passing a variable as a parameter and not effecting it

Question

some times in typescript when i pass a variable as a parameter in a function and modify it in the function the original variable also modify can any one explain to me what's happening

let mybrew :any[] = [
    {
        id:5,
        materials:[-1,-1,-1,-1],
        price:16
        },

]
console.log(mybrew);
function bestBrew(abrow:any[]){
    abrow.forEach(
        e=>{
            for (let i = 0; i < e.materials.length; i++) {
            e.materials[0] += e.materials[i];
            }
        }
    )
}
bestBrew(mybrew);
console.log(mybrew);

output

[ { id: 5, materials: [ -1, -1, -1, -1 ], price: 16 } ]
[ { id: 5, materials: [ -5, -1, -1, -1 ], price: 16 } ]

i don't want the variable in the global scoop (mybrew) to change — Ëł Řęÿ NôùrËddîne Tôùzânî, Nov 17 '20 at 09:47
So if you do not want the variable in the global scope to change you have to clone the parameter's value that you are passing into the function. — Shyam Pillai, Nov 17 '20 at 09:52
Since it's an array that you want to pass in, just do `newBrew = Array.from(abrow)` and pass in `newBrew` . Read about pass by value and pass by reference. That'll help. ` — Shyam Pillai, Nov 17 '20 at 09:55

score 1 · Answer 1 · answered Nov 17 '20 at 09:46

1

You can do some deep clone on the object.

lent newbrew = JSON.parse(JSON.stringify(mybrew));
bestBrew(newbrew);

answered Nov 17 '20 at 09:46

York Chen

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passing a variable as a parameter and not effecting it

1 Answers1