如何从闭包函数调用中返回相同的值 n 次？

Question

What I would like to do is have a closure return the same value n amount of times.

let value = getCurrentValue([0, 315, 270, 225, 180, 135, 90], 3);

I should then get:

value() // returns 0
value() // returns 0
value() // returns 0
value() // returns 315
value() // returns 315
value() // returns 315
value() // returns 270
value() // returns 270
value() // returns 270

This is what I tried!

  function getCurrentValue(values, num) {
    let index = -1;
    let l = values.length;

    function increment() {
      ++index;
      if (index < l) {
        return values[index]
      } else {
        index = -1;
        ++index;
        return values[index]
      }
    }

    if (num)  {
     for (var i = 0; i < num; i++){
      return increment         
     }
   }


    return increment;
  }

Any help would be appreciated!

Answer 1

Here is my generator solution & closure solution

 // generator solution function* getCurrentValueIter(iter, times) { for (let i = 0; i < iter.length * times; i++) { const cur = Math.floor(i / times); yield iter[cur] } } console.log('generate') const valueIter = getCurrentValueIter([0, 1], 2) for (let value of valueIter) { console.log(value) } // you can also use valueIter manually // console.log(valueIter.next().value) // closure solution function getCurrentValue(iter, times) { let i = 0 function value() { while (i < iter.length * times) { const cur = Math.floor(i / times); i += 1 return iter[cur] } } return value } const value = getCurrentValue([0, 1], 2) console.log('closure', value()) console.log(value()) console.log(value()) console.log(value())

Answer 2

Simply use an iterator function.

 function* getCurrentValue(arr, count) { for (let item of arr) { for (let j = 0; j < count; j++) { yield item; } } } for (let value of getCurrentValue([0, 315, 270, 225, 180, 135, 90], 3)) { console.log(value); }

Answer 3

Using just a closure - keep track of the index and the number of repeats outside the returned function, so the values persist through calls:

 function getCurrentValue(arr, n) { let index = 0; let repeat = 0; return function() { if (repeat < n) { repeat++; } else { repeat = 1; index++; } return arr[index]; } } let value = getCurrentValue([0, 315, 270, 225, 180, 135, 90], 3); console.log(value()); // returns 0 console.log(value()); // returns 0 console.log(value()); // returns 0 console.log(value()); // returns 315 console.log(value()); // returns 315 console.log(value()); // returns 315 console.log(value()); // returns 270 console.log(value()); // returns 270 console.log(value()); // returns 270 console.log(value()); // returns 225

 .as-console-wrapper { max-height: 100% !important; }

The if / else section can be shortened by using % to increase the repeat value and cap it to the maximum:

 function getCurrentValue(arr, n) { let index = -1; let repeat = -1; return function() { repeat = (repeat + 1) % n; if (repeat === 0) index++; return arr[index]; } } let value = getCurrentValue([0, 315, 270, 225, 180, 135, 90], 3); console.log(value()); // returns 0 console.log(value()); // returns 0 console.log(value()); // returns 0 console.log(value()); // returns 315 console.log(value()); // returns 315 console.log(value()); // returns 315 console.log(value()); // returns 270 console.log(value()); // returns 270 console.log(value()); // returns 270 console.log(value()); // returns 225

 .as-console-wrapper { max-height: 100% !important; }

Answer 4

In Javascript:

const values = [0, 315, 270, 225, 180, 135, 90];
const rNum = 3;

const getCurrentValue = (values, num) => {
    if (num <= 0) {
        return console.log("Repeater is invalid");
    } else {
        values.map((number) => {
            let repeatingCount = 0;
            while (repeatingCount != num) {
                console.log(number);
                repeatingCount++;
            }
        })
    }
}
getCurrentValue(values, rNum);

Answer 5

 const value = getCurrentValue([0, 315, 270, 225, 180, 135, 90], 3); function getCurrentValue(arr = [], n = 0) { // Return if the array is empty or `n` is 0 if (!arr.length || !n) return; let index = 0; let count = 0; return function () { // Have we reached the end of the array? if (index < arr.length) { // No? Check to see if our count is less than `n` // increment it if it is, and log the number if (count < n) { ++count; console.log(arr[index]); // Otherwise increase the array index // and set `count` to zero } else { ++index; count = 0; } } } } value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value(),value();

Answer 6

If I understood well this is what you want

    let res = [];
    
    function getCurrentValue(arr, x) {
    res = [];
      for (val of arr) {
        for (let i = 0; i < x; i++) {
          res.push(val);
        }
      }
    }
    getCurrentValue([0, 42, 4], 3)