Skip type check on unused parameters

问题

When I compile my typescript project, I'm using the noImplicitAny option so that I won't forget to specify the types on my variables and arguments.

However sometimes you have arguments that you don't use. For example:

jQuery.ajaxTransport("+*", function (options: JQueryAjaxSettings) {
  return {
    abort: function (_, callback: JQueryCallback) { 

I am not interested in the first argument of the abort function, so I ignore it by naming it _.

Is that the proper way to do that in TypeScript? I couldn't find it in the guide. I suspect that it isn't the proper way, because I can only name one argument _.

Typescript raises the following error:

error TS7006: Parameter '_' implicitly has an 'any' type.

I could just type _:any but that seems a bit overkill for an argument that I don't use.

回答1:

I was having the same problem. Using say express and routing you would often only want the res parameter.

router.get('/', function (req, res) { res.end('Bye.'); });

Your idea of using _ works here, but I've also found doing this works too.

function (_1, _2, _3, onlyThis) { console.log(onlyThis); }

This seems better, as only doing '_' I think might make using lodash/underscore a bit confusing, and it also makes it obvious it's the 4th parameter your interested in.

回答2:

I may be late, but I got stuck with the other solutions and this one work all the time for me:

function ({}={}, {}={}, {}={}, onlyThis) { console.log(onlyThis); }

comparison

When using the _0, _1, ... solution, I faced difficulties with scooped function like:

function parent(_0, _1, _2) {
  function child(_0, _1, _2) {
    // TypeScript crying about shallowed variables
  }
}

but with the empty object it work well:

function parent({}, {}, {}) {
  function child({}, {}, {}) {
    // :-)
  }
}

Even if the parameters are typed like:

function parent({}: number, {}: string, {}: any) {
  function child({}: number, {}: string, {}: any) {
    // :-) x2
  }
}

EDIT:

And as written here, setting a default value avoid error throwing if the given parameter is undefined or null.

function parent({}={}, {}={}, {}={}) {
  function child({}={}, {}={}, {}={}) {
    // :-)
  }
}

来源：https://stackoverflow.com/questions/38835001/skip-type-check-on-unused-parameters