<?php
function sayHello(string $name): never
{
echo "Hello, $name";
exit(); // if we comment this line, php throws fatal error
}
sayHello("John"); // result: "Hello, John"
never es únicamente un tipo de retorno que indica que la función no se termina. Esto significa que llama a exit(), lanza una excepción, o es un bucle infinito. Por lo tanto, no puede formar parte de una declaración de type union Disponible a partir de PHP 8.1.0.
never es, en el lenguaje de la teoría de tipos, el tipo vacío. Esto significa que es el subtipo de todos los otros tipos y que puede reemplazar cualquier otro tipo de retorno durante la herencia.
<?php
function sayHello(string $name): never
{
echo "Hello, $name";
exit(); // if we comment this line, php throws fatal error
}
sayHello("John"); // result: "Hello, John"I think the description should be corrected from return-only to non-return function since the context is now misleadingNever cannot be used in a union type because, being the bottom type, it is already automatically a subtype of every other type. "A|never" is equivalent to "A".
When one type is "obviously" a subtype of another (i.e., it doesn't require loading the class definitions of all the types involved), the former is redundant in union types, and PHP flags the union of both as an error.
Similarly for intersection types, where "A&never" means the same thing as "never". It "obviously" doesn't make sense to mention A there, so PHP won't allow doing so.