Keep original array after JavaScript function [duplicate]

Question

I'm sure there is an easy answer for this, though I only really have experience with PHP. Why does the "pizza" array change when I pass it to my function as "my_pizza" and only make changes to "my_pizza"? How do I keep my original array I pass to the function outside of the function? Once the function is finished running, the pizza array should have not changed. I noticed if I change my string variable (pie) it will stay the same after the function runs, unlike the array.

In short, I want the first set of results to be identical to the second.

var pizza = [];
pizza.push('crust');
pizza.push('ham');

var pie = "apple"

function bake_goods(my_pizza, my_pie){
    console.log(my_pizza);
    console.log(my_pie);

    delete my_pizza['1'];
    my_pie = "peach";

    console.log(my_pizza);
    console.log(my_pie);
}

//first run

bake_goods(pizza, pie);
//console logs
//['crust','ham']
//apple

//['crust']
//peach

//second run

bake_goods(pizza, pie);
//console logs
//['crust']
//apple

//['crust']
//peach

Answer 1

you should clone (create a copy of) your array in your function

function bake_goods(my_pizza, my_pie){
    var innerPizza = my_pizza.slice(0);
    console.log(innerPizza);
    console.log(my_pie);

    delete innerPizza ['1'];
    my_pie = "peach";

    console.log(innerPizza );
    console.log(my_pie);
}

Answer 2

Arrays and objects are passed as pointers to the original object. If you don't want to modify the original, you need to make a copy first.

function bake_goods(my_pizza, my_pie) {
    my_pizza = my_pizza.slice(0);
    delete my_pizza[1];
}

Answer 3

The "pizza" array changes because the function is apparently call by reference. The function manipulates the parameters through the same location in memory that the variable outside the function is initialized in. You can avoid these unwanted changes by creating a copy of the my_pizza array and its elements, and working with that.

Answer 4

use this, to force javascript to edit the object in the local context and not in the global context, also youll need to clone objects.

function bake_goods(my_pizza, my_pie){
    this.my_pizza = my_pizza.slice(0);

    console.log(this.my_pizza);
    console.log(this.my_pie);

    delete this.my_pizza['1'];
    this.my_pie = "peach";

    console.log(this.my_pizza);
    console.log(this.my_pie);
}