I have several modules (or bridges) that follow the same format as below:
export interface Bridge {
foo: (a: string, b: boolean, c: string) => number;
bar: (a: number, b: number, c: string, d: string) => string;
x: (a: string) => boolean;
y: () => null;
}
As a result, I need to create a layer than does some if-else processing to import the respective module. However, the functions within these modules will require a fallback mechanism that is made to be generic.
export type BridgeFunctions = keyof Bridge;
/* Called when a particular bridge throws an error */
export async function fallback(fnName: BridgeFunctions, ...args: any[]): Promise<any> {
const fallbackBridge = await import('./fallbackBridge');
const fn = fallbackBridge[fnName];
return fn(...args);
}
However, VSCode returns me this error:
A spread argument must either have a tuple type or be passed to a rest parameter.ts(2556)
Tried using call and apply, which led to this error instead:
The 'this' context of type ... is not assignable to method's 'this' of type '(this: any[]) => Promise<any>'.
Based on my understanding, in vanilla Javascript, passing args straight into fn should work. How do I emulate the same in TypeScript?
EDIT:
The Bridge modules are used as follows:
function getBridge(): Bridge {
if (inIOS()) {
return await import('./ios');
}
// ...
return await import ('./web');
}
export function foo(a: string, b: boolean, c: string): number {
try {
const bridge = await getBridge();
return bridge.foo(a, b, c);
} catch (error: any) {
// ...
if (error.bridge.failed) {
return await fallback('foo', a, b, c);
}
}
}
This is how I would do it.
Basically, you want to make sure the user passes valid arguments to your function. To do this, you set the args to be the parameters of the passed in function.
export async function fallback<FN extends keyof Bridge>(fnName: FN, args: Parameters<Bridge[FN]>){
Then, to bypass the error, just type fn to any:
const fn = fallbackBridge[fnName] as any;
return fn(...args);
Using any here is fine since you already know the types are compatible.
The problem here is that your fallbackBridge functions require a certain number of parameters but you give them the type " any[] " that could have any length.
You could bypass this error with return fn(...(args as Parameters<typeof fn>));
Related
I am trying to write a function that accepts an API interface (I've created a sample here), and wraps a Proxy around it, so that any calls to the that API's methods get intercepted, and I can do some logging, custom error handling etc. I am having a terrible time with the types. This is similar to another question I have asked (Writing wrapper to third party class methods in TS), but uses a completely different approach than that one, based on some feedback I got.
Currently I am getting
Element implicitly has an 'any' type because expression of type 'string | symbol' can't be used to index type 'API'. No index signature with a parameter of type 'string' was found on type 'API'. which makes sense given that sayHello is not strictly a string as far as typescript is concerned, but I do not know the best way to be able to get methods on this class without uses the property accessor notation.
class API {
sayHello(name: string) {
console.log(“hello “ + name)
return name
}
}
export default <T extends API>(
api: T,
) =>
new Proxy(api, {
get(target, prop) {
if (typeof target[prop] !== "function") { // type error here with "prop"
return target[prop]; // and here
}
return async (...args: Parameters<typeof target[prop]>) => {
try {
const res = await target[prop](...args); // and here
// do stuff
return res
} catch (e) {
// do other stuff
}
};
},
});
Is this possible in TS?
TypeScript doesn't currently model the situation when a Proxy differs in type from its target object. There's a longstanding open feature request for this at microsoft/TypeScript#20846, and I don't know if or when the situation will change.
For now if you want to do this you'll need to manually describe the expected return type of your proxy function, and use type assertions liberally inside the implementation in order to suppress any errors. This means you'll need to verify that your function is type safe yourself; the compiler won't be able to help much.
Here's one possible approach:
const prox = <T extends API>(api: T) => new Proxy(api, {
get(target: any, prop: string) {
if (typeof target[prop] !== "function") {
return target[prop];
}
return async (...args: any) => {
try {
const res = await target[prop](...args);
return res;
} catch (e) { }
};
},
}) as any as { [K in keyof T]: AsyncifyFunction<T[K]> };
type AsyncifyFunction<T> = T extends (...args: infer A) => infer R ?
