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>));
I am trying to set the type of a function to be an overload type that is defined as the property of an interface, but am getting Binding element 'text' implicitly has an 'any' type.ts(7031) for the function arguments. The overloads all have the same typing for the args so I'm not sure why it's erroring.
interface Args {
text: string
}
interface Utils {
toUpperCase?(args: Args): string
toUpperCase?(args: Args): any
}
export const toUpperCase: Utils['toUpperCase'] = ({ text }) => text.toUpperCase()
It works fine with either overload individually, so if I remove one of them (doesn't matter which), then the error is gone, e.g.
interface Utils {
toUpperCase?(args: Args): string
// toUpperCase?(args: Args): any
}
Adding an :Args explicit typing to the function parameter seems to eliminate the compiler error.
interface Args {
text: string
}
interface Utils {
toUpperCase?(args: Args): string
toUpperCase?(args: Args): any
}
export const toUpperCase: Utils['toUpperCase'] = ({ text }: Args) => text.toUpperCase()
Looks like the inference isn't or maybe can't be made that the function on the right fulfils the signature of Utils['toUpperCase'] without it.
Below decorator has been used only as a simple example. Do not try to figure out whether it has sense or not, please.
export function ToString(target: Object, key: string) {
let parsedValue: string;
Object.defineProperty(target, key, {
get: () => parsedValue,
set: (value: number) => parsedValue = number.toString()
});
}
As you see, above decorator takes 'number' as a parameter and in final step returns the same value but parsed to string.
set type -> number, get type -> string
unfortunately below notation is wrong
class TestClass {
#ToString parsedValue: string = 5; // number is not assignable to string
}
We know that ToString decorator will parse the value but we can't type decorated property as string.
The workaround would be to pass the setter value to the decorator like this:
export function ToString(value: number) {
return (target: Object, key: string) {
let parsedValue: string;
Object.defineProperty(target, key, {
get: () => parsedValue,
set: (value: number) => parsedValue = number.toString()
});
}
}
class TestClass {
#ToString(5) parsedValue: string; // it's okay because we don't use setter here
}
Everything would be okay unless we want to use this context while passing value to the decorator.
class TestClass {
#ToString(this.someService.value) parsedValue: string;
constructor(private someService = {value: 5}) {
}
}
this is undefined at this stage. Is there a way to achieve expected result without using any type? :)
Why doesn't Typescript warn me that the function I am defining does not match the interface declaration, but it does warn me if I try to invoke the function.
interface IFormatter {
(data: string, toUpper : boolean): string;
};
//Compiler does not flag error here.
var upperCaseFormatter: IFormatter = function (data: string) {
return data.toUpperCase();
}
upperCaseFormatter("test"); //but does flag an error here.
The interface ensures that all callers of functions that implement the interface supply the required arguments - data and toUpper.
Because TypeScript understands that JavaScript doesn't mind if you pass arguments that aren't used, it cleverly allows this in implementations.
Why is this okay? Because it means you can substitute any implementation of the interface without affecting calling code.
Example: You can substitute either IFormatter implementation and the code works.
interface IFormatter {
(data: string, toUpper: boolean): string;
};
var upperCaseFormatter: IFormatter = function (data: string) {
return data.toUpperCase();
}
var variableCaseFormatter: IFormatter = function (data: string, toUpper: boolean) {
if (toUpper) {
return data.toUpperCase();
}
return data.toLowerCase();
}
// Switch between these at will
//var formatter = upperCaseFormatter;
var formatter = variableCaseFormatter;
formatter("test", true);
If TypeScript didn't do this, your upperCaseFormatter would have to have to have a parameter called toUpper that wasn't used anywhere in the function - which makes the code less readable.
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>;
}