Question

When you have noImplicityAny true, typescript will show error, the value implicitly has an any type Why to use noImplicitAny?

a.true
b. false

Answer 1

The noImplicitAny flag in TypeScript, when set to true, results in compiler errors for variables with an implicit 'any' type, promoting type safety and code maintainability.

Step-by-step explanation:

When the noImplicitAny flag is set to true in TypeScript, the compiler will show an error if a variable's type is implicitly inferred as any. The reason to use noImplicitAny is to encourage better type safety in your code. By avoiding the any type, developers are forced to explicitly declare variable types or to provide appropriate type annotations, which can prevent runtime errors and improve code maintainability.

Using noImplicitAny helps in catching potential bugs during compile time rather than at runtime. It also makes the code more readable and easier to understand, as the types of variables are clear. Therefore, the answer to the question is true.