PHP 7.1中的隐式void返回？

I found in here the new spec: https://wiki.php.net/rfc/void_return_type

function lacks_return(): void {
    // valid
}
function returns_nothing(): void {
    return; // valid
}
function returns_void(): void {
    return void; // valid
}

Ask: Do you know what happens behind the scene. Will the lacks_return function return actually void?

You could have tested this yourself pretty easily:

function lacks_return(): void {
}

function returns_nothing(): void {
    return;
}

echo gettype(lacks_return()); // NULL
echo gettype(returns_nothing()); // NULL

So the answer is yes - there is an implicit empty (null) return so you could either use an empty return or skip it completely. Which kind of makes sense - returning nothing is the same as not returning anything?

Behind the scenes, PHP checks for return statements in void functions and, if they specify a return value, throws a compile-time error:

/* `return ...;` is illegal in a void function (but `return;` isn't) */
if (return_info->type_hint == IS_VOID) {
    if (expr) {
        if (expr->op_type == IS_CONST && Z_TYPE(expr->u.constant) == IS_NULL) {
            zend_error_noreturn(E_COMPILE_ERROR,
                "A void function must not return a value "
                "(did you mean \"return;\" instead of \"return null;\"?)");
        } else {
            zend_error_noreturn(E_COMPILE_ERROR, "A void function must not return a value");
        }
    }
    /* we don't need run-time check */
    return;
}

Otherwise, compilation of void functions works like normal. return without a value implicitly returns NULL:

if (!expr_ast) {
    expr_node.op_type = IS_CONST;
    ZVAL_NULL(&expr_node.u.constant);

And every function is compiled with an implicit return at the end:

zend_emit_final_return(0);

Whose return value is NULL:

zn.op_type = IS_CONST;
if (return_one) {
    ZVAL_LONG(&zn.u.constant, 1);
} else {
    ZVAL_NULL(&zn.u.constant);
}