polymorphic
this
, but here you have mutually recursive type parameters so you can't. The compiler can't verify that A
is equivalent to Animal<Farmer<A>>
(it only knows that A
is a subtype of that) so it balks at new Animal(this)
without those type assertions.– jcalz
Nov 11 '18 at 20:27
Wow, usually in TypeScript you can sidestep F-bounded polymorphism with polymorphic
this
, but here you have mutually recursive type parameters so you can't. The compiler can't verify that A
is equivalent to Animal<Farmer<A>>
(it only knows that A
is a subtype of that) so it balks at new Animal(this)
without those type assertions.– jcalz
Nov 11 '18 at 20:27
The reason why the last solution "works" is because
Farmer<A>
no longer depends on A
structurally, which is all TypeScript really cares about. That is, Farmer<A>
and Farmer<B>
are equivalent types, and there is no more mutual recursion to care about.– jcalz
Nov 11 '18 at 20:28
The reason why the last solution "works" is because
Farmer<A>
no longer depends on A
structurally, which is all TypeScript really cares about. That is, Farmer<A>
and Farmer<B>
are equivalent types, and there is no more mutual recursion to care about.– jcalz
Nov 11 '18 at 20:28
I'm not sure if someone has a cleaner solution that doesn't involve type assertions.. if not I'll turn these comments into an answer at some point. Cheers
– jcalz
Nov 11 '18 at 20:29
I'm not sure if someone has a cleaner solution that doesn't involve type assertions.. if not I'll turn these comments into an answer at some point. Cheers
– jcalz
Nov 11 '18 at 20:29
@jcalz, useful points. thanks. I edited the question adding another case.
– Andrea
Nov 12 '18 at 7:29
@jcalz, useful points. thanks. I edited the question adding another case.
– Andrea
Nov 12 '18 at 7:29
add a comment |
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Wow, usually in TypeScript you can sidestep F-bounded polymorphism with polymorphic
this
, but here you have mutually recursive type parameters so you can't. The compiler can't verify thatA
is equivalent toAnimal<Farmer<A>>
(it only knows thatA
is a subtype of that) so it balks atnew Animal(this)
without those type assertions.– jcalz
Nov 11 '18 at 20:27
The reason why the last solution "works" is because
Farmer<A>
no longer depends onA
structurally, which is all TypeScript really cares about. That is,Farmer<A>
andFarmer<B>
are equivalent types, and there is no more mutual recursion to care about.– jcalz
Nov 11 '18 at 20:28
I'm not sure if someone has a cleaner solution that doesn't involve type assertions.. if not I'll turn these comments into an answer at some point. Cheers
– jcalz
Nov 11 '18 at 20:29
@jcalz, useful points. thanks. I edited the question adding another case.
– Andrea
Nov 12 '18 at 7:29