Typescript How to define type for a function callback as any function type not universal any used in a method parameter

Question

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

User · Answer

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;

}

User · Answer

Hopefully  this will help    interface Param       title  string      callback   error  Error  data  string    gt  void      Or in a Function  let myfunction    title  string  callback   error  Error  data  string    gt  void   string   gt         callback new Error  Error Message Here      quot This is callback data  quot        return title

User · Answer

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        gt  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      gt  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      gt  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      gt  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

User · Answer

Following from Ryan s answer  I think that the interface you are looking for is defined as follows   interface Param       title  string      callback       gt  void

User · Answer

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    gt  void     function property    interface ParamMethod       callback a  Animal   void     method    There is an important typing difference since TS 2 6  You get stronger   quot sound quot   types in --strict or --strictFunctionTypes mode  when a function property is declared  Let s take an example  const animalCallback    a  Animal   void   gt         Animal is the base type for Dog const dogCallback    d  Dog   void   gt           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

User · Answer

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        gt  void  export declare type fEmptyReturn        gt  any  export declare type fArgVoid       args  any      gt  void  export declare type fArgReturn       args  any      gt  any    like this   public isValid  fEmptyReturn       boolean   gt  true  public setStatus  fArgVoid    status  boolean   void   gt  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   gt  true  public setStatus  fFunction    status  boolean   void   gt  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

User · Answer

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    gt  returnType and we can use this type definition in another function s parameter list  function runCallback callback   sum  number    gt  void  a  number  b  number   void       callback a   b         Another way of writing the function would be     let logSum   sum  number    gt  void   function sum  number   void          console log sum         function logSum sum  number   void       console log  The sum is   sum         runCallback logSum  2  2

User · Answer

Here s an example of a function that accepts a callback  const sqk    x  number  callback       number    gt  number    number   gt         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    gt  void    void   gt         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

User · Answer

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      gt  void matches all functions taking no arguments   Update with details and improvements  The answer and question here has way too many updoots not to mention the only-slightly more complex yet best-practice method  Solution to your question interface Easy Fix Solution       title  string      callback  Function     NOTE  Please do not use Function as you most likely do not want any callback function  It s okay to be a little specific  Improvement to that solution interface Safer Easy Fix       title  string      callback       gt  void    interface Alternate Syntax 4 Safer Easy Fix       title  string      callback    void     NOTE  The original author is correct  accepting no arguments and returning void is better   it s much safer  it tells the consumer of your interface that you will not be doing anything with their return value  and that you will not be passing them any parameters  And better yet Use generics  This interface would also work for the same      gt  void function types mentioned before  interface Better still safe but way more flexible fix       title  string      callback   lt T   unknown  R   unknown gt  args   T    gt  R    interface Alternate Syntax 4 Better still safe but way more flexible fix       title  string      callback lt T   unknown  R   unknown gt  args   T   R     NOTE  If you aren t 100  sure about the callback signature right now  please choose the void option above  or this generic option if you think you may extend functionality going forward  Callbacks usually receive some arguments of some sort  and sometimes the callback orchestrator even does something with the return value  And a slightly more advanced usecase you shouldn t use unless you need it This allows any number of arguments  of any type in T  More details here  interface Alternate Syntax 4 Advanced       title  string      callback lt T extends unknown    R   unknown gt     args   T   R

[javascript] Typescript: How to define type for a function callback (as any function type, not universal any) used in a method parameter

Examples related to javascript

Examples related to typescript