TypeScript tries to find the best common type when a type inference is made from several expressions.
Let’s experiment a bit with following types. Square
is a subtype of Shape
and Circle
is also a subtype of Shape
.
class Shape { color: string; } class Square extends Shape { sideLength: number; } class Circle extends Shape { radius: number; }
The type of xs1
is Shape[]
because Shape
is the common supertype of Shape
, Square
and Circle
. However, surprisingly the type of xs2
is not Shape[]
though Shape
is the common supertype of Square
and Circle
. Here the type of xs2
is (Circle | Square)[]
because there is no object that is strictly of type Shape
in the array.
var xs1 = [new Shape(), new Square(), new Circle()]; var xs2 = [new Square(), new Circle()];
You can override this type by annotating the type explicitly with Shape[]
.
var xs3: Shape[] = [new Square(), new Circle()];
Now let’s see how the return type of a function is inferred when the function returns two values with different types.
function createShape1(square: boolean) { if (square) { return new Square(); } else { return new Shape();; } }
The return type of createShape1
function is Shape
as expected because Shape
is the common supertype of Square
and Shape
.
function createShape2(square: boolean) { // Type Error if (square) { return new Square(); } else { return new Circle(); } }
However, if we change the return value of else branch to new Circle()
, createShape2
function no longer type checks. TypeScript complains about “error TS2354: No best common type exists among return expressions.” It fails to find the best common type though it can be inferred as either Square|Circle
or Shape
.
We can satisfy the type checker by giving an explicit type. The return type can be either Square | Circle
or Shape
.
function createShape3(square: boolean): Square | Circle { if (square) { return new Square(); } else { return new Circle(); } } function createShape4(square: boolean): Shape { if (square) { return new Square(); } else { return new Circle(); } }
From the examples above, we can see that the type inference rule of TypeScript is not orthogonal. When you mix multiple types in an array, TypeScript finds the best common type that is the union of the types of the element expressions. But when a function mixes multiple return types in return values, it finds the best common type that is the first supertype of each of the others.
You can see the difference in the type inference rules taken from the TypeScript 1.4 specification.
Array Type
- If the array literal is empty, the resulting type is an array type with the element type Undefined.
- Otherwise, if the array literal is contextually typed by a type that has a property with the numeric name ‘0’, the resulting type is a tuple type constructed from the types of the element expressions.
- Otherwise, the resulting type is an array type with an element type that is the union of the types of the element expressions.
Function Return Type
- If there are no return statements with expressions in f’s function body, the inferred return type is Void.
- Otherwise, if f’s function body directly references f or references any implicitly typed functions that through this same analysis reference f, the inferred return type is Any.
- Otherwise, if f is a contextually typed function expression (section 4.9.3), the inferred return type is the union type (section 3.4) of the types of the return statement expressions in the function body, ignoring return statements with no expressions.
- Otherwise, the inferred return type is the first of the types of the return statement expressions in the function body that is a supertype (section 3.10.3) of each of the others, ignoring return statements with no expressions. A compile-time error occurs if no return statement expression has a type that is a supertype of each of the others.