below is a simple function. It's adds all number in the array and store in a variable. The problem is, the loop execute just once even though the condition for exiting the loop was not met. Am I missing something here?
const numChecker = (...args) => {
let x = args;
let y;
let i;
for (i = 0; i < x.length - 1; i++) {
if ((typeof x[i]) === "number") {
y += x[i];
}
return y;
}
}
console.log(numChecker("A", "B", "C", 100, 300, 200));There are 3 issues with your code :-
let y;.i < x.length - 1.return y at wrong place.In the first issue, when you declare a variable like this let y, then the typeof(y) is undefined. Here, you want to store the sum of numbers, so the variable y must be of type number. So, you must declare and initialize it like let y = 0;
In 2nd one, in the loop, the array will be parsed to 2nd last element. x[4] in your case. So, the condition in the loop should be like i < x.length.
In the last, you must return y outside the loop, so that the sum will be printed on the console will be of all the numbers in the array.
Check the Snippet below.
const numChecker = (...args) => {
let x = args;
let y = 0;
let i;
for (i = 0; i < x.length; i++) {
if ((typeof x[i]) === "number") {
y += x[i];
}
}
return y;
}
console.log(numChecker("A", "B", "C", 100, 300, 200));Hope, it will solve your issue.
i < x.length - 1??