How can i modify multiple variables using a single function in JS?

Question

im a c programmer and im starting with JS. I have a date and i want to make a function that: -Recieves a month and a year and checks if the date is invalid, and corrects it (month will never be bigger than 24).

const check_month = (month, year) => {
  if(month > 12){
   month = month - 12;
   year = year + 1;
  }
}

The problem is that when i exit the fuction, month and year variables are not actually modified outside that function, so i understand that im passing it copies of my variables, not the actual reference of the variable. I thought about giving it a return but i cant return multiple values at once (i would have to return year and month), so im a little bit lost. Maybe returning

date = {month, year};

return date;

?

Could you help me? Thank you

Why is this surprising to you? You said you're a C programmer, and the same thing would happen in C. Parameters are passed by value, and assigning the local variable doesn't affect the caller's variable. — Barmar
– Barmar, Commented Sep 1, 2022 at 20:35
JS doesn't have pointers, although you can use containers (objects or arrays) for similar purposes. — Barmar
– Barmar, Commented Sep 2, 2022 at 2:44

Barmar · Accepted Answer · 2022-09-01 20:36:59Z

1

If you return the object, you can use destructuring.

const check_month = (month, year) => {
  while(month > 12){
   month = month - 12;
   year = year + 1;
  }
  return {month, year}
};

{month, year} = check_month(month, year);

I also changed if to while so that this will work for months above 24 as well.

answered Sep 1, 2022 at 20:36

Barmar

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Comments

Carsten Massmann · Accepted Answer · 2022-09-01 21:06:55Z

0

Or you can pass an object as the argument. Objects are always passed by reference:

const adjust_month = o => {
  o.year+=Math.floor(o.month/12)
  o.month=o.month%12;
  return o;
};
const monthYear={month:25,year:2020};
adjust_month(monthYear);
console.log(monthYear);

console.log(adjust_month({month:63,year:2017}));

I added the return o; into the function, so the modified object o can be assigned to another variable or be used directly in an expression.

edited Sep 1, 2022 at 21:06

answered Sep 1, 2022 at 21:00

Carsten Massmann

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