ES6其余参数参数示例

Question

I don't really understand the below JavaScript example.

function containsAll(arr) {
  for (let k = 1; k < arguments.length; k++) {
    let num = arguments[k];
    if (arr.indexOf(num) === -1) {
      return false;
    }
  }
  return true;
}
let x = [2, 4, 6, 7];
console.log(containsAll(x, 2, 4, 7));
console.log(containsAll(x, 6, 4, 9));

The output is 1 and 0 in the console.

I'm trying to envision how it should work.

1. In this console.log(containsAll(x, 2, 4, 7)), the x should be replaced, changing it to console.log(containsAll(2, 4, 6, 7, 2, 4, 7)).
2. The function containsAll get those numbers (2, 4, 6, 7, 2, 4, 7).

3. In this line, for(let k = 1; k < arguments.length; k++) , k should be a new array [1, 2, 3, 4, 5, 6], with the argument length which is the length of arr (in this case, 7 is the length), right?

4. In next step, which is let num = arguments[k]; if (arr.indexOf(num) === -1) {return false;} let num = arguments[k]; if (arr.indexOf(num) === -1) {return false;} let num should be num = 1 , right?

5. Then, in the if statement, it tests if 1 is in the array, arr = [2, 4, 6, 7, 2, 4, 7]. And if no match is found, it returns false. And it should repeat for the next number in the arr array, correct?

I'm just trying to figure out why it output 1 for console.log(containsAll(x, 2, 4, 7)); when it should return something like [false, true, false, true, false, true] .

Answer 1

1. In this console.log(containsAll(x, 2, 4, 7)), the x should be replaced, changing it to console.log(containsAll(2, 4, 6, 7, 2, 4, 7)).

No, it stays [2, 4, 6, 7], 2 ,4 ,7

2. The function containsAll get those numbers (2, 4, 6, 7, 2, 4, 7).

It gets [2, 4, 6, 7], 2 ,4 ,7

3. In this line, for(let k = 1; k < arguments.length; k++) , k should be a new array [1, 2, 3, 4, 5, 6], with the argument length which is the length of arr (in this case, 7 is the length), right?

k is just a number representing the index inside arguments not an array. The length is just 4, the array is just one element

4. In next step, which is let num = arguments[k]; if (arr.indexOf(num) === -1) {return false;} let num = arguments[k]; if (arr.indexOf(num) === -1) {return false;} let num should be num = 1 , right?

num iterates over the remaining elements of arguments ( 2 ,4 ,7 ). Notice the k=1 in the for loop to skip the first element.

5. Then, in the if statement, it tests if 1 is in the array, arr = [2, 4, 6, 7, 2, 4, 7]. And if no match is found, it returns false. And it should repeat for the next number in the arr array, correct?

It returns already if it finds the first match.

I'm just trying to figure out why it output 1 for console.log(containsAll(x, 2, 4, 7)); when it should return something like [false, true, false, true, false, true] .

When return is called the execution inside the method stops.

Answer 2

First, your code currently outputs `true` and `false` not 1 and 0.
The reason you don't have an array like this,[false, true, false, true, false, true], is because of the `return` keyword used in the conditional `if` statement inside your `for` loop.

The code below is one way to do it.

function containsAll(arr) {
  let result = []
  for (let k = 1; k < arguments.length; k++) {
    let num = arguments[k];
    if (arr.indexOf(num) === -1) {
      result.push(false);
    } else{
        result.push(true);
    }
  }
    return result;
}
let x = [2, 4, 6, 7];
console.log(containsAll(x, 2, 4, 7)); //Output [true, true, true]
console.log(containsAll(x, 6, 4, 9)); //Output [true, true, false]