Declaring Function type with TypeScript

Question

I know I have asked this before, and I cannot find the question nor remember the answer.

I have an object with several methods with the same signature:

{

  foo: function(){
    return [];   // (return Array<T>)
  },

  bar: function(){
    return [];   // (return Array<T>)
  },

  baz: function(){
    return [];   // (return Array<T>)
  }

}

When I declare the interface for this object:

interface SomeObj {

   foo: Function,
   bar: Function,
   baz: Function

}

but I want to declare a type for these functions, something like this:

  type TestSuiteGetterFn <T>() => Array<T>;

  interface SomeObj {

       foo: TestSuiteGetterFn,
       bar: TestSuiteGetterFn,
       baz: TestSuiteGetterFn

    }

but this does not compile.

I have rarely found something so difficult to Google as this one.

@peeskillet thanks, maybe that works, but I think gyre's answer is more ideal? — Alexander Mills
– Alexander Mills, Commented Mar 5, 2017 at 8:15
@Chris I got a red squiggly in my IDE (Webstorm), I didn't need to transpile to know there would be an error, but doubt it would be a meaningful error. — Alexander Mills
– Alexander Mills, Commented Mar 5, 2017 at 9:14

gyre · Accepted Answer · 2017-03-05 08:13:57Z

13

You simply forgot the equals sign when declaring your function type.

TypeScript Playground Permalink

type TestSuiteGetterFn<T> = () => Array<T>;

interface SomeObj {
     foo: TestSuiteGetterFn<string>,
     bar: TestSuiteGetterFn<number>,
     baz: TestSuiteGetterFn<string>
}

answered Mar 5, 2017 at 8:13

gyre

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4 Comments

gyre Over a year ago

Glad to help! Learned something new about TypeScript myself, too.

Alexander Mills Over a year ago

ditto if you think it's a well framed question

mesqueeb Over a year ago

Guys what is that <T> ?

Alexander Mills Over a year ago

<T> is the generic type, it must be passed in, see the 3 times it is passed

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