Typescript apply+concat with a function argument

Question

The following code compiles.

function myFunction(arg1:string, arg2:(msg:string)=>void){ /* do stuff */ }
var args = ["hello", function(msg){ /* do stuff */ }];
myFunction.apply(myFunction, args);

But the following code does not compile, even though the variable args is of the same type as above.

function myFunction(arg1:string, arg2:(msg:string)=>void){ /* do stuff */ }
var args = ["hello"].concat(function(msg){ /* do stuff */ });
myFunction.apply(myFunction, args);

It throws the following error.

>> src/workspace.ts(20,29): error TS2345: Argument of type '(msg: any) => void' is not assignable to parameter of type 'string'.

Any idea why? Is this a bug in my code or a bug in the TypeScript compiler?

Fenton · Accepted Answer · 2015-07-18 22:18:14Z

3

In TypeScript, when you create an array, the type is inferred from the array members. The following snippet is an Array<string>, or string[]

var a = ["hello"];

If you try to add another value to this array, it will have to be a string:

a.concat("A string");

If you added a function, it would warn you that it isn't compatible:

a.concat(() => { return 5; });

And this also applies in your example, even though you are doing it immediately, as you can tell by adding an allowable value:

["hello"].concat("A string");

You can get around this by widening the type of the array:

var args = (<any[]>["hello"]).concat(function(msg){ /* do stuff */ });

answered Jul 18, 2015 at 22:18

Fenton

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1 Comment

sffc Over a year ago

Thanks, that makes sense. Since it returns a new array, I don't know if I agree with TypeScript's design decision of .concat() requiring its arguments to be of the same time as the array it was called on; it makes sense for something like .push() which modifies the existing array. The workaround gets the job done though.

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