When taking values from an object, does unshift() change the source object?

Question

This question sparked when I was setting the base case of the following code:

function withoutReverse(str, arrayFor = []){
    arrayFor.unshift(str[0]);
    if (str.length === 1) 
    return arrayFor.join("");
    else return withoutReverse(str.slice(1), arrayFor)  }

  let hola = "hello";  
  withoutReverse(hola);//-->olleh

I thought that the base should be str.length === 0 , cause I assumed that unshift literally took each element from the source object to put it into arrayFor . Then, I realized that that was what I was using slice for.

I didn't found a conclusive information in mdn. Am I right? unshift() doesn't really take those values (when stored in an object)? if so, how is that handled in memory?

Thank you

Answer 1

When you get to the 5th call:

console.log(str) // "o"
arrayFor.unshift(str[0]); // str[0] === "o" 
console.log(str.length) // "1" because accessing the property here does not change it.

Therefore your exit condition is now true.

For functions in JS:

• numbers/strings/booleans are passed as values
• arrays/objects are passed as references

Extra info about strings:

Unlike some programming languages (such as C), JavaScript strings are immutable. This means that once a string is created, it is not possible to modify it.

Source MDN

return 'cat'[1] // returns "a"

When using bracket notation for character access, attempting to delete or assign a value to these properties will not succeed. The properties involved are neither writable nor configurable. (See Object.defineProperty() for more information.)

Source MDN