How to resolve "must be an instance of string, string given" prior to PHP 7?

Question

Here is my code:

function phpwtf(string $s) {
    echo "$s\n";
}
phpwtf("Type hinting is da bomb");

Which results in this error:

Catchable fatal error: Argument 1 passed to phpwtf() must be an instance of string, string given

It's more than a little Orwellian to see PHP recognize and reject the desired type in the same breath. There are five lights, damn it.

What is the equivalent of type hinting for strings in PHP? Bonus consideration to the answer that explains exactly what is going on here.

Answer 1

Prior to PHP 7 type hinting can only be used to force the types of objects and arrays. Scalar types was not type-hintable. In this case an object of the class string is expected, but you're giving it a (scalar) string. The error message may be funny, but it's not supposed to work to begin with. Given the dynamic typing system, this actually makes some sort of perverted sense.

Your best bet is to upgrade the PHP version, because even PHP7 is long gone now.

In case it's impossible, you can simply remove that typehint altogether.

If you still want to check the parameter type, you can do it manually:

function foo($string) {
    if (!is_string($string)) {
        throw new RuntimeException(
            'Argument must be of type string '.gettype($type).' given'
        );
    }
    ...
}

Answer 2

From the PHP manual at the time of writing this answer:

Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn't supported.

So you have it. The error message is not really helpful, I give you that though.

** 2017 Edit **

PHP7 introduced more function data type declarations, and the aforementioned link has been moved to Function arguments : Type declarations. From that page :

Valid types

• Class/interface name : The parameter must be an instanceof the given class or interface name. (since PHP 5.0.0)
• self : The parameter must be an instanceof the same class as the one the method is defined on. This can only be used on class and instance methods. (since PHP 5.0.0)
• array : The parameter must be an array. (since PHP 5.1.0)
• callable : The parameter must be a valid callable. (since PHP 5.4.0)
• bool : The parameter must be a boolean value. (since PHP 7.0.0)
• float : The parameter must be a floating point number. (since PHP 7.0.0)
• int : The parameter must be an integer. (since PHP 7.0.0)
• string : The parameter must be a string. (since PHP 7.0.0)
• iterable : The parameter must be either an array or an instanceof Traversable. (since PHP 7.1.0)

Warning

Aliases for the above scalar types are not supported. Instead, they are treated as class or interface names. For example, using boolean as a parameter or return type will require an argument or return value that is an instanceof the class or interface boolean, rather than of type bool:

<?php
  function test(boolean $param) {}
  test(true);
?>

The above example will output:

 Fatal error: Uncaught TypeError: Argument 1 passed to test() must be an instance of boolean, boolean given, called in - on line 1 and defined in -:1

The last warning is actually significant to understand the error "Argument must of type string, string given"; since mostly only class/interface names are allowed as argument type, PHP tries to locate a class name "string", but can't find any because it is a primitive type, thus fail with this awkward error.

Answer 3

^{(originally posted by leepowers in his question)}

The error message is confusing for one big reason:

Primitive type names are not reserved in PHP

The following are all valid class declarations:

class string { }
class int { }
class float { }
class double { }

My mistake was in thinking that the error message was referring solely to the string primitive type - the word 'instance' should have given me pause. An example to illustrate further:

class string { }
$n = 1234;
$s1 = (string)$n;
$s2 = new string();
$a = array('no', 'yes');
printf("\$s1 - primitive string? %s - string instance? %s\n",
        $a[is_string($s1)], $a[is_a($s1, 'string')]);
printf("\$s2 - primitive string? %s - string instance? %s\n",
        $a[is_string($s2)], $a[is_a($s2, 'string')]);

Output:

$s1 - primitive string? yes - string instance? no

$s2 - primitive string? no - string instance? yes

In PHP it's possible for a string to be a string except when it's actually a string. As with any language that uses implicit type conversion, context is everything.

Answer 4

As of PHP 7.0 type declarations allow scalar types, so these types are now available: self, array, callable, bool, float, int, string. The first three were available in PHP 5, but the last four are new in PHP 7. If you use anything else (e.g. integer or boolean) that will be interpreted as a class name.

See the PHP manual for more information.

Answer 5

I got this error when invoking a function from a Laravel Controller to a PHP file.

After a couple of hours, I found the problem: I was using $this from within a static function.

Answer 6

Maybe not safe and pretty but if you must:

class string
{
    private $Text;
    public function __construct($value)
    {
        $this->Text = $value;
    }

    public function __toString()
    {
        return $this->Text;
    }
}

function Test123(string $s)
{
    echo $s;
}

Test123(new string("Testing"));

Answer 7

PHP allows "hinting" where you supply a class to specify an object. According to the PHP manual, "Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn't supported." The error is confusing because of your choice of "string" - put "myClass" in its place and the error will read differently: "Argument 1 passed to phpwtf() must be an instance of myClass, string given"

Answer 8

As others have already said, type hinting currently only works for object types. But I think the particular error you've triggered might be in preparation of the upcoming string type SplString.

In theory it behaves like a string, but since it is an object would pass the object type verification. Unfortunately it's not yet in PHP 5.3, might come in 5.4, so haven't tested this.

Answer 9

I think typecasting on php on inside block, String on PHP is not object as I know:

<?php
function phpwtf($s) {
    $s = (string) $s;
    echo "$s\n";
}
phpwtf("Type hinting is da bomb");