Why do parentheses affect type narrowing in TypeScript?

Question

In typescript:

let str: string = 'abc';
let val: unknown = 'test';

if (typeof val === 'string') {
    str = val;
} 
// this code does not report any error, everything works fine.

but, if I change the code a little bit:

if ((typeof val) === 'string') { 
    str = val; 
} 
// add the () to hold typeof val;
// error report in typescript in this line: str = val !

TS Playground link

This is really confusing me, who can help to explain what happened here.

Answer 1

TypeScript's typeof type guards walk a fine line. typeof val is a string, and you can do arbitrary string operations to it, but typeof val === "string" is a special construction that narrows the type of val when the expression is true. Consequently, TypeScript is explicitly programmed to match typeof ${reference} ${op} ${literal} and ${literal} ${op} typeof ${reference} (for op = ==, !=, ===, and !==), but typeof ${reference} has no tolerance built in for parentheses (which is a SyntaxKind.ParenthesizedExpression and not a SyntaxKind.TypeOfExpression), string manipulation, or anything else.

TypeScript lead Ryan Cavanaugh describes this in microsoft/TypeScript#42203, "typeof type narrowing acts differently with equivalent parentheses grouping", with gratitude to jcalz for the link:

Narrowings only occur on predefined syntactic patterns, and this isn't one of them. I could see wanting to add parens here for clarity, though -- we should detect this one too.

It sounds like this is a candidate for a future fix, though even if the pattern were added, you would still be somewhat limited in the complexity of typeof expressions that work as type guards.

---

From compiler source microsoft/TypeScript main/src/compiler/checker.ts, comments mine:

function narrowTypeByBinaryExpression(type: Type, expr: BinaryExpression, assumeTrue: boolean): Type {
  switch (expr.operatorToken.kind) {
    // ...
    case SyntaxKind.EqualsEqualsToken:
    case SyntaxKind.ExclamationEqualsToken:
    case SyntaxKind.EqualsEqualsEqualsToken:
    case SyntaxKind.ExclamationEqualsEqualsToken:
        const operator = expr.operatorToken.kind;
        const left = getReferenceCandidate(expr.left);
        const right = getReferenceCandidate(expr.right);
        // Check that the left is typeof and the right is a string literal...
        if (left.kind === SyntaxKind.TypeOfExpression && isStringLiteralLike(right)) {
            return narrowTypeByTypeof(type, left as TypeOfExpression, operator, right, assumeTrue);
        }
        // ...or the opposite...
        if (right.kind === SyntaxKind.TypeOfExpression && isStringLiteralLike(left)) {
            return narrowTypeByTypeof(type, right as TypeOfExpression, operator, left, assumeTrue);
        }
        // ...or skip it and move on. Don't bother trying to remove parentheses
        // or doing anything else clever to try to make arbitrary expressions work.
        if (isMatchingReference(reference, left)) {
            return narrowTypeByEquality(type, operator, right, assumeTrue);
        }
        if (isMatchingReference(reference, right)) {
            return narrowTypeByEquality(type, operator, left, assumeTrue);
        }
        // ...
  }
  return type;
}