Let's say I am working on creating an array of type CoolObject. What would be the better approach if some objects have the property format, while others do not?
// Option 1
export interface CoolObject {
name: string;
color: string;
format? string;
}
or
// Option 2
export interface CoolObject {
name: string;
color: string;
format: string;
}
With the second option, I can easily check whether the format property has a value or not.
The alternate approach informs the compiler that the property will always exist, making it a poor choice for indicating optionality. To convey that a property is optional, one should use the '?' symbol:
export interface CoolObject {
name: string;
color: string;
format? string;
}
If you enable strictNullChecks in TypeScript, the compiler will require you to verify that the property is defined before accessing it.
When the property
formatis not included in every object, you can implement the following solution:
export interface CoolObject {
name : string;
color : string;
format ?: string;
}
Just like our colleagues mentioned,
export interface CoolObject {
name : string;
color : string;
format ?: string;
}
is the way to go that makes the most sense.
If it's a highly demanding application in terms of performance, I would take it a step further with
export interface CoolObject {
name : string;
color : string;
format : string;
}
and assign format = undefined if it's not utilized.
The reason being, the Javascript Compiler optimizes for "equal structured Objects." For example, the objects:
{
name: 'LALALA'
color: 'blue'
}
{
name: 'James Blond'
color: 'rose'
}
are structurally equal.
{
name: 'LALALA'
color: 'blue'
}
{
name: 'James Blond'
color: 'rose'
format: 'lowlevel'
}
are not.
However, we are discussing optimizations that are necessary for very few specific Use Cases.
That's why I recommend sticking with format ?: string, as it is much easier to implement and understand.
Best regards
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