(...args: A) => Promise<Awaited<R>> : T;
The idea is that prox(api) returns a version of api where every non-function property is the same, but every function property has been changed to an async version that returns a Promise. So if api is of type T, then prox(api) is of mapped type { [K in keyof T]: AsyncifyFunction<T[K]> }, where AsyncifyFunction<T> is a conditional utility type that represents the transformation of each property type. If X is a function type with argument tuple A and return type R, then AsyncifyFunction<X> is (...args: A) => Promise<Awaited<R>>, using the Awaited<T> utility type to deal with any nested promises (we don't want a Promise<Promise<number>> to come out of this, for example).
Okay, let's test it:
class Foo extends API {
str = "abc";
double(x: number) { return x * 2 };
promise() { return Promise.resolve(10) };
}
const y = prox(new Foo());
/* const y: {
str: string;
double: (x: number) => Promise<number>;
promise: () => Promise<number>;
sayHello: (name: string) => Promise<void>;
} */
So, according to the compiler, y has a string-valued str property, and the rest of its properties are asynchronous methods that return promises. Notice that the promise() method returns Promise<number> and not Promise<Promise<number>>.
Let's make sure it works as expected:
console.log(y.str.toUpperCase()) // "ABC"
y.sayHello("abc") // "helloabc"
y.double(123).then(s => console.log(s.toFixed(2))) // "246.00"
y.promise().then(s => console.log(s.toFixed(2))) // "10.00"
Looks good!
Playground link to code
You can try add index signature to your class:
class API {
/* -- index signature -- */
[index:string|symbol]: any;
sayHello(name: string) {
console.log("hello" + name)
}
}
export default <T extends API>(
api: T,
) =>
new Proxy(api, {
get(target, prop) {
if (typeof target[prop] !== "function") {
return target[prop];
}
/* -- had to store it in a variable -- */
const data = target[prop];
return async (...args: Parameters<typeof data>) => {
try {
const res = await target[prop](...args);
// do stuff
return res.data;
} catch (e) {
// do other stuff
}
};
},
});
I want to write a call function:
function call<F extends (...arg: any) => any, P extends Parameters<F>>(
fn?: F,
...arg: P
): ReturnType<F> | undefined {
if (fn) return fn(...arg) // Type 'P' must have a '[Symbol.iterator]()' method that returns an iterator
}
const fn = (a: string) => {}
// automatically infer `fn` arguments type
call(fn, )
what should I do？
This is a known bug in TypeScript; see microsoft/TypeScript#36874. It looks like the problem is triggered by F extends (...arg: any) => any instead of F extends (...arg: any[]) => any. That is, your arg rest parameter is of the anything-at-all any type instead of the array-of-anything any[] type. The obvious workaround is therefore to change any to any[], which shouldn't be a problem because function rest parameters are essentially always arraylike:
function call<F extends (...arg: any[]) => any, P extends Parameters<F>>(
fn?: F, ...arg: P): ReturnType<F> | undefined {
return fn?.(...arg) // okay (I used optional chaining here to deal with missing fn)
}
Looks good!
That's the answer to the question as asked, although it is useful to note that generally speaking, the preferred way to do this is to make call() generic in P (the rest parameter type) and R (the return type), and not use the Parameters<T> and the ReturnType<T> utility types:
function call<P extends any[], R>(fn?: (...arg: P) => R, ...arg: P): R | undefined {
return fn?.(...arg); // okay
}
They are similar, but your approach ends up being unable to properly represent what happens when fn is itself generic:
const pair = <T, U>(left: T, right: U): [T, U] => [left, right];
const ret = call(pair, "abc", 123); // old version of call
// const ret: [any, any] | undefined ☹
whereas the approach I recommend can take advantage of the support for higher order type inference from generic functions:
const ret = call(pair, "abc", 123); // new version of call
// const ret: [string, number] | undefined 👍
If you can rewrite generics using conditional types like Parameters<T> and ReturnType<T>, it will sometimes improve the compiler's ability to produce desirable results.
