Imagine having the given class and code snippet:
class myClass{
x: number;
}
const obj = new myClass();
obj.y = 7; // produces a compile error. Property 'y' does not exist on type myClass.
obj['y'] = 7; // compiles without any issues.
What sets apart using dot notation from utilizing square brackets? Are they essentially interchangeable?
When using square brackets like o['b'], the o object is treated as an any type in Typescript, which means no type checks will be applied.
To receive warnings, you can set the noImplicitAny: true attribute in your tsconfig.json file. This will result in an error stating
Element implicitly has an 'any' type because type 'MyClass' has no index signature.o['b']. To work around this issue, you must explicitly declare that o should be treated as an any type like so: (o as any)['b'].
An exception to this rule is if you define an attribute in MyClass that already exists. For example:
export class MyClass{
a: number;
"Weirdly-Named-Attribute": number;
}
const o = new MyClass();
o["a"] = 4; // compiles
o["Weirdly-Named-Attribute"] = 5; // compiles
o["b"] = 6; // Gives `Element implicitly has an 'any' type` error if `noImplicityAny` is enabled in tsconfig.
o.b is interpreted by the compiler as specifying: "I am initializing this attribute (b in this case) of the MyClass class." This results in a compilation error.
On the other hand, o['b'] creates a new attribute b within the instance o belonging to the type MyClass, allowing it to compile successfully.
Essentially, the distinction between using a dot and square brackets lies in the fact that the dot attempts to access the desired property of the instance, while square brackets will either assign a value to an existing property within the class or create the attribute and assign the value if it does not already exist.
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