TS2365: The '!== 'operator is not compatible with the types ""("" and "")""

Question

TS2365: The '!== 'operator is not compatible with the types ""("" and "")""

function myFunction(identifier: string) {
  identifier = "(";
};

let identifier: string = ")";
if (identifier !== '(')
  throw "Expected '(' in function";

myFunction(identifier);

if (identifier !== ')')
  throw "Expected ')' in function";

If I make the change directly within the code, the error mentioned below does not occur

error TS2365: Operator '!==' cannot be used with types '"("' and '")"'.

The actual strings being compared are interchangeable. Any arbitrary strings like a, basdasd, cqwe, etc. can replace ( and )

typescript

Answer 1

Answer №1

When the code runs through the first if statement, a process known as type narrowing occurs. This narrows the type of id to the string literal ")" in order to prevent an exception from being thrown.

function myFunc(id: string) {
  id = "(";
};
// It's important to note that this function doesn't actually do anything significant

let id: string = ")";

if (id !== '(')
  throw "Expected '(' in prototype";
// The type of id is narrowed to "(" at this point

myFunc(id);

// As a result, id and ')' no longer have the same type, leading to an error
if (id !== ')')
  throw "Expected ')' in prototype";

Answer 2

When the code runs through the first if statement, a process known as type narrowing occurs. This narrows the type of id to the string literal ")" in order to prevent an exception from being thrown.

function myFunc(id: string) {
  id = "(";
};
// It's important to note that this function doesn't actually do anything significant

let id: string = ")";

if (id !== '(')
  throw "Expected '(' in prototype";
// The type of id is narrowed to "(" at this point

myFunc(id);

// As a result, id and ')' no longer have the same type, leading to an error
if (id !== ')')
  throw "Expected ')' in prototype";

TS2365: The '!== 'operator is not compatible with the types ""("" and "")""

Answer №1

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