As I delve into learning Typescript, a question arises in my mind. I find myself pondering the purpose of the any type. It seems redundant to specify it when every variable essentially acts as an "any" type by default.
Consider this scenario where the output will always be 'yes' regardless of the values of x and y:
var x = "d";
var y: any = 5;
alert(y);
y = x;
alert(y);
if (x === y)
alert("yes");
else
alert("no");
The any type in Typescript serves as a workaround when there is no clear way to specify what you are doing, allowing you to bypass type checking temporarily.
It's important to note that types may not default to any if type annotations are omitted. Typescript will often infer a type and enforce it. For example, declaring let x = 2; will automatically infer the type of x as number, leading to an error if you later assign x = "cat";.
In certain scenarios where explicit typing is necessary, such as with generic types, using any becomes essential:
type Pair<A, B> = [A, B];
declare function needsSomethingPairedWithString(value: Pair<any, string>): void;
When defining functions like needsSomethingPairedWithString, specifying only the required type while leaving room for flexibility can be achieved by using Pair<any, string>.
While unknown is preferable to any for cases where information is both unknown and irrelevant, any remains appropriate for handling constraints effectively.
Consider enabling the --noImplicitAny flag in the compiler settings to identify instances where any is inferred and potentially lead to errors. This practice encourages explicit declaration of any, promoting code readability and safety across projects implementing Typescript from the start.
Assuming all variables are of type "any", unless specified
While it is true that in TypeScript, if no type is specified for a variable it defaults to any, using the flag --noImplicitAny changes this behavior. In such cases:
function foo(x) {
// x is considered as any
}
In --noImplicitAny mode, the above code snippet would trigger an error. If you explicitly want x to be of type any, you must declare it as such:
function foo(x: any) {
// x is now specifically defined as any
}
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