Issues with the "noImplicitAny" setting in TS compilation

Question

Issues with the "noImplicitAny" setting in TS compilation

Having the following code snippet:

let z;
z = 50;
z = 'z';

paired with the configuration in my tsconfig.json file:

{
  "compilerOptions": {
    "target": "es5",
    "module": "commonjs",
    "sourceMap": false,
    "noEmitOnError": true,
    "strict": true,
    "noImplicitAny": true
  }
}

Why am I not getting any exceptions during the compilation process to JavaScript?

Warm regards, Crova

javascript typescript tsconfig

Answer 1

Answer №1

Since the variable z is never explicitly declared as any, its type is determined by the value assigned to it.

According to the release notes:

With TypeScript 2.1, rather than defaulting to the any type, TypeScript will now infer types based on the assigned values.

For example:

let x;

// You can assign any value to 'x'.
x = () => 42;

// TypeScript 2.1 infers that 'x' has type '() => number' after the assignment.
let y = x();

// As a result, it will correctly flag an error when you try to add a number to a function.
console.log(x + y);
//          ~~~~~
// Error! Operator '+' cannot be used on types '() => number' and 'number'.

// TypeScript still allows assigning any value to 'x'.
x = "Hello world!";

// It now knows that 'x' is a 'string'.
x.toLowerCase();

Therefore, in your scenario:

let z;
z = 50;
let y = z * 10; // `z` is treated as a number, no error
z = 'z';
z.replace("z", "")// `z` is considered a string, no error

Answer 2

Since the variable z is never explicitly declared as any, its type is determined by the value assigned to it.

According to the release notes:

With TypeScript 2.1, rather than defaulting to the any type, TypeScript will now infer types based on the assigned values.

For example:

let x;

// You can assign any value to 'x'.
x = () => 42;

// TypeScript 2.1 infers that 'x' has type '() => number' after the assignment.
let y = x();

// As a result, it will correctly flag an error when you try to add a number to a function.
console.log(x + y);
//          ~~~~~
// Error! Operator '+' cannot be used on types '() => number' and 'number'.

// TypeScript still allows assigning any value to 'x'.
x = "Hello world!";

// It now knows that 'x' is a 'string'.
x.toLowerCase();

Therefore, in your scenario:

let z;
z = 50;
let y = z * 10; // `z` is treated as a number, no error
z = 'z';
z.replace("z", "")// `z` is considered a string, no error

Answer 3

Answer №2

noImplicitAny is a TypeScript setting that requires the compiler to display an error whenever it cannot infer a specific type and defaults to using 'any'.

This setting ensures that TypeScript avoids using 'any' when determining types in your code

For instance, in the scenario mentioned above, the compiler can easily determine the type of z, allowing it to verify if the methods or properties called on z are permitted or not.

Answer 4

noImplicitAny is a TypeScript setting that requires the compiler to display an error whenever it cannot infer a specific type and defaults to using 'any'.

This setting ensures that TypeScript avoids using 'any' when determining types in your code

For instance, in the scenario mentioned above, the compiler can easily determine the type of z, allowing it to verify if the methods or properties called on z are permitted or not.

Issues with the "noImplicitAny" setting in TS compilation

Answer №1

Answer №2

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