CODEWITHSUNDEEP
javascriptarrays
sourlemonaid
sourlemonaid

Reputation: 504

performing math on an array of strings(with numbers inside) with a loop of math operators

I have an array of strings and the strings hold numbers inside of them. I also have a set of operations i need to perform on those numbers. They follow the order of +, -,*, /. if the amount of numbers is more than 4 then it goes back to the plus sign.

  function calculate(n, operator, b) {
    switch (operator) {
      case 1:
        n += b;
        return n;
      case 2:
        n -= b;
        return n;
      case 3:
        n *= b;
        return n;
      case 4:
        n /= b;
        return n;
    }
}
numberArray = ["23", "44", "99", "324", "19"]
 let n=0;
let c = 1;
for (var i = 0; i < numberArray.length; i++) {

    calculate(n, c, parseInt(numberArray[i]));
    if ( c < 4) {
      c++;
    } else {
      c=1;
    }
  }
console.log(n);

for some reason the switch statement doesn't run correctly. it doesn't perform the operations and return the n back so i can get the total from the calculations.

Upvotes: 0

Views: 1156

Answers (2)

Nina Scholz
Nina Scholz

Reputation: 386654

You could use a more compact version without the switch structure, but with an array for the functions.

It works with the index and is zero based, so no hassle with the first or last item, just take the modulo operator % with the length of the array and get the right index.

let array = ["23", "44", "99", "324", "19"],
    fn = [(a, b) => a + b, (a, b) => a - b, (a, b) => a * b, (a, b) => a / b],
    n = array.reduce((n, a, i) => fn[i % fn.length](n, a), 0);

console.log(n);

Upvotes: 0

T.J. Crowder
T.J. Crowder

Reputation: 1074495

i thought the n += b; in the function would update the value of n.

It does, but that n isn't the n in your code calling calculate. It's the n that's an argument to calculate. When you do:

calculate(n, c, parseInt(numbersArray[i]));

the value of n is passed into calculate, not something that refers to the variable itself. (This is called "passing by value." JavaScript always passes the value of a variable.) The argument is like a local variable; updating it has no effect outside the function.

Simply use the result:

n = calculate(n, c, parseInt(numbersArray[i]));

Updated snippet:

function calculate(n, operator, b) {
    switch (operator) {
      case 1:
        n += b;
        return n;
      case 2:
        n -= b;
        return n;
      case 3:
        n *= b;
        return n;
      case 4:
        n /= b;
        return n;
    }
}
numberArray = ["23", "44", "99", "324", "19"]
 let n=0;
let c = 1;
for (var i = 0; i < numberArray.length; i++) {

    n = calculate(n, c, parseInt(numberArray[i]));
    if ( c < 4) {
      c++;
    } else {
      c=1;
    }
  }
console.log(n);

Of course, this means your switch can be simpler:

switch (operator) {
  case 1:
    return n + b;
  case 2:
    return n - b;
  case 3:
    return n *= b;
  case 4:
    return n / b;
}

Just FWIW, there's a clever way to update c as well, to avoid the if:

c = (c % 4) + 1;

That'll wrap around for you.

Upvotes: 3

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