How do I assign the right type signature to this curried function in typescript?
I am new to typescript & generics and working with a Maybe monad. I have created this small utility function using Ramda:
const fromEmpty = R.ifElse(hasLength, Just, Maybe.zero);
The closest signature I can infer from this is:
type EmptyFunc<T> = (val: T) => Maybe<T>;
It's a function takes in an array and returns a function that returns a Maybe of that array.
I tried doing
const fromEmpty(<U extends EmptyFunc<U>) = R.ifElse(hasLength, Just, Maybe.zero);
But that doesn't work. It returns error TS1005: ',' expected.
What is the right way to work with curried functions in typescript?
typescript functional-programming currying ramda.js
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
add a comment |
I am new to typescript & generics and working with a Maybe monad. I have created this small utility function using Ramda:
const fromEmpty = R.ifElse(hasLength, Just, Maybe.zero);
The closest signature I can infer from this is:
type EmptyFunc<T> = (val: T) => Maybe<T>;
It's a function takes in an array and returns a function that returns a Maybe of that array.
I tried doing
const fromEmpty(<U extends EmptyFunc<U>) = R.ifElse(hasLength, Just, Maybe.zero);
But that doesn't work. It returns error TS1005: ',' expected.
What is the right way to work with curried functions in typescript?
typescript functional-programming currying ramda.js
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
add a comment |
I am new to typescript & generics and working with a Maybe monad. I have created this small utility function using Ramda:
const fromEmpty = R.ifElse(hasLength, Just, Maybe.zero);
The closest signature I can infer from this is:
type EmptyFunc<T> = (val: T) => Maybe<T>;
It's a function takes in an array and returns a function that returns a Maybe of that array.
I tried doing
const fromEmpty(<U extends EmptyFunc<U>) = R.ifElse(hasLength, Just, Maybe.zero);
But that doesn't work. It returns error TS1005: ',' expected.
What is the right way to work with curried functions in typescript?
typescript functional-programming currying ramda.js
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
I am new to typescript & generics and working with a Maybe monad. I have created this small utility function using Ramda:
const fromEmpty = R.ifElse(hasLength, Just, Maybe.zero);
The closest signature I can infer from this is:
type EmptyFunc<T> = (val: T) => Maybe<T>;
It's a function takes in an array and returns a function that returns a Maybe of that array.
I tried doing
const fromEmpty(<U extends EmptyFunc<U>) = R.ifElse(hasLength, Just, Maybe.zero);
But that doesn't work. It returns error TS1005: ',' expected.
What is the right way to work with curried functions in typescript?
typescript functional-programming currying ramda.js
typescript functional-programming currying ramda.js
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
asked Nov 11 '18 at 8:46
Amit ErandoleAmit Erandole
4,022164982
4,022164982
add a comment |
add a comment |
1 Answer
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The declaration
type EmptyFunc<T> = (val: T) => Maybe<T>;
declares a family of different function types EmptyFunc<number>, EmptyFunc<string>, etc., each of which works for only the specified type T. You probably meant
type EmptyFunc = <T>(val: T) => Maybe<T>;
which declares the type EmptyFunc of a single generic function that works for all types T. Then to apply this type to fromEmpty, just write:
const fromEmpty: EmptyFunc = R.ifElse(hasLength, Just, Maybe.zero);
(I'm unable to test this myself since you didn't give the definitions of hasLength, Just, and Maybe.) If that wasn't what you were asking, please clarify the question.
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
The declaration
type EmptyFunc<T> = (val: T) => Maybe<T>;
declares a family of different function types EmptyFunc<number>, EmptyFunc<string>, etc., each of which works for only the specified type T. You probably meant
type EmptyFunc = <T>(val: T) => Maybe<T>;
which declares the type EmptyFunc of a single generic function that works for all types T. Then to apply this type to fromEmpty, just write:
const fromEmpty: EmptyFunc = R.ifElse(hasLength, Just, Maybe.zero);
(I'm unable to test this myself since you didn't give the definitions of hasLength, Just, and Maybe.) If that wasn't what you were asking, please clarify the question.
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
add a comment |
The declaration
type EmptyFunc<T> = (val: T) => Maybe<T>;
declares a family of different function types EmptyFunc<number>, EmptyFunc<string>, etc., each of which works for only the specified type T. You probably meant
type EmptyFunc = <T>(val: T) => Maybe<T>;
which declares the type EmptyFunc of a single generic function that works for all types T. Then to apply this type to fromEmpty, just write:
const fromEmpty: EmptyFunc = R.ifElse(hasLength, Just, Maybe.zero);
(I'm unable to test this myself since you didn't give the definitions of hasLength, Just, and Maybe.) If that wasn't what you were asking, please clarify the question.
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
add a comment |
The declaration
type EmptyFunc<T> = (val: T) => Maybe<T>;
declares a family of different function types EmptyFunc<number>, EmptyFunc<string>, etc., each of which works for only the specified type T. You probably meant
type EmptyFunc = <T>(val: T) => Maybe<T>;
which declares the type EmptyFunc of a single generic function that works for all types T. Then to apply this type to fromEmpty, just write:
const fromEmpty: EmptyFunc = R.ifElse(hasLength, Just, Maybe.zero);
(I'm unable to test this myself since you didn't give the definitions of hasLength, Just, and Maybe.) If that wasn't what you were asking, please clarify the question.
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
The declaration
type EmptyFunc<T> = (val: T) => Maybe<T>;
declares a family of different function types EmptyFunc<number>, EmptyFunc<string>, etc., each of which works for only the specified type T. You probably meant
type EmptyFunc = <T>(val: T) => Maybe<T>;
which declares the type EmptyFunc of a single generic function that works for all types T. Then to apply this type to fromEmpty, just write:
const fromEmpty: EmptyFunc = R.ifElse(hasLength, Just, Maybe.zero);
(I'm unable to test this myself since you didn't give the definitions of hasLength, Just, and Maybe.) If that wasn't what you were asking, please clarify the question.
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
answered Nov 11 '18 at 14:03
Matt McCutchenMatt McCutchen
13.5k819
13.5k819
add a comment |
add a comment |
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