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