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Implementing an interface with a call signature and method returning “this”

How to make a class implement a call signature in Typescript?TypeScript: Implement an interface with both call signature and indexing signatureTypeScript: Implement interface with both call signature and constructor signatureTypescript with jQuery UI Widgets - Implementing call signaturesHow to implement ng.IDirectiveFactory in TypeScriptAnonymous/inline interface implementation in TypeScriptTypescript not checking function argument types declared by interfacesUnion types and “currying” / partial application style APIsWriting TypeScript DefinitionTypescript: Interfaces vs TypesTypeScript compiler not enforcing type parameter in implementation of generic interface?Enumerate all classes implementing an interface in Typescripthow is implicit index signature type checked in typescript?

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2

0

Preface: Our team is working on a library built on top of d3. Since we are using TypeScript, we are also using d3's types from DefinitelyTyped. The following question arises when trying to work with interfaces such as ScaleOrdinal and many others from there.

---

Suppose we have an interface containing both a call signature and additional properties:

export interface Foo<T> 
 // Let's pretend this will be the identity function
 (arg: T): T;

 // Let's pretend that this will be a no-op function
 // Note that this returns "this" 
 doFoo(): this;

How can we implement such an interface correctly & in a type-safe manner[1]? Upon research I've found the following related questions, all of which are slightly different and/or fairly old. I'd like to get an idea of whether we're missing something or whether to raise an issue with the TypeScript team here:

1. How to make a class implement a call signature in Typescript?


2. TypeScript: Implement interface with both call signature and constructor signature


3. TypeScript: Implement an interface with both call signature and indexing signature

Note that the interface is external to us, thus implementing it is our only option.

---

¹ For the sake of the question, I would like the implementation to explicitly restate all type annotations.

typescript definitelytyped

share|improve this question

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

add a comment | 

2

0

Preface: Our team is working on a library built on top of d3. Since we are using TypeScript, we are also using d3's types from DefinitelyTyped. The following question arises when trying to work with interfaces such as ScaleOrdinal and many others from there.

---

Suppose we have an interface containing both a call signature and additional properties:

export interface Foo<T> 
 // Let's pretend this will be the identity function
 (arg: T): T;

 // Let's pretend that this will be a no-op function
 // Note that this returns "this" 
 doFoo(): this;

How can we implement such an interface correctly & in a type-safe manner[1]? Upon research I've found the following related questions, all of which are slightly different and/or fairly old. I'd like to get an idea of whether we're missing something or whether to raise an issue with the TypeScript team here:

1. How to make a class implement a call signature in Typescript?


2. TypeScript: Implement interface with both call signature and constructor signature


3. TypeScript: Implement an interface with both call signature and indexing signature

Note that the interface is external to us, thus implementing it is our only option.

---

¹ For the sake of the question, I would like the implementation to explicitly restate all type annotations.

typescript definitelytyped

share|improve this question

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

add a comment | 

Preface: Our team is working on a library built on top of d3. Since we are using TypeScript, we are also using d3's types from DefinitelyTyped. The following question arises when trying to work with interfaces such as ScaleOrdinal and many others from there.

---

Suppose we have an interface containing both a call signature and additional properties:

export interface Foo<T> 
 // Let's pretend this will be the identity function
 (arg: T): T;

 // Let's pretend that this will be a no-op function
 // Note that this returns "this" 
 doFoo(): this;

How can we implement such an interface correctly & in a type-safe manner[1]? Upon research I've found the following related questions, all of which are slightly different and/or fairly old. I'd like to get an idea of whether we're missing something or whether to raise an issue with the TypeScript team here:

1. How to make a class implement a call signature in Typescript?


2. TypeScript: Implement interface with both call signature and constructor signature


3. TypeScript: Implement an interface with both call signature and indexing signature

Note that the interface is external to us, thus implementing it is our only option.

---

¹ For the sake of the question, I would like the implementation to explicitly restate all type annotations.

typescript definitelytyped

share|improve this question

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

export interface Foo<T> 
 // Let's pretend this will be the identity function
 (arg: T): T;

 // Let's pretend that this will be a no-op function
 // Note that this returns "this" 
 doFoo(): this;

share|improve this question

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

share|improve this question

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

asked Mar 27 at 11:13

Ingo BürkIngo Bürk

11.8k55 gold badges3838 silver badges7777 bronze badges

1 Answer
1

active

oldest

votes

1

In recent versions of typescript (3.2 or 3.3 not sure which) when you declare a function, you can also assign extra properties to the function and typescript will consider these as the definition for those properties and not complain about them not having been defined:

export interface Foo<T> 
 (arg: T): T; 
 doFoo(): this;


function foo(arg: number) : number 
 return arg

foo.doFoo = function <TThis extends typeof foo>(this: TThis): TThis // no polymorphic this in simple functions
 return this


let o: Foo<number> = foo; // foo is compatible with Foo<number>

The old was of doing it, which still works is using Object.assign to create the function with extra properties:

let o: Foo<number> = Object.assign(function (arg: number): number 
 return arg
, 
 doFoo: function <TThis>(this: TThis): TThis 
 return this
 
)

share|improve this answer

edited Mar 27 at 11:29

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! The trick of using this: TThis in the signature is what we seem to have been missing.
  