Playground link to code
This is a tricky one, but the solution is not.
You have to ensure the typescript that args is iterable. You can do that like manually saying to him - Hey this will be an array, believe me
Working playground
function call<F extends (...arg: any) => any, P extends Parameters<F>>(
fn?: F,
...args: P
): ReturnType<F> | undefined {
if (fn){
// here is the ugly hint to typescript
return fn(...(args as any[]))
}
return undefined;
}
Have you considered using apply as a workaround?
function call<F extends (...arg: any) => any, P extends Parameters<F>>(
fn?: F,
...arg: P,
): ReturnType<F> | undefined {
if (fn) return fn.apply(undefined, arg);
return undefined;
}
Playground
Is there a way to set up a function based on an initial interface? I have a bunch of listOfFunctions and a an equal amount of interfaces that contain types of the functions. I would like to create a factory that returns a custom hook. The idea is to save a ton of code and make editing and adding features to all the hooks a breeze.
FunctionInterface Is all of the function names and their returns.
listOfFunctions is a setup function (the real code has a config) that returns a object with a list of functions.
theChosenFunction is a function that will be returned in the hook that allows me to interact with one of the listOfFunction functions. This means parameters (if needed). The functions in the real app are Promises that will set state of the hook. In this example, I just return a value and setState.
A) Is this possible to do with funcObject[funcName]() and create a function to invoke with typescript? This would need that ability to add params or not.
B) Is there a way to get the return value of a function type? So in this example:
type CanThisWork = () => string; I want to extract string; The idea is not to do a refactor on all of the listOfFunctions and FunctionInterfaces. There are a lot of them. If I have to create one complicated Factory, then I find this more economical.
interface FunctionInterface {
noParamsNoReturn: () => void;
noParamsNumberReturn: () => number;
propsAndNoReturn: (id: string) => void;
propsAndReturn: (id: string) => string;
}
const listOfFunctions = (): FunctionInterface => {
return {
noParamsNoReturn: () => void,
noParamsNumberReturn: () => 1,
propsAndNoReturn: (id: string) => console.log(id),
propsAndReturn: (id: string) => id,
}
}
function runThis<T>(funcName: keyof T, funcObject: T): void {
const [value, setValue] = React.useState();
// Can I have typescript set up the params here?
const theChosenFunction = ();
const theChosenFunction = (props) => {
// Can I have typescript invoke the function properly here?
const result = funcObject[funcName](); // funcObject[funcName](props)
if (result) {
setValue(result);
}
}
return {
theChosenFunction
}
}
const result = runThis<FunctionInterface>('noParamsNoReturn', listOfFunctions());
result.theChosenFunction()
// OR
result.theChosenFunction('someId');
There are built-in Parameters<T> and ReturnType<T> utility types that take a function type T and use conditional type inference to extract the tuple of parameters and the return type, respectively:
type SomeFuncType = (x: string, y: number) => boolean;
type SomeFuncParams = Parameters<SomeFuncType>; // [x: string, y: number]
type SomeFuncRet = ReturnType<SomeFuncType>; // boolean
As for your runThis() function,
I'd be inclined to give it the following typings and implementation, assuming you want the minimal changes that compile and run:
function runThis<K extends PropertyKey, F extends Record<K, (...args: any[]) => any>>(
funcName: K, funcObject: F
) {
const [value, setValue] = React.useState();
const theChosenFunction = (...props: Parameters<F[K]>): void => {
const result = funcObject[funcName](...props);
if (result) {
setValue(result);
}
}
return {
theChosenFunction
}
}
I've given the function two generic type parameters: K, corresponding to the type of funcName, and F, corresponding to the type of funcObject. The constraints K extends PropertyKey and F extends Record<K, (...args: any[])=>any> guarantee that funcName will be of a key-like type (string, number, or symbol), and that funcObject will have a function-valued property at that key.
