Skip type check on unused parameters Skip type check on unused parameters typescript typescript

Skip type check on unused parameters


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.


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 shadowed 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({}={}, {}={}, {}={}) {    // :-)  }}