Typescript3.1.3 + Generics, error in type assignment










2















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.










share|improve this question
























  • 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











  • 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
















2















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.










share|improve this question
























  • 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











  • 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














2












2








2


1






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.










share|improve this question
















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








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 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











  • 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












  • 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











  • @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













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