The contrast among the outcomes of each example unveils how the 'Generic Type' functions and what distinguishes using 'extends' independently?
type NameOrId<T> = T extends number ? 'add' : 'plus';
type foo = string | number;
type x = NameOrId<foo>
// x is 'add' | 'plus'
type bar = foo extends number ? 'add' : 'plus'
// bar is 'plus'
type zoo = foo extends number | string ? 'add' : 'plus'
// zoo is 'add'
The concept is detailed in the TypeScript documentation under Conditional Types:
When conditional types interact with a generic type and are presented with a union type, they exhibit distributive behavior.
The documentation also provides insights on how to prevent this distribution:
Typically, distributing behavior is expected. To circumvent it, you can encapsulate each side of the extends keyword in square brackets.
As illustrated:
// By enclosing T in square brackets, you avoid the distribution on the union type's members
type NameOrId<T> = [T] extends number ? 'add' : 'plus';
type y = NameOrId<foo>;
// The output for y is 'plus'
You can experiment with it online.
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