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Specifying the return type based on optional parameters

A Typed array of functionsHow do you specify that a class property is an integer?Are strongly-typed functions as parameters possible in TypeScript?Creating String Literal Type with constSpecify return type in TypeScript arrow functionTypeScript type inference on function type parameter based on its argument typeHow can I check that a string is a property a particular interface in TypeScriptTypescript type inference of return dict keys based on supplied parametersTypes of optional and default parameters in typescriptAbstract function return type in typescript

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1

function f<T>(defaultValue?: T) return defaultValue; 

const definitelyUndefined = f<string>(); // type: string | undefined
const definitelyString = f<string>('foobar'); // type: string | undefined

Is it possible to define f() such that definitelyUndefined is implicitly undefined and definitelyString is implicitly string?

Background

My real world use case is a function I'm working with and want to improve. It is function f<T>(o: [key: string]: T , key: string, defaultValue?: T) and it returns key[o] if it exists, otherwise defaultValue. When I provide it a defaultValue, I'm guaranteed to get T back, but Typescript considers it T | undefined.

typescript

share|improve this question

edited Mar 24 at 20:53

Mark

asked Mar 24 at 20:44

MarkMark

456310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  My first thought was adding function f<T>(defaultValue: T): T; as an overload definition, but it doesn't work.
  
  – Mark
  Mar 24 at 20:53
  

  
  
  
  

add a comment | 

1

function f<T>(defaultValue?: T) return defaultValue; 

const definitelyUndefined = f<string>(); // type: string | undefined
const definitelyString = f<string>('foobar'); // type: string | undefined

Is it possible to define f() such that definitelyUndefined is implicitly undefined and definitelyString is implicitly string?

Background

My real world use case is a function I'm working with and want to improve. It is function f<T>(o: [key: string]: T , key: string, defaultValue?: T) and it returns key[o] if it exists, otherwise defaultValue. When I provide it a defaultValue, I'm guaranteed to get T back, but Typescript considers it T | undefined.

typescript

share|improve this question

edited Mar 24 at 20:53

Mark

asked Mar 24 at 20:44

MarkMark

456310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  My first thought was adding function f<T>(defaultValue: T): T; as an overload definition, but it doesn't work.
  
  – Mark
  Mar 24 at 20:53
  

  
  
  
  

add a comment | 

function f<T>(defaultValue?: T) return defaultValue; 

const definitelyUndefined = f<string>(); // type: string | undefined
const definitelyString = f<string>('foobar'); // type: string | undefined

Is it possible to define f() such that definitelyUndefined is implicitly undefined and definitelyString is implicitly string?

Background

My real world use case is a function I'm working with and want to improve. It is function f<T>(o: [key: string]: T , key: string, defaultValue?: T) and it returns key[o] if it exists, otherwise defaultValue. When I provide it a defaultValue, I'm guaranteed to get T back, but Typescript considers it T | undefined.

typescript

share|improve this question

edited Mar 24 at 20:53

Mark

asked Mar 24 at 20:44

MarkMark

456310

function f<T>(defaultValue?: T) return defaultValue; 

const definitelyUndefined = f<string>(); // type: string | undefined
const definitelyString = f<string>('foobar'); // type: string | undefined

share|improve this question

edited Mar 24 at 20:53

Mark

asked Mar 24 at 20:44

MarkMark

456310

share|improve this question

edited Mar 24 at 20:53

Mark

asked Mar 24 at 20:44

MarkMark

456310

asked Mar 24 at 20:44

MarkMark

456310

asked Mar 24 at 20:44

MarkMark

456310

1 Answer
1

active

oldest

votes

3

For your first case, I immediately thought of using an overload, for example:

function f<T>(): undefined;
function f<T>(value: T): T;
function f<T>(value?: T) return value; 

const definitelyUndefined = f(); // type: undefined
const definitelyString = f('foobar'); // type: "foobar"

