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JavaScript arithmetic operations between strings of numbers

I'm wondering what happens behind the scenes when we either add, subtract or multiply two strings of numbers. Here is an example:

let strNum1 = "300";
let strNum2 = "22";

let multiply = function(num1, num2) {
    let product = num1 * num2;
    
    return `${product}`
      
};

multiply(strNum1, strNum2); //this will return ==> "6600"

Does the JS engine turns these into integers first and then performs the operations or does it know "magically" that they are numbers even though it's in a string form? The reason I'm asking is because of the long multiplication algorithm. For numbers bigger than 8 chars it becomes funky when multiplying with the operator vs using the algorithm.

This is a leetcode question btw.

7 months ago · Juan Pablo Isaza
3 answers
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0

You can parse your values before and after operation:

let multiply = function(num1 = '', num2 = '') {
    const product = Number(num1) * Number(num2);
    
    return `${product}` // Or String(product)
};
7 months ago · Juan Pablo Isaza Report

0

To answer your question: The JS Engine turns those strings into integers before doing the arithmetic operation. It is called Implicit coercion.

You can read more about that in this You Don't Know JS Chapter.

7 months ago · Juan Pablo Isaza Report

0

In case of addition (+), When a number is added to a string, JavaScript converts the number to a string before concatenation but in case of other arithmetic operations like *, -, / JS engine implicitly convert the string into integer.

Demo :

let result;

// numeric string used with + gives string type

result = '3' + '2'; 
console.log(result, typeof result) // "32", "string"

// numeric string used with - , / , * results number type

result = '3' * '2'; 
console.log(result, typeof result) // 6, "number"

result = '3' - '2'; 
console.log(result, typeof result) // 1, "number"

result = '3' / '2'; 
console.log(result, typeof result) // 1.5, "number"

7 months ago · Juan Pablo Isaza Report
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