Parameter X implicitly has an 'any' type

This error occurs when you fail to add a type annotation to a parameter in a function.

const addTwoNumbers = (a, b) => {
Parameter 'b' implicitly has an 'any' type.7006
Parameter 'b' implicitly has an 'any' type.
Parameter 'a' implicitly has an 'any' type.7006
Parameter 'a' implicitly has an 'any' type.  return a + b;
#
Why Is It Happening?
This is happening because of a setting that most TypeScript projects have turned on - strict mode.
// tsconfig.json
  "compilerOptions": {
    "strict": true
Strict mode tells TypeScript to add more checks to your code to make sure it's safe. One of these checks is to make sure that all parameters have a type annotation.
Pretty much all modern TypeScript projects use strict mode, so you'll see this error a lot.
#
Why Is This Error Part Of Strict Mode?
TypeScript can't infer the type of a parameter if you don't give it a type annotation. In the example above, TypeScript can't infer from usage what the type of a and b should be.
Adding two parameters together could be a number, sure. But it could be that you want to concatenate two strings together, which is also possible with the + operator.
const concatTwoStrings = (a, b) => {
Parameter 'b' implicitly has an 'any' type.7006
Parameter 'b' implicitly has an 'any' type.
Parameter 'a' implicitly has an 'any' type.7006
Parameter 'a' implicitly has an 'any' type.  return a + b;
So, in strict mode TypeScript demands that you add a type annotation to all parameters. This is to make sure your function is called correctly:
const addTwoNumbers = (a: number, b: number) => {
  return a + b;
addTwoNumbers(1, 2); // OK
addTwoNumbers("1", "2"); // Error!
Argument of type 'string' is not assignable to parameter of type 'number'.2345
Argument of type 'string' is not assignable to parameter of type 'number'.
#
What Is An Implicit Any?
If you don't have strict mode turned on, TypeScript will automatically assign the any type to parameters that don't have a type annotation.
const addTwoNumbers = (a, b) => {
  return a + b;
const result = addTwoNumbers(1, 2);
        
const result: any

This is bad, because any disables type checking on anything it's applied to. We can begin really messing up if we're not careful:
const addTwoNumbers = (a, b) => {
  return a + b;