However, for the more complex use case, I think perhaps you can solve it with overloads and more complex generics, for example:

function f<T, K extends string & keyof T>(o: T, key: K, defaultValue?: T[K]): T[K];
function f<T, V = undefined>(o: T, key: string, defaultValue?: V): V;
function f<T, V = undefined>(o: T, key: string, defaultValue?: V) defaultValue;


const obj = foo: "123" ;

const definitelyUndefined = f(obj, "bar"); // type: undefined
const definitelyNumber = f(obj, "bar", 123); // type: number
const definitelyString = f(obj, "foo"); // type: string

I don't know if this will work for every conceivable scenario (because the return type is determined based on the generic type argument, not the actual function argument), but I think it gets pretty close.

share|improve this answer

edited Mar 24 at 21:28

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's weird your "more complex" example does not run on typescriptlang.org/play UPD: it actually does not run in ts 3.3.3333 locally either.
  
  – zerkms
  Mar 24 at 21:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @zerkms I see the issue. Odd -- I didn't get the error earlier in the playground (maybe the parser was just a little slow). I'll see if I can correct it.
  
  – p.s.w.g
  Mar 24 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This line (function f<T>(): undefined;) cleared it up for me. Without it TS was complaining that An argument for 'defaultValue' was not provided. Once I added that it worked like a charm. Your second example is great for me to learn from, but my actual use case has a few caveats I left out so I wouldn't be able to use it (see getFieldValue here).
  
  – Mark
  Mar 24 at 21:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @zerkms Updated, but I'm still not 100% happy with the provided solution. It still conflates 'type space' and 'value space' but I'm not sure if that can be avoided without more restrictions on T.
  
  – p.s.w.g
  Mar 24 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @p.s.w.g it looks good to me - I was half-way writing a similar answer right when you posted it :-)
  
  – zerkms
  Mar 24 at 21:49
  

  
  
  
  

add a comment | 

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3

For your first case, I immediately thought of using an overload, for example:

function f<T>(): undefined;
function f<T>(value: T): T;
function f<T>(value?: T) return value; 

const definitelyUndefined = f(); // type: undefined
const definitelyString = f('foobar'); // type: "foobar"

However, for the more complex use case, I think perhaps you can solve it with overloads and more complex generics, for example:

function f<T, K extends string & keyof T>(o: T, key: K, defaultValue?: T[K]): T[K];
function f<T, V = undefined>(o: T, key: string, defaultValue?: V): V;
function f<T, V = undefined>(o: T, key: string, defaultValue?: V) defaultValue;


const obj = foo: "123" ;

const definitelyUndefined = f(obj, "bar"); // type: undefined
const definitelyNumber = f(obj, "bar", 123); // type: number
const definitelyString = f(obj, "foo"); // type: string

I don't know if this will work for every conceivable scenario (because the return type is determined based on the generic type argument, not the actual function argument), but I think it gets pretty close.

share|improve this answer

edited Mar 24 at 21:28

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's weird your "more complex" example does not run on typescriptlang.org/play UPD: it actually does not run in ts 3.3.3333 locally either.
  
  – zerkms
  Mar 24 at 21:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @zerkms I see the issue. Odd -- I didn't get the error earlier in the playground (maybe the parser was just a little slow). I'll see if I can correct it.
  
  – p.s.w.g
  Mar 24 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This line (function f<T>(): undefined;) cleared it up for me. Without it TS was complaining that An argument for 'defaultValue' was not provided. Once I added that it worked like a charm. Your second example is great for me to learn from, but my actual use case has a few caveats I left out so I wouldn't be able to use it (see getFieldValue here).
  
  – Mark
  Mar 24 at 21:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @zerkms Updated, but I'm still not 100% happy with the provided solution. It still conflates 'type space' and 'value space' but I'm not sure if that can be avoided without more restrictions on T.
  
