function() must be an instance of string, string given" prior to PHP 7?
Here is my code:
function phpwtf(string $s) {
echo "$sn";
}
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.
Summary
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? %sn",
$a[is_string($s1)], $a[is_a($s1, 'string')]);
printf("$s2 - primitive string? %s - string instance? %sn",
$a[is_string($s2)], $a[is_a($s2, 'string')]);
Output:
$s1 - primitive string? yes - string instance? no
$s2 - primitive string? no - string instance? yes
In PHP it's possible for a string to be a string except when it's actually a string . As with any language that uses implicit type conversion, context is everything.
Prior to PHP 7 type hinting can only be used to force the types of objects and arrays. Scalar types are not type-hintable. In this case an object of the class string is expected, but you're giving it a (scalar) string . The error message may be funny, but it's not supposed to work to begin with. Given the dynamic typing system, this actually makes some sort of perverted sense.
You can only manually "type hint" scalar types:
function foo($string) {
if (!is_string($string)) {
trigger_error('No, you fool!');
return;
}
...
}
From PHP's 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.
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
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.
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"
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