CODEWITHSUNDEEP
javascript
Omer Zano
Omer Zano

Reputation: 33

Checking if multiple inputs are of number type

Say I'm creating a function to add some numbers together, and I want to verify they are all actually numbers.

I am doing this:

function addNumbers(x, y) {     
    if (typeof x == 'number' || typeof y == 'number') {
        // do something..
    }        
}

It seems unpractical if I had more than two numbers. What would be a better way to check for multiple numbers?

Upvotes: 3

Views: 932

Answers (5)

Dennis Vash
Dennis Vash

Reputation: 53874

I think its more readable form, use Array.prototype.every, Number.isInteger and Array.prototype.reduce.

I'm not sure what you want to do with errors, so we just logging them (and the result);

// consider floating point
const isNumber = n => typeof n == 'number' && !isNaN(n)

const addNumbers = (...args) => {
  const isValid = args.every(isNumber);
  // const isValid = args.every(Number.isInteger);
  if (!isValid) {
    console.log('Error');
    return;
  }
  const sum = args.reduce((sum, curr) => sum + curr, 0);
  console.log(sum);
  return sum;
}

addNumbers(5, 6);
addNumbers(5, 6, NaN);
addNumbers(5, '6');
addNumbers('5', '6');
addNumbers('5', 6);
addNumbers(5.5, 6);
addNumbers(5.5, 6, 6.4, 65);
addNumbers(5.5, 6, {});

Upvotes: 1

Djaouad
Djaouad

Reputation: 22766

You can put them in an array and use a Boolean flag to check if all are numbers, using Array.prototype.every (and check for NaNs, because typeof NaN === 'number'):

function addNumbers(...args) {
  var all_numbers = args.every(a => typeof a == 'number' && !isNaN(a));
  if (all_numbers) {
    var sum = 0;
    args.forEach(n => sum += n);
    console.log(sum);
  } else {
    console.log('something is not right!');
  }
}

addNumbers(5, 6);
addNumbers(5, 6.2);
addNumbers(5, 6, NaN);
addNumbers(5, 6, []);
addNumbers(5, 6, {});
addNumbers(5, '6');
addNumbers('5', 6);

Upvotes: 3

Fullstack Guy
Fullstack Guy

Reputation: 16908

You can use get your inputs as an array using the ... rest parameter syntax then use Array.prototype.reduce to sum them and while doing so you can convert the elements to numbers using the + operator and add them: 

function addNumbers(...nums) {     
    return nums.reduce((sum, num) => sum + +num)
}
console.log(addNumbers(1, 2, "3", 4));

Or if you want to skip the non-numbers (which will produce NaN if you use the first code snippet) just check the type before adding, if number you're good else replace that with a 0:

function addNumbers(...nums) {     
  return nums.reduce((sum, num) => sum + (!(typeof(num) === "number") ? 0 : +num));
}
console.log(addNumbers(1, 2, "3", 4, "non-number"));

Upvotes: 1

ibrahim tanyalcin
ibrahim tanyalcin

Reputation: 6491

If performance not the issue here:

function testN(){
    if(!Array.from(arguments).every((d)=>typeof d === "number")){
        return;
    }
    console.log("pass");
    //do something;
}

replace "do something" with whatever you want to do. Can use negated "some" method, that'll terminate earlier. Whichever fits you.

Upvotes: 0

user11804236
user11804236

Reputation:

You can push all your numbers into and array and run a loop over that array and add them.

let arr = [1,2,3,4]; #be it your array

function addNumbers(arr) {
let sum=0;
arr.map(item => {
if(typeof item === 'number') {
sum=sum+item;
}
else {
return item+' is not a number in given array';
}
return sum;
});

Upvotes: 0

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