After activating the 'noImplicitAny' feature in the typescript playground, I proceeded to input the following code:
async function a () {
const b = JSON.parse('{"a":"x"}');
console.log(b)
}
Upon inspecting the second line where "b" is referenced, I noticed that its type was inferred as any. Despite this, there was no error generated. Could it be that my understanding of the purpose of 'noImplicitAny' is flawed, or is this possibly a glitch in the system?
When using the --noImplicitAny compiler option, warnings are generated only where type inference is unsuccessful, leading to a fallback or default to any. For instance, an implicit-any error occurs when a function parameter cannot be contextually typed:
const f = (arg) => arg + 1; // error!
// ------> ~~~ implicit any
This error also appears if a variable that should be auto-typed (microsoft/TypeScript#11263) cannot be inferred by the compiler through control flow, like in cases where the variable is referenced in a separate function scope:
let w; // error!
// >~ implicit any
w = 2;
function foo() { w } // implicit any
In these instances, the compiler indicates uncertainty by assigning an any type.
In contrast, calling a function with a return type of any results in a value of type any. While this may seem like an "implicit" assignment, it does not stem from a failure of type inference. When an assigned variable is successfully inferred as any, similar to how x being a number in const x = 1 + 2 demonstrates successful inference, there are no implicit-any errors present.
Moreover, since the TypeScript library typings for JSON.parse() appear as:
interface JSON {
parse(text: string, reviver?: (this: any, k: string, v: any) => any): any;
}
the code snippet:
const b = JSON.parse('{"a":"x"}');
successfully infers any for the type of b without triggering a compiler warning.
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