Then, we can make theChosenFunction use spread and rest syntax to allow the function to be called with a variadic list of parameters. So (...props) => funcObject[funcName](...props) will accept any number of parameters (including zero) and pass them to the called function. TypeScript represents such lists-of-parameters as a tuple type. Therefore, having theChosenFunction's call signature look like (...props: Parameters<F[K]>) => void means that it will accept the same parameters as the entry in funcObject at the key funcName, and that it will not output anything (because your implementation doesn't output anything).
Let's see if it works:
const result = runThis('noParamsNoReturn', listOfFunctions());
result.theChosenFunction(); // okay
result.theChosenFunction('someId'); // error! Expected 0 arguments, but got 1.
const anotherResult = runThis('propsAndReturn', listOfFunctions());
anotherResult.theChosenFunction(); // error! Expected 1 arguments, but got 0.
anotherResult.theChosenFunction("someId"); // okay
runThis(listOfFunctions(), 'someId'); // error! '
// FunctionInterface' is not assignable to 'string | number | symbol'.
runThis('foo', listOfFunctions()); // error!
// Property 'foo' is missing in type 'FunctionInterface'
runThis(
'bar',
{ bar: (x: string, y: number, z: boolean) => { } }
).theChosenFunction("hey", 123, true); // okay
Looks good to me.
Playground link to code
I encounter two case about this problem when I try to refactor a es6 project in typescript, and one is about Object.keys() and the other is about import * as xxx.
Case 1:
const SUPPORTED_VALUES = {
min_s: 'Mininum similarity',
max_rc: 'Maximum result count'
}
const UNSUPPORTED_MSG =
'Configurable values:\n' +
Object.keys(SUPPORTED_VALUES)
.map(k => `${k}: ${SUPPORTED_VALUES[k]}`)
.join('\n')
The k in the map is guaranteed to be a key in SUPPORTED_VALUES, but typescript compiler doesn't know about this. How should I fix this without disabling noImplicitAny?
Case 2:
I have a file called cmd.ts:
export async function cmd1(args){}
export async function cmd2(args){}
It is used is another file like this:
import * as cmdHandlers from './cmd'
// some code...
if (cmd in cmdHandlers) {
await cmdHandlers[cmd](bot, msg, ...args)
}
This is also guaranteed that cmd exists in cmdHandlers, but typescript compiler can't handle this.
In case 1, you need define exactly type for SUPPORTED_VALUES like {[key: string]: string}
const SUPPORTED_VALUES: {[key: string]: string} = {
min_s: 'Minimum similarity',
max_rc: 'Maximum result count'
}
const UNSUPPORTED_MSG =
'Configurable values:\n' +
Object.keys(SUPPORTED_VALUES)
.map(k => `${k}: ${SUPPORTED_VALUES[k]}`)
.join('\n')
Case 2, The same case 1, cmd can be anything, but cmdHandlers only includes cmd1 and cmd2. You can define your type for cmd.ts module to make cmdHandlers is a object with any property name.
I think it is ok, because you have a condition before call a function
// cmd.ts
async function cmd1(...args: any) { }
async function cmd2(...args: any) { }
const myExport: { [key: string]: (...args: any) => void } = {
cmd1,
cmd2,
}
export default myExport;
//
import cmdHandlers from './cmd'
// some code...
if (cmd in cmdHandlers) {
await cmdHandlers[cmd](bot, msg, ...args)
}
Currently I have type definition as:
interface Param {
title: string;
callback: any;
}
I need something like:
interface Param {
title: string;
callback: function;
}
but the 2nd one is not being accepted.
The global type Function serves this purpose.
Additionally, if you intend to invoke this callback with 0 arguments and will ignore its return value, the type () => void matches all functions taking no arguments.
