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Typescript function can return Promise as Promise?

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2

I'm creating a type-safe wrapper for the fetch-api, and I noticed that the typescript compiler classifies this as perfectly valid code:

function foo(response: Response): Promise<Bar> //response received from a fetch() call
 const json: Promise<any> = response.json();
 return json; //Promise<any> can be returned as Promise<Bar>?

Why is it possible to directly return a Promise<any> as Promise<Bar>? Shouldn't this require some kind of type assertion?

typescript promise fetch-api

share|improve this question

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

add a comment | 

2

I'm creating a type-safe wrapper for the fetch-api, and I noticed that the typescript compiler classifies this as perfectly valid code:

function foo(response: Response): Promise<Bar> //response received from a fetch() call
 const json: Promise<any> = response.json();
 return json; //Promise<any> can be returned as Promise<Bar>?

Why is it possible to directly return a Promise<any> as Promise<Bar>? Shouldn't this require some kind of type assertion?

typescript promise fetch-api

share|improve this question

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

add a comment | 

I'm creating a type-safe wrapper for the fetch-api, and I noticed that the typescript compiler classifies this as perfectly valid code:

function foo(response: Response): Promise<Bar> //response received from a fetch() call
 const json: Promise<any> = response.json();
 return json; //Promise<any> can be returned as Promise<Bar>?

Why is it possible to directly return a Promise<any> as Promise<Bar>? Shouldn't this require some kind of type assertion?

typescript promise fetch-api

share|improve this question

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

function foo(response: Response): Promise<Bar> //response received from a fetch() call
 const json: Promise<any> = response.json();
 return json; //Promise<any> can be returned as Promise<Bar>?

share|improve this question

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

share|improve this question

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

asked Mar 25 at 20:29

superbadcodemonkeysuperbadcodemonkey

8288 bronze badges

2 Answers
2

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oldest

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6

Always avoid any. any by definition is assignable to anything and assignable from anything without any sort of type assertions. When used as a generic type argument, the resuting type (ex Promise<any>) will generally be assignable to any other instantiation with any other type argument on that position (ex Promise<Bar>).

Since 3.0 typescript introduced unknown (read here for more) which is similar to any in that you can assign anything to it, it is not assignable to anything else without a type assertion. So this is an error:

let u: Promise<unknown>
let n: Promise<number> = u; // error

You might aslo look into tslint rules no-unsafe-any and no-any that prevent usage of any in a project, depending on how far you want to go with banning any, and like I said in the beginning I would ban it completely and add exceptions for the few cases where it is absolutely necessary.

share|improve this answer

edited Mar 25 at 20:43

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes a lot of sense. Is there a reason why Body.json() returns Promise<any> instead of Promise<unknown>?
  
  – superbadcodemonkey
  Mar 25 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @superbadcodemonkey It would break a lot of code if it did. There is a GH discussion to add an option to use unknown instead of any in lib.d.ts but I can't find it right now and I don't think it will be implemented any time soon.
  
  – Titian Cernicova-Dragomir
  Mar 25 at 20:49
  

  
  
  
  

add a comment | 

4

If something can be anything it can also be a Bar. If you look at the Typescript documentation for Basic Types it says (emphasis mine)

Any

We may need to describe the type of variables that we do not know when we are writing an application. These values may come from dynamic content, e.g. from the user or a 3rd party library. In these cases, we want to opt-out of type-checking and let the values pass through compile-time checks. To do so, we label these with the any type:

Which means that the any type completely ignores compile-time type checks.

share|improve this answer

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges

add a comment | 

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6

Always avoid any. any by definition is assignable to anything and assignable from anything without any sort of type assertions. When used as a generic type argument, the resuting type (ex Promise<any>) will generally be assignable to any other instantiation with any other type argument on that position (ex Promise<Bar>).

Since 3.0 typescript introduced unknown (read here for more) which is similar to any in that you can assign anything to it, it is not assignable to anything else without a type assertion. So this is an error:

let u: Promise<unknown>
let n: Promise<number> = u; // error

You might aslo look into tslint rules no-unsafe-any and no-any that prevent usage of any in a project, depending on how far you want to go with banning any, and like I said in the beginning I would ban it completely and add exceptions for the few cases where it is absolutely necessary.

share|improve this answer

edited Mar 25 at 20:43

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes a lot of sense. Is there a reason why Body.json() returns Promise<any> instead of Promise<unknown>?
  
