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When I make any property of an interface optional, and while assigning its member to some other variable like this:
interface Person {
name?: string,
age?: string,
gender?: string,
occupation?: string,
function getPerson() {
let person = <Person>{name:"John"};
return person;
let person: Person = getPerson();
let name1: string = person.name; // <<< Error here
I get an error like the following:
TS2322: Type 'string | undefined' is not assignable to type 'string'.
Type 'undefined' is not assignable to type 'string'.
How do I get around this error?
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You can now use the non-null assertion operator that is here exactly for your use case.
It tells TypeScript that even though something looks like it could be null, it can trust you that it's not:
let name1:string = person.name!;
// ^ note the exclamation mark here
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I know this is a kinda late, but another way besides yannick's answer to use ! is to cast it as string thus telling TypeScript: I am sure this is a string, thus converting it.
let name1:string = person.name;//<<<Error here
let name1:string = person.name as string;
This will make the error go away, but if by any chance this is not a string you will get a run-time error... which is one of the reasons we are using TypeScript to ensure that the type matches and avoid such errors at compile time.
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As of TypeScript 3.7 you can use nullish coalescing operator ??. You can think of this feature as a way to “fall back” to a default value when dealing with null or undefined
let name1:string = person.name ?? '';
The ?? operator can replace uses of || when trying to use a default value and can be used when dealing with booleans, numbers, etc. where || cannot be used.
As of TypeScript 4 you can use ??= assignment operator as a ??= b which is an alternative to a = a ?? b;
By your definition Person.name can be null but name1 cannot.
there are two scenarios:
Person.name is never null
tell the compiler your are sure the name is not null by using !
let name1: string = person.name!;
Person.name can be null
specify a default value in case name is null
let name1: string = person.name ?? "default name";
A more production-ready way to handle this is to actually ensure that name is present. Assuming this is a minimal example of a larger project that a group of people are involved with, you don't know how getPerson will change in the future.
if (!person.name) {
throw new Error("Unexpected error: Missing name");
let name1: string = person.name;
Alternatively, you can type name1 as string | undefined, and handle cases of undefined further down. However, it's typically better to handle unexpected errors earlier on.
You can also let TypeScript infer the type by omitting the explicit type: let name1 = person.name This will still prevent name1 from being reassigned as a number, for example.
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You were saying to TypeScript it was optional. Nevertheless, when you did:
let name1 : string = person.name; //<<<Error here
You did not leave it a choice. You needed to have a Union on it reflecting the undefined type:
let name1 : string | undefined = person.name; //<<<No error here
Using your answer, I was able to sketch out the following which is basically, an Interface, a Class and an Object. I find this approach simpler, never mind if you don't.
// Interface
interface iPerson {
fname? : string,
age? : number,
gender? : string,
occupation? : string,
get_person?: any
// Class Object
class Person implements iPerson {
fname? : string;
age? : number;
gender? : string;
occupation? : string;
get_person?: any = function () {
return this.fname;
// Object literal
const person1 : Person = {
fname : 'Steve',
age : 8,
gender : 'Male',
occupation : 'IT'
const p_name: string | undefined = person1.fname;
// Object instance
const person2: Person = new Person();
person2.fname = 'Steve';
person2.age = 8;
person2.gender = 'Male';
person2.occupation = 'IT';
// Accessing the object literal (person1) and instance (person2)
console.log('person1 : ', p_name);
console.log('person2 : ', person2.get_person());
try to find out what the actual value is beforehand. If person has a valid name, assign it to name1, else assign undefined.
let name1: string = (person.name) ? person.name : undefined;
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Solution 1: Remove the explicit type definition
Since getPerson already returns a Person with a name, we can use the inferred type.
function getPerson(){
let person = {name:"John"};
return person;
let person = getPerson();
If we were to define person: Person we would lose a piece of information. We know getPerson returns an object with a non-optional property called name, but describing it as Person would bring the optionality back.
Solution 2: Use a more precise definition
type Require<T, K extends keyof T> = T & {
[P in K]-?: T[P]
function getPerson() {
let person = {name:"John"};
return person;
let person: Require<Person, 'name'> = getPerson();
let name1:string = person.name;
Solution 3: Redesign your interface
A shape in which all properties are optional is called a weak type and usually is an indicator of bad design. If we were to make name a required property, your problem goes away.
interface Person {
name:string,
age?:string,
gender?:string,
occupation?:string,
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if you want to have nullable property change your interface to this:
interface Person {
name?:string | null,
age?:string | null,
gender?:string | null,
occupation?:string | null,
if being undefined is not the case you can remove question marks (?) from in front of the property names.
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You trying to set variable name1, witch type set as strict string (it MUST be string) with value from object field name, witch value type set as optional string (it can be string or undefined, because of question sign). If you really need this behavior, you have to change type of name1 like this:
let name1: string | undefined = person.name;
And it'll be ok;
type T0 = NonNullable<string | number | undefined>; // string | number
type T1 = NonNullable<string[] | null | undefined>; // string[]
Docs.
This was the only solution I found to check if an attribute is undefined that does not generate warnings
type NotUndefined<T, K extends keyof T> = T & Record<K, Exclude<T[K], undefined>>;
function checkIfKeyIsDefined<T, K extends keyof T>(item: T, key: K): item is NotUndefined<T, K> {
return typeof item === 'object' && item !== null && typeof item[key] !== 'undefined';
usage:
interface Listing {
order?: string
obj = {..., order: 'pizza'} as Listing
if(checkIfKeyIsDefined(item: obj, 'order')) {
order.toUpperCase() //no undefined warning O.O
original answer
If you remove the <Person> casting from your getPerson function, then TypeScript will be smart enough to detect that you return an object which definitely has a name property.
So just turn:
interface Person {
name?: string,
age?: string,
gender?: string,
occupation?: string,
function getPerson() {
let person = <Person>{name: 'John'};
return person;
let person: Person = getPerson();
let name1: string = person.name;
Into:
interface Person {
name?: string,
age?: string,
gender?: string,
occupation?: string,
function getPerson() {
let person = {name: 'John'};
return person;
let person = getPerson();
let name1: string = person.name;
If you cannot do that, then you will have to use the "definite assignment assertion operator" as @yannick1976 suggested:
let name1: string = person.name!;
I think to use Require as mentioned by Karol Majewski is quite nice. Another way to achieve the same would be to use intersection types(which is actually used internally by Require)
function getPerson(): Person & {name: string} {
const person = {name:"John"};
return person;
const person = getPerson();
const name1: string = person.name;
The advantage of using Require or intersection types is that we don't overrule the typescript compiler as it happens for the non-null assertion operator.
One thing that was happening to me in my particular situation (but it is not related with the specific problem asked here) was that I was importing the wrong type in my file, given that the type was called exactly the same, but defining different properties.
Once I imported the correct type, all the issues dissapeared.
I hope this can help someone that's facing the same as me.
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I find out that react-scrips add "strict": true to tsconfig.json.
After I removed it everything works great.
Need to warn that changing this property means that you:
not being warned about potential run-time errors anymore.
as been pointed out by PaulG in comments! Thank you :)
Use "strict": false only if you fully understand what it affects!
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