The mixed type of PHP 8
Up until now, or precisely before PHP 8, whenever you were unsure about which type to use for the propperty or return type, you’d leave it without assigning any type. And if you’re using an IDE such as PhpStorm, the docBlock would mark those properties as mixed type like so.
/**
* Handle an incoming request.
*
* @param mixed $request
* @return mixed
*/
public function handle($request)
{
// processing request
return $next($request);
}
This is fine as long as you want to give some context as to what would be the type of that property but it essentially does not really do type check by the PHP compiler.
So, to mitigate this issue, PHP 8 has introduced a dedicated mixed type that can be used to assign a mixed type to the class properties and return type of class methods.
Essentially, A type of mixed would be equivalent to array or bool or callable or int or float or null or object or resource or string. So, whenever you would be unsure about the type of the property, you would use mixed for the same. The same goes for the return type as well.
Now, the above example could be rewritten with mixed type like so.
/**
* Handle an incoming request.
*
* @param mixed $request
* @return mixed
*/
public function handle(mixed $request): mixed
{
// processing request
return $next($request);
}
As you can see, it’s now more explicit and now PHP’s compiler know what to do with these properties.
The introduction of mixed types comes with a certain set of rules that one needs to conform to.
Mixed types can’t be nullable
As the mixed type also include null type, it would be ridiculous to make it nullable as it would only add duplication of information as ?mixed would be always be equivalent to be mixed. A fatal error would be thrown if mixed is used as a nullable.
function foo(?mixed $arg) {}
// Fatal error: Mixed types cannot be nullable, null is already part of the mixed type.
function bar(): ?mixed {}
// Fatal error: Mixed types cannot be nullable, null is already part of the mixed type.
Explicit return for mixed type
When using mixed as a return type, a value must be explicitly returned from the function, otherwise a TypeError will be thrown.
function foo(): mixed {}
foo();
// Uncaught TypeError: Return value of foo() must be of
// the type mixed, none returned
Property types can be “widen” in inheritance
This essentially means if a type of the property is anything from array or bool or callable or int or float or null or object or resource or string in base class, it can be widen by using mixed type in the child class. So, the following is valid.
class A
{
public function foo(int $value) {}
}
class B extends A
{
// Parameter type was widened from int to mixed, this is allowed
public function foo(mixed $value) {}
}
But the vice-versa would be invalid. That is one can not “narrow down” the property type in the child class. So, following would be invalid.
class A
{
public function foo(mixed $value) {}
}
class B extends A
{
// Parameter type cannot be narrowed from mixed to int
// Fatal error thrown
public function foo(int $value) {}
}
Return types can be “narrowed” in inheritance
he mixed return type could be narrowed in a subclass as this is covariant. So, the following would be valid.
class A
{
public function bar(): mixed {}
}
class B extends A
{
// return type was narrowed from mixed to int, this is allowed
public function bar(): int {}
}
But the vice-versa, i.e. a return type can not be “widen” in the child class. So, the following would be invalid.
class C
{
public function bar(): int {}
}
class D extends C
{
// return type cannot be widened from int to mixed
// Fatal error thrown
public function bar(): mixed {}
}
Assume mixed type when no type is specified
When no type is present for a function parameter, the signature checks for inheritance are done as if the parameter had a mixed type. So, you can use mixed type for the property in inheritance or leave it without any type whatsoever.
class A
{
// no type is specified, mixed type is assumed
public function foo($value) {}
}
class B extends A
{
// mixed type is explicitly specified, and is invariant to
// type in parent class
public function foo(mixed $value) {}
}
class C extends B
{
// no type is specified, mixed type is assumed which is
// invariant to type in parent class
public function foo($value) {}
}
class D extends B
{
public function foo(mixed $value = null) {}
}
Signature checking of function when no return type present
When no type is present for a function return, the signature checks for inheritance are done as if the parameter had a mixed or void type. So, the following is valid.
class A
{
// no return type is specified, mixed|void is assumed
public function foo() {}
}
class B extends A
{
// mixed type is explicitly specified. The type 'mixed' is
// covariant to 'mixed|void' and so is allowed to be declared
// for this function.
public function foo(): mixed {}
}
class C extends B
{
// INVALID - no type is specified, mixed|void is assumed.
// 'mixed|void' is not covariant to 'mixed' and so this breaks LSP.
// Fatal error is thrown
public function foo() {}
}
But void can’t be used if mixed used as a return type in base class. So, the following would be invalid.
class A
{
public function foo(): mixed {}
}
class B extends A
{
// INVALID - as void is not subtype of mixed, Fatal error is thrown
public function foo(): void {}
}
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