Weird behavior of JS function "every()"-CodePudding

Please, can someone explain me behavior of function every.

I have an array and I want to apply function every to it. Function every is taking a predicate.

My simple predicate:

function test() {
  return function(arg) {
    console.log(arg);
    
    return true;
  }
}

I called function every on the array with my predicate:

const predicate = test();

[1, 2, 3].every(predicate);

Result:

Output:

1
2
3

BUT when I changed my predicate to this:

function test() {
  return function(...args) {
    console.log(...args);
    
    return true;
  }
}

The result is so weird for me:

Output:

1 0 [ 1, 2, 3 ] 
2 1 [ 1, 2, 3 ]
3 2 [ 1, 2, 3 ]

Why so? I expection somethin like this:

[1]
[2]
[3]

function test() {
  return function(arg) {
    console.log(arg);
    
    return true;
  }
}

function test2() {
  return function(...args) {
    console.log(...args);
    
    return true;
  }
}

const predicate = test();
const predicate2 = test2();
[1, 2, 3].every(predicate);
[1, 2, 3].every(predicate2);

CodePudding user response：

...arg is like a passing of all the arguments that .every method accepts.

i.e. An array element, index of the element and a whole array.

Hence, to print values as an array. You can simply return the values in a square bracket.

Demo :

[1, 2, 3].every(function(arg) {
  console.log([arg]);
  return true;
});

CodePudding user response：

every(), like map() and forEach(), calls function with two arguments: value and index.