1

In one of my procedures I need to periodically reset a custom type, but I noticed that in TypeScript (and I think in JavaScript too maybe), when I assign a variable of a custom type to another variable of the same type, the assignment is made by reference and not by value.

For example this code:

type testing = {
    a: string;
    b: number;
};

let v1: testing = { a: "one", b: 1 };
let v2: testing = { a: "two", b: 2 };

console.log(v1);
console.log(v2);

v1 = v2;
v1.a = "Edited";

console.log(v1);
console.log(v2);

generates this output

{ a: 'one', b: 1 }
{ a: 'two', b: 2 }
{ a: 'Edited', b: 2 }
{ a: 'Edited', b: 2 }

Is there a way to assign it by value without assigning every property of the type?

(in my example I need my v2 variable to remain equals to { a: "two", b: 2 })

Mark Rotteveel
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ArkoMo93
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1 Answers1

2

For simple cases like this, you can make use of the spread operator to create a shallow copy of an object.

type testing = {
    a: string;
    b: number;
};

let v1: testing = { a: "one", b: 1 };
let v2: testing = { a: "two", b: 2 };

console.log(v1);
console.log(v2);
// copy all of the top-level keys and values from v2 
// into a new object and assign that to v1
v1 = { ...v2 }; 
v1.a = "Edited";

console.log(v1);
console.log(v2);

Note there are some caveats here. As indicated by the link in Tobias S' comment, this will only do a shallow copy. If you have more complex objects, this will cause issues.

let v1: testing = { a: "one", b: 1, c: { d: 1 } };
let v2: testing = { a: "two", b: 2, c: { d: 2 } };

v2 = { ...v1 };
// v1.c is the same object as v2.c because we only shallowly
// copied, so this assignment is reflected in both
v2.c.d = 60;

console.log(v1);
console.log(v2);
CollinD
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