  – p.s.w.g
  Mar 24 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @p.s.w.g it looks good to me - I was half-way writing a similar answer right when you posted it :-)
  
  – zerkms
  Mar 24 at 21:49
  

  
  
  
  

add a comment | 

3

For your first case, I immediately thought of using an overload, for example:

function f<T>(): undefined;
function f<T>(value: T): T;
function f<T>(value?: T) return value; 

const definitelyUndefined = f(); // type: undefined
const definitelyString = f('foobar'); // type: "foobar"

However, for the more complex use case, I think perhaps you can solve it with overloads and more complex generics, for example:

function f<T, K extends string & keyof T>(o: T, key: K, defaultValue?: T[K]): T[K];
function f<T, V = undefined>(o: T, key: string, defaultValue?: V): V;
function f<T, V = undefined>(o: T, key: string, defaultValue?: V) defaultValue;


const obj = foo: "123" ;

const definitelyUndefined = f(obj, "bar"); // type: undefined
const definitelyNumber = f(obj, "bar", 123); // type: number
const definitelyString = f(obj, "foo"); // type: string

I don't know if this will work for every conceivable scenario (because the return type is determined based on the generic type argument, not the actual function argument), but I think it gets pretty close.

share|improve this answer

edited Mar 24 at 21:28

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's weird your "more complex" example does not run on typescriptlang.org/play UPD: it actually does not run in ts 3.3.3333 locally either.
  
  – zerkms
  Mar 24 at 21:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @zerkms I see the issue. Odd -- I didn't get the error earlier in the playground (maybe the parser was just a little slow). I'll see if I can correct it.
  
  – p.s.w.g
  Mar 24 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This line (function f<T>(): undefined;) cleared it up for me. Without it TS was complaining that An argument for 'defaultValue' was not provided. Once I added that it worked like a charm. Your second example is great for me to learn from, but my actual use case has a few caveats I left out so I wouldn't be able to use it (see getFieldValue here).
  
  – Mark
  Mar 24 at 21:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @zerkms Updated, but I'm still not 100% happy with the provided solution. It still conflates 'type space' and 'value space' but I'm not sure if that can be avoided without more restrictions on T.
  
  – p.s.w.g
  Mar 24 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @p.s.w.g it looks good to me - I was half-way writing a similar answer right when you posted it :-)
  
  – zerkms
  Mar 24 at 21:49
  

  
  
  
  

add a comment | 

For your first case, I immediately thought of using an overload, for example:

function f<T>(): undefined;
function f<T>(value: T): T;
function f<T>(value?: T) return value; 

const definitelyUndefined = f(); // type: undefined
const definitelyString = f('foobar'); // type: "foobar"

However, for the more complex use case, I think perhaps you can solve it with overloads and more complex generics, for example:

function f<T, K extends string & keyof T>(o: T, key: K, defaultValue?: T[K]): T[K];
function f<T, V = undefined>(o: T, key: string, defaultValue?: V): V;
function f<T, V = undefined>(o: T, key: string, defaultValue?: V) defaultValue;


const obj = foo: "123" ;

const definitelyUndefined = f(obj, "bar"); // type: undefined
const definitelyNumber = f(obj, "bar", 123); // type: number
const definitelyString = f(obj, "foo"); // type: string

I don't know if this will work for every conceivable scenario (because the return type is determined based on the generic type argument, not the actual function argument), but I think it gets pretty close.

share|improve this answer

edited Mar 24 at 21:28

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260

function f<T>(): undefined;
function f<T>(value: T): T;
function f<T>(value?: T) return value; 

const definitelyUndefined = f(); // type: undefined
const definitelyString = f('foobar'); // type: "foobar"

function f<T, K extends string & keyof T>(o: T, key: K, defaultValue?: T[K]): T[K];
function f<T, V = undefined>(o: T, key: string, defaultValue?: V): V;
function f<T, V = undefined>(o: T, key: string, defaultValue?: V) defaultValue;


const obj = foo: "123" ;

const definitelyUndefined = f(obj, "bar"); // type: undefined
const definitelyNumber = f(obj, "bar", 123); // type: number
const definitelyString = f(obj, "foo"); // type: string

share|improve this answer

edited Mar 24 at 21:28

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260

answered Mar 24 at 21:02

p.s.w.gp.s.w.g

122k19218260