偶 奇 javascript 浏览数组

Question

how to browse an array to find out if all the numbers in the array are even or odd?

thanks you.

script

 function evenOdd() {
    let tab = [10, 5, 8, 1];
    if (tab % (2 = 0)) {
        console.log('even');
    } else {
        console.log('odd');
    }
}
console.log(evenOdd());

Answer 1

You need to iterate the values from the array and check the value with the comparison operator === instead of an assignment = .

 function evenOdd() { let tab = [10, 5, 8, 1]; for (const value of tab) { if (value % 2 === 0) { console.log(value, 'even'); } else { console.log(value, 'odd'); } } } evenOdd(); // no console.log because no return value

Answer 2

The trick is to assume, that your list only contains even numbers until an odd number is encountered and that it only contains odd numbers until an even number is encountered

 function evenOdd(tab) { let isEven = true; let isOdd = true; for (let value of tab) { // note: no comparision to 0 or 1 needed, as value % 2 is already // truthy for odd value and falsy for even value if (value % 2) { isEven = false; } else { isOdd = false; } } return {isEven, isOdd}; } let tab = [10, 5, 8, 1]; let {isEven, isOdd} = evenOdd(tab);

You will also need to return two distinct booleans or an integer, as there are 3 different return options: all even, all odd, or even and odd mixed.

Answer 3

First, I would call the function outside of a console.log(). This is because you already are consoling the result in the if statement and there is no return value.

Second, you need to make sure that you are using some method to map over the array. In this case, I would use a.forEach().

Last, you want to check it the item divided by 2 equals 0. To do this it looks like:

item % 2 === 0

For the final result I would do this:

function evenOdd() {
    let tab = [10, 5, 8, 1]

    tab.forEach((item) => {
        item % 2 === 0
            ? console.log(item, "even")
            : console.log(item, "odd")
    })
}

evenOdd()