I want to create callback function with some parameters, but I want that one of the values within never change and be set when the function is created!
P.e.
let a = 2;
let y = (b) => { let staticValue = a; return staticValue * b; }
y(1) => 2 -> OK
then
a = 3
y(1) => 3 -> Not OK -> I want the function to have the original value of a.
I know, this happens because the variable a is only evaluated when the function is called. But is there any workaround?
You can accomplish this with what is called an IIFE (immediately invoked function expression)
let a = 2;
const y = (staticValue => b => staticValue * b)(a);
a = 44;
console.log(y(1));What's happening here is that you have two functions. The first one, staticValue => ... is immediately invoked, and staticValue is given the value of a. Since a is given as an argument of the IIFE, once the function is invoked staticValue won't change, even if the value of a is changed.
The return value of the first function is a second function, b=> ....
let a = 2
let makeY = () => {
let staticValue = a;
return (b)=> staticValue * b;
}
let y = makeY();
or
let a = 2
let y = ((staticValue, b) => {
return staticValue * b;
}).bind(null, a)
you can use closure
function first (a) {
function second (b) {
return a * b;
}
return second;
}
let x = first(2) // the inner function second will close around the variable a which is 2 in this case
console.log(x(3))
let y = first(5) // the inner function second will close around the variable a which is 5 in this case
console.log(y(3))I think you could apply a Functional programming approach.
const getYFn = (a , b) => {
return a * b;
}
const a = 2
const y = getYFn.bind(null, a);
// the constant y now is a function that partially applied first param
console.log(y(1)); // 2
console.log(y(3)); // 6
localStorageand all that mess - but why bother? Just hard code the value into the function