Do not mark callable-owned template types as arguments when entering a function

Templates declared by callable<T>(...)/Closure<T>(...) types in a function
signature were only recognized as owned by the callable when used as a bare
parameter or return type. When the template occurred only inside another type,
like callable<T of object>(class-string<T> $class): list<T>, it was marked
as an argument type and invoking the callable could not resolve it.

Such templates always have the anonymous function template scope, which the
entered function's own templates never have, so they are recognized by scope.

Co-Authored-By: Claude Fable 5 <noreply@anthropic.com>

Author: ondrejmirtes

src/Type/Generic/TemplateTypeHelper.php

@@ -131,6 +131,12 @@ public static function toArgument(Type $type): Type
 			}
 
 			if ($type instanceof TemplateType) {
+				// templates declared by a callable<T>(...)/Closure<T>(...) type in the signature
+				// belong to the callable value, not to the entered function
+				if ($type->getScope()->equals(TemplateTypeScope::createWithAnonymousFunction())) {
+					return $traverse($type);
+				}
+
 				return $traverse($type->toArgument());
 			}
 

tests/PHPStan/Analyser/nsrt/generic-callables.php

@@ -78,3 +78,11 @@ function testNestedClosures(Closure $closure, string $str, int $int): void
 	$result = $closure1($int);
 	assertType('int|string', $result);
 }
+
+/**
+ * @param callable<T of object>(class-string<T> $class): list<T> $factory
+ */
+function templateOnlyInsideParameterType(callable $factory): void
+{
+	assertType('list<DateTimeImmutable>', $factory(\DateTimeImmutable::class));
+}