  – Ingo Bürk
  Mar 27 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @IngoBürk also added a constraint on it to make foo members accessible
  
  – Titian Cernicova-Dragomir
  Mar 27 at 11:30
  

  
  
  
  

add a comment | 

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1

In recent versions of typescript (3.2 or 3.3 not sure which) when you declare a function, you can also assign extra properties to the function and typescript will consider these as the definition for those properties and not complain about them not having been defined:

export interface Foo<T> 
 (arg: T): T; 
 doFoo(): this;


function foo(arg: number) : number 
 return arg

foo.doFoo = function <TThis extends typeof foo>(this: TThis): TThis // no polymorphic this in simple functions
 return this


let o: Foo<number> = foo; // foo is compatible with Foo<number>

The old was of doing it, which still works is using Object.assign to create the function with extra properties:

let o: Foo<number> = Object.assign(function (arg: number): number 
 return arg
, 
 doFoo: function <TThis>(this: TThis): TThis 
 return this
 
)

share|improve this answer

edited Mar 27 at 11:29

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! The trick of using this: TThis in the signature is what we seem to have been missing.
  
  – Ingo Bürk
  Mar 27 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @IngoBürk also added a constraint on it to make foo members accessible
  
  – Titian Cernicova-Dragomir
  Mar 27 at 11:30
  

  
  
  
  

add a comment | 

1

In recent versions of typescript (3.2 or 3.3 not sure which) when you declare a function, you can also assign extra properties to the function and typescript will consider these as the definition for those properties and not complain about them not having been defined:

export interface Foo<T> 
 (arg: T): T; 
 doFoo(): this;


function foo(arg: number) : number 
 return arg

foo.doFoo = function <TThis extends typeof foo>(this: TThis): TThis // no polymorphic this in simple functions
 return this


let o: Foo<number> = foo; // foo is compatible with Foo<number>

The old was of doing it, which still works is using Object.assign to create the function with extra properties:

let o: Foo<number> = Object.assign(function (arg: number): number 
 return arg
, 
 doFoo: function <TThis>(this: TThis): TThis 
 return this
 
)

share|improve this answer

edited Mar 27 at 11:29

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! The trick of using this: TThis in the signature is what we seem to have been missing.
  
  – Ingo Bürk
  Mar 27 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @IngoBürk also added a constraint on it to make foo members accessible
  
  – Titian Cernicova-Dragomir
  Mar 27 at 11:30
  

  
  
  
  

add a comment | 

In recent versions of typescript (3.2 or 3.3 not sure which) when you declare a function, you can also assign extra properties to the function and typescript will consider these as the definition for those properties and not complain about them not having been defined:

export interface Foo<T> 
 (arg: T): T; 
 doFoo(): this;


function foo(arg: number) : number 
 return arg

foo.doFoo = function <TThis extends typeof foo>(this: TThis): TThis // no polymorphic this in simple functions
 return this


let o: Foo<number> = foo; // foo is compatible with Foo<number>

The old was of doing it, which still works is using Object.assign to create the function with extra properties:

let o: Foo<number> = Object.assign(function (arg: number): number 
 return arg
, 
 doFoo: function <TThis>(this: TThis): TThis 
 return this
 
)

share|improve this answer

edited Mar 27 at 11:29

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges

export interface Foo<T> 
 (arg: T): T; 
 doFoo(): this;


function foo(arg: number) : number 
 return arg

foo.doFoo = function <TThis extends typeof foo>(this: TThis): TThis // no polymorphic this in simple functions
 return this


let o: Foo<number> = foo; // foo is compatible with Foo<number>

let o: Foo<number> = Object.assign(function (arg: number): number 
 return arg
, 
 doFoo: function <TThis>(this: TThis): TThis 
 return this
 
)

share|improve this answer

edited Mar 27 at 11:29

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges

answered Mar 27 at 11:23

Titian Cernicova-DragomirTitian Cernicova-Dragomir

92.1k55 gold badges7474 silver badges9292 bronze badges