Typescript from v1.4 has the type keyword which declares a type alias (analogous to a typedef in C/C++). You can declare your callback type thus:
type CallbackFunction = () => void;
which declares a function that takes no arguments and returns nothing. A function that takes zero or more arguments of any type and returns nothing would be:
type CallbackFunctionVariadic = (...args: any[]) => void;
Then you can say, for example,
let callback: CallbackFunctionVariadic = function(...args: any[]) {
// do some stuff
};
If you want a function that takes an arbitrary number of arguments and returns anything (including void):
type CallbackFunctionVariadicAnyReturn = (...args: any[]) => any;
You can specify some mandatory arguments and then a set of additional arguments (say a string, a number and then a set of extra args) thus:
type CallbackFunctionSomeVariadic =
(arg1: string, arg2: number, ...args: any[]) => void;
This can be useful for things like EventEmitter handlers.
Functions can be typed as strongly as you like in this fashion, although you can get carried away and run into combinatoric problems if you try to nail everything down with a type alias.
Following from Ryan's answer, I think that the interface you are looking for is defined as follows:
interface Param {
title: string;
callback: () => void;
}
You can define a function type in interface in various ways,
general way:
export interface IParam {
title: string;
callback(arg1: number, arg2: number): number;
}
If you would like to use property syntax then,
export interface IParam {
title: string;
callback: (arg1: number, arg2: number) => number;
}
If you declare the function type first then,
type MyFnType = (arg1: number, arg2: number) => number;
export interface IParam {
title: string;
callback: MyFnType;
}
Using is very straight forward,
function callingFn(paramInfo: IParam):number {
let needToCall = true;
let result = 0;
if(needToCall){
result = paramInfo.callback(1,2);
}
return result;
}
You can declare a function type literal also , which mean a function can accept another function as it's parameter. parameterize function can be called as callback also.
export interface IParam{
title: string;
callback(lateCallFn?:
(arg1:number,arg2:number)=>number):number;
}
Here's an example of a function that accepts a callback
const sqk = (x: number, callback: ((_: number) => number)): number => {
// callback will receive a number and expected to return a number
return callback (x * x);
}
// here our callback will receive a number
sqk(5, function(x) {
console.log(x); // 25
return x; // we must return a number here
});
If you don't care about the return values of callbacks (most people don't know how to utilize them in any effective way), you can use void
const sqk = (x: number, callback: ((_: number) => void)): void => {
// callback will receive a number, we don't care what it returns
callback (x * x);
}
// here our callback will receive a number
sqk(5, function(x) {
console.log(x); // 25
// void
});
Note, the signature I used for the callback parameter ...
const sqk = (x: number, callback: ((_: number) => number)): number
I would say this is a TypeScript deficiency because we are expected to provide a name for the callback parameters. In this case I used _ because it's not usable inside the sqk function.
However, if you do this
// danger!! don't do this
const sqk = (x: number, callback: ((number) => number)): number
It's valid TypeScript, but it will interpreted as ...
// watch out! typescript will think it means ...
const sqk = (x: number, callback: ((number: any) => number)): number
Ie, TypeScript will think the parameter name is number and the implied type is any. This is obviously not what we intended, but alas, that is how TypeScript works.
So don't forget to provide the parameter names when typing your function parameters... stupid as it might seem.
There are four abstract function types, you can use them separately when you know your function will take an argument(s) or not, will return a data or not.
export declare type fEmptyVoid = () => void;
export declare type fEmptyReturn = () => any;
export declare type fArgVoid = (...args: any[]) => void;
export declare type fArgReturn = (...args: any[]) => any;
like this:
public isValid: fEmptyReturn = (): boolean => true;
public setStatus: fArgVoid = (status: boolean): void => this.status = status;
For use only one type as any function type we can combine all abstract types together, like this:
export declare type fFunction = fEmptyVoid | fEmptyReturn | fArgVoid | fArgReturn;
then use it like:
public isValid: fFunction = (): boolean => true;
public setStatus: fFunction = (status: boolean): void => this.status = status;
In the example above everything is correct. But the usage example in bellow is not correct from the point of view of most code editors.
