The line use type Bar\{Bar}; imports the type Bar\Bar as a named type. However, the type Bar\Bar is actually an abstract class. When the compiler sees the line const Bar\my_type CONFIG = shape('a' => 'b');, it expects the type Bar\my_type to be a concrete type. However, since Bar\Bar is an abstract class, it cannot be a concrete type. This is why the compiler is throwing an error.
Steps to Reproduce:
Run the following code:
php
namespace Foo;
use namespace Bar;
use type Bar{Bar};
final class FooBar extends Bar {
const Bar\my_type CONFIG = shape('a' => 'b');
}
namespace Bar;
abstract class Bar {
const my_type CONFIG = shape('a' => 'b');
}
type my_type = shape('a' => string);
2. Run the command `hh_client`.
**Expected Behavior:**
The code should compile without any errors.
**Actual Behavior:**
The code will compile with the following error:
File "/root/foo.hack", line 6, characters 28-30:
Some members in class Foo\FooBar are incompatible with those declared in type Bar\Bar (Typing[4348])
File "/root/foo.hack", line 13, characters 9-15:
Expected shape('a' => string)
File "/root/foo.hack", line 7, characters 10-20:
But got Bar\Bar\my_type
Workaround:
The workaround for this bug is to use the line use type Bar\Bar; instead of use type Bar\{Bar};. This line imports the abstract class Bar\Bar as a reference type. This means that the compiler will not try to instantiate the class Bar\Bar as a concrete type.
Suggested Fix:
The suggested fix for this bug is to change the line use type Bar\{Bar}; to use type Bar\Bar;. This will import the abstract class Bar\Bar as a reference type, which will prevent the compiler from throwing an error.
Additional Information:
The bug was found on Docker nightly.
The HipHop VM version is 6.33.0-dev (rel) (non-lowptr).
The compiler version is 1676345911_402707174.
The repo schema is 960a4d092e2e8175b8f0a3b2caecffb7c980c672.
lexidor
commented
10 months ago
Title: Bug in use type Bar{Bar};
Description:
The line
use type Bar\{Bar};imports the typeBar\Baras a named type. However, the typeBar\Baris actually an abstract class. When the compiler sees the lineconst Bar\my_type CONFIG = shape('a' => 'b');, it expects the typeBar\my_typeto be a concrete type. However, sinceBar\Baris an abstract class, it cannot be a concrete type. This is why the compiler is throwing an error.Steps to Reproduce:
php namespace Foo;
use namespace Bar; use type Bar{Bar};
final class FooBar extends Bar { const Bar\my_type CONFIG = shape('a' => 'b'); }
namespace Bar;
abstract class Bar { const my_type CONFIG = shape('a' => 'b'); }
type my_type = shape('a' => string);
File "/root/foo.hack", line 6, characters 28-30: Some members in class
Foo\FooBarare incompatible with those declared in typeBar\Bar(Typing[4348]) File "/root/foo.hack", line 13, characters 9-15: Expectedshape('a' => string)File "/root/foo.hack", line 7, characters 10-20: But gotBar\Bar\my_typeWorkaround:
The workaround for this bug is to use the line
use type Bar\Bar;instead ofuse type Bar\{Bar};. This line imports the abstract classBar\Baras a reference type. This means that the compiler will not try to instantiate the classBar\Baras a concrete type.Suggested Fix:
The suggested fix for this bug is to change the line
use type Bar\{Bar};touse type Bar\Bar;. This will import the abstract classBar\Baras a reference type, which will prevent the compiler from throwing an error.Additional Information: