Types: Type Inferencing

While certain kinds of variables must have their type declared explicitly, others can have their type inferred by having the implementation look at the context in which those variables are used. For example:

function foo(int $i): void {
    $v = 100;

As we can see, $v is implicitly typed as int, and $i is explicitly typed.

In Hack,

  • Types must be declared for properties and for the parameters and the return type of named functions.
  • Types must be inferred for local variables, which includes function statics and parameters.
  • Types can be declared or inferred for constants and for the parameters and return type of unnamed functions.

The process of type inferencing does not cross function boundaries.

Here's an example involving a local variable:

function f(): void {
  $v = 'acb';       // $v has type string
  // ...
  $v = true;        // $v has type bool
  // ...
  $v = dict['red' => 10, 'green' => 15]; // $v has type dict<string, int>
  // ...
  $v = new C();     // $v has type C

For each assignment, the type of $v is inferred from the type of the expression on the right-hand side, as shown in the comments. The type of function statics is inferred in the same manner, as are function parameters. For example:

function g(int $p1 = -1): void {
  // on entry to the function, $p1 has the declared type int
  // ...
  $p1 = 23.56;      // $p1 has type float
  // ...

As a parameter, $p1 is required to have a declared type, in this case, int. However, when used as an expression, $p1's type can change, as shown.

In the case of a class constant, if the type is omitted, it is inferred from the initializer:

class C {
  const C1 = 10;            // type int inferred from initializer
  const string C2 = "red";  // type string declared

Let's consider types in closures:

$doubler = $p ==> $p * 2;

The type of the parameter $p and the function's return type have been omitted. These types are inferred each time the anonymous function is called through the variable $doubler. When 3 is passed, as that has type int, that is inferred as the type of $p. The literal 2 also has type int, so the type of the value returned is the type of $p * 2, which is int, and that becomes the function's return type.

We can add partial explicit type information; the following all result in the same behavior:

$doubler = (int $p) ==> $p * 2;
$doubler = ($p = 0) ==> $p * 2;
$doubler = ($p): int ==> $p * 2;

In the first case, as $p has the declared type int, and int * int gives int, the return type is inferred as int. In the second case, as the default value 0 has type int, $p is inferred to also have that type, and int * int gives int, so the return type is inferred as int. In the third case, as the return type is declared as int, and $p * 2 must have that type, the type of $p is inferred as int, so that must also be the type of the parameter.

While all three of these cases allow a call such as $doubler(3), none of them allow a call such as $doubler(4.2). So, the fact that type information can be provided explicitly in these cases doesn't mean it's necessarily a good idea to do so.

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