  – superbadcodemonkey
  Mar 25 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @superbadcodemonkey It would break a lot of code if it did. There is a GH discussion to add an option to use unknown instead of any in lib.d.ts but I can't find it right now and I don't think it will be implemented any time soon.
  
  – Titian Cernicova-Dragomir
  Mar 25 at 20:49
  

  
  
  
  

add a comment | 

6

Always avoid any. any by definition is assignable to anything and assignable from anything without any sort of type assertions. When used as a generic type argument, the resuting type (ex Promise<any>) will generally be assignable to any other instantiation with any other type argument on that position (ex Promise<Bar>).

Since 3.0 typescript introduced unknown (read here for more) which is similar to any in that you can assign anything to it, it is not assignable to anything else without a type assertion. So this is an error:

let u: Promise<unknown>
let n: Promise<number> = u; // error

You might aslo look into tslint rules no-unsafe-any and no-any that prevent usage of any in a project, depending on how far you want to go with banning any, and like I said in the beginning I would ban it completely and add exceptions for the few cases where it is absolutely necessary.

share|improve this answer

edited Mar 25 at 20:43

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes a lot of sense. Is there a reason why Body.json() returns Promise<any> instead of Promise<unknown>?
  
  – superbadcodemonkey
  Mar 25 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @superbadcodemonkey It would break a lot of code if it did. There is a GH discussion to add an option to use unknown instead of any in lib.d.ts but I can't find it right now and I don't think it will be implemented any time soon.
  
  – Titian Cernicova-Dragomir
  Mar 25 at 20:49
  

  
  
  
  

add a comment | 

Always avoid any. any by definition is assignable to anything and assignable from anything without any sort of type assertions. When used as a generic type argument, the resuting type (ex Promise<any>) will generally be assignable to any other instantiation with any other type argument on that position (ex Promise<Bar>).

Since 3.0 typescript introduced unknown (read here for more) which is similar to any in that you can assign anything to it, it is not assignable to anything else without a type assertion. So this is an error:

let u: Promise<unknown>
let n: Promise<number> = u; // error

You might aslo look into tslint rules no-unsafe-any and no-any that prevent usage of any in a project, depending on how far you want to go with banning any, and like I said in the beginning I would ban it completely and add exceptions for the few cases where it is absolutely necessary.

share|improve this answer

edited Mar 25 at 20:43

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

let u: Promise<unknown>
let n: Promise<number> = u; // error

share|improve this answer

edited Mar 25 at 20:43

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

answered Mar 25 at 20:36

Titian Cernicova-DragomirTitian Cernicova-Dragomir

86.2k55 gold badges7070 silver badges8383 bronze badges

4

If something can be anything it can also be a Bar. If you look at the Typescript documentation for Basic Types it says (emphasis mine)

Any

We may need to describe the type of variables that we do not know when we are writing an application. These values may come from dynamic content, e.g. from the user or a 3rd party library. In these cases, we want to opt-out of type-checking and let the values pass through compile-time checks. To do so, we label these with the any type:

Which means that the any type completely ignores compile-time type checks.

share|improve this answer

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges

add a comment | 

4

If something can be anything it can also be a Bar. If you look at the Typescript documentation for Basic Types it says (emphasis mine)

Any

We may need to describe the type of variables that we do not know when we are writing an application. These values may come from dynamic content, e.g. from the user or a 3rd party library. In these cases, we want to opt-out of type-checking and let the values pass through compile-time checks. To do so, we label these with the any type:

Which means that the any type completely ignores compile-time type checks.

share|improve this answer

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges

add a comment | 

If something can be anything it can also be a Bar. If you look at the Typescript documentation for Basic Types it says (emphasis mine)

Any

We may need to describe the type of variables that we do not know when we are writing an application. These values may come from dynamic content, e.g. from the user or a 3rd party library. In these cases, we want to opt-out of type-checking and let the values pass through compile-time checks. To do so, we label these with the any type:

Which means that the any type completely ignores compile-time type checks.

share|improve this answer

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges

share|improve this answer

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges

answered Mar 25 at 20:36

DanielDaniel

6,36588 gold badges2323 silver badges5151 bronze badges