Implicit void return in PHP 7.1?

Question

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?

`Is this function also returning implicit void?` - yes – scrowler Nov 14 '16 at 01:01 — scrowler, Nov 14 '16 at 01:01
NULL http://ideone.com/AXErws or https://3v4l.org/RJRoR – Nov 14 '16 at 01:04 — , Nov 14 '16 at 01:04

score 5 · Answer 1 · answered Dec 04 '16 at 15:51

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);
}

So the compile time error will prevent bad practices returning non `void` if the function is `void`? Right? — prosti, Dec 04 '16 at 18:47

score 4 · Accepted Answer · answered Nov 14 '16 at 01:05

4

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?

answered Nov 14 '16 at 01:05

Denis Mysenko

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4

`var_dump` would be better than `echo` for this example – scrowler Nov 14 '16 at 01:12
please also consider the new void type http://stackoverflow.com/questions/29792827/void-as-return-type. This looks like the main course. I was really looking for some debugging, something from the symbol table or PHP AST explanation. But this is also great bacuska. – prosti Nov 16 '16 at 10:13
@RobbieAverill gettype() returns a string, why var_dump() a string? :) – Denis Mysenko Nov 17 '16 at 00:13
Sorry - I posted this when I thought it was `echo lacks_return()`, did you edit it? I may have been dreaming :) – scrowler Nov 17 '16 at 00:14
I originally accepted your answer @DenisMysenko, I realized it is not fair to alter this since I experienced that over time. – prosti Feb 04 '17 at 01:26

Implicit void return in PHP 7.1?

2 Answers2