// you can call this function with any type of function as argument
public callArgument(callback: fFunction) {
// but you will get editor error if call callback argument like this
callback();
}
Correct call for editors is like this:
public callArgument(callback: fFunction) {
// pay attention in this part, for fix editor(s) error
(callback as fFunction)();
}
There are various ways to define function types; however, some are better than others.
Although it is possible to use Function, the JavaScript function object, don't do that.
Source: TypeScript ESLint plugin recommended rule ban-types
Avoid the Function type, as it provides little safety for the following reasons:
It provides no type safety when calling the value, which means it's easy to provide the wrong arguments.
It accepts class declarations, which will fail when called, as they are called without the new keyword.
TypeScript supports multiple other ways. Most commonly, function type expressions are used. This method highly resembles arrow functions.
If arguments and return values are known to be of a certain form, then they should be typed.
For example,
interface Param {
callback: (foo: string, bar: number) => void
}
Note that these types can be as complex as needed, for example using object types or types created from other types.
If the types are truly unknown then prefer unknown over any.
Source: TypeScript ESLint plugin recommended rule no-explicit-any
When any is used, all compiler type checks around that value are ignored.
From the TS docs,
unknown is the type-safe counterpart of any.
Thus, using spread syntax,
interface Params {
callback: (...args: unknown[]) => unknown
}
Hopefully, this will help...
interface Param {
title: string;
callback: (error: Error, data: string) => void;
}
Or in a Function
let myfunction = (title: string, callback: (error: Error, data: string) => void): string => {
callback(new Error(`Error Message Here.`), "This is callback data.");
return title;
}
Typescript: How to define type for a function callback used in a method parameter?
You can declare the callback as 1) function property or 2) method:
interface ParamFnProp {
callback: (a: Animal) => void; // function property
}
interface ParamMethod {
callback(a: Animal): void; // method
}
There is an important typing difference since TS 2.6:
You get stronger ("sound") types in --strict or --strictFunctionTypes mode, when a function property is declared. Let's take an example:
const animalCallback = (a: Animal): void => { } // Animal is the base type for Dog
const dogCallback = (d: Dog): void => { }
// function property variant
const param11: ParamFnProp = { callback: dogCallback } // error: not assignable
const param12: ParamFnProp = { callback: animalCallback } // works
// method variant
const param2: ParamMethod = { callback: dogCallback } // now it works again ...
Technically spoken, methods are bivariant and function properties contravariant in their arguments under strictFunctionTypes. Methods are still checked more permissively (even if not sound) to be a bit more practical in combination with built-in types like Array.
Summary
There is a type difference between function property and method declaration
Choose a function property for stronger types, if possible
Playground sample code
I've just started using Typescript and I've been trying to solve a similar problem like this; how to tell the Typescript that I'm passing a callback without an interface.
After browsing a few answers on Stack Overflow and GitHub issues, I finally found a solution that may help anyone with the same problem.
A function's type can be defined with (arg0: type0) => returnType and we can use this type definition in another function's parameter list.
function runCallback(callback: (sum: number) => void, a: number, b: number): void {
callback(a + b);
}
// Another way of writing the function would be:
// let logSum: (sum: number) => void = function(sum: number): void {
// console.log(sum);
// };
function logSum(sum: number): void {
console.log(`The sum is ${sum}.`);
}
runCallback(logSum, 2, 2);
In typescript 4.8, Function type gives error. Instead we can write the type explicitly as fn: () => void.
If you want to use args as well,
function debounce(fn: (...args: any[]) => void, ms = 300) {
If you are looking for input callback function like on change event then use the following
type Props = {
callBack: ChangeEventHandler<HTMLInputElement>;
}