say I have this simple situation:
const f = function(){
const fn = f.bind(null, arguments);
}
I am trying to implement a "retry" mechanism - this code checks out according to the static analysis and seems to the signature of Function.prototype.bind, but my question is:
is arguments going to be applied as simply the first argument to f or will it be spread out, as in, f.apply(null, arguments)? I am having trouble finding an example of this online.
.bind works similar to .call, not .apply - the second argument will be treated as just that, the second argument. So in your example:
f(1,2,3) would produce fn([1,2,3])
arguments will be passed as simply the first argument to f. Moreover, if you then call the bound function with more arguments, those arguments will come after the bound one (i.e. you cannot overwrite that first argument).
read more here
Yeah, so Function.prototype.bind has similar signature to Function.prototype.call
so you need to do:
fn.bind(null, ...arguments);
or
fn.bind(null, ...Array.from(arguments));
const fn = f.bind(null, ...arguments);to spreadf.apply.bind(f, null, arguments)orf.bind.apply(f, [null, ...arguments])function f(...args) { const fn = () => f(...args); }...Array.from(arguments)...not sure why.