Sketch out static polymorphism in C++

static_polymorphism
Bradlee Speice 2020-08-06 01:09:19 -04:00
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---
layout: post
title: "Static Polymorphism"
description: "Emulating Traits in C++"
category:
tags: [python]
---
# Simple Example
Accept parameter types, return known type.
# Generic return
Same parameter signature, but return different types - `AsRef`
# Associated types
`.as_iter()`, and the iterator item types
# Arbitrary `self`
[`std::enable_shared_from_this`](https://en.cppreference.com/w/cpp/memory/enable_shared_from_this)
`enable_unique_from_this` doesn't make a whole lot of sense, but Rust can do it:
```rust
struct MyStruct {}
impl MyStruct {
fn my_function(self: &Box<Self>) {}
}
fn main() {
let unboxed = MyStruct {};
// error[E0599]: no method named `my_function` found for struct `MyStruct` in the current scope
// unboxed.my_function();
let boxed = Box::new(MyStruct {});
boxed.my_function();
boxed.my_function();
}
```
Interestingly enough, can't bind `static` version using equality:
```c++
#include <iterator>
#include <vector>
#include <concepts>
std::uint64_t free_get_value() {
return 24;
}
class MyClass {
public:
// <source>:11:47: error: invalid pure specifier (only '= 0' is allowed) before ';' token
std::uint64_t get_value() = free_get_value;
};
int main() {
auto x = MyClass {};
}
```
---
Turns out the purpose of `enable_shared_from_this` is so that you can create new shared instances of
yourself from within yourself, it doesn't have anything to do with enabling extra functionality
depending on whether you're owned by a shared pointer. _At best_, you could have other runtime
checks to see if you're owned exclusively, or as part of some other smart pointer, but the type
system can't enforce that. And if you're _not_ owned by that smart pointer, what then? Exceptions?
UFCS would be able to help with this - define new methods like:
```c++
template<>
void do_a_thing(std::unique_ptr<MyType> value) {}
```
In this case, the extension is actually on `unique_ptr`, but the overload resolution applies only to
pointers of `MyType`. Note that `shared_ptr` and others seem to work by overloading `operator ->` to
proxy function calls to the delegates; you could inherit `std::shared_ptr` and specialize the
template to add methods for specific classes I guess? But it's still inheriting `shared_ptr`, you
can't define things directly on it.
Generally, "you can just use free functions" seems like a shoddy explanation. We could standardize
overload `MyClass_init` as a constructor, etc., but the language is designed to assist us so we
don't have to do crap like that. I do hope UFCS becomes a thing.
# Default implementation
First: example of same name, different arguments. Not possible in Rust.
Can you bind a free function in a non-static way? Pseudocode:
```c++
template<typename T>
concept DoMethod = requires (T a) {
{ a.do_method(std::declval<std::uint64_t>() } -> std::same_as<std::uint64_t>;
{ a.do_method() } -> std::same_as<std::uint64_t>;
}
template<typename T> requires DoMethod<T>
std::uint64_t free_do_method(T& a) {
a.do_method(0);
}
class MyClass {
public:
std::uint64_t do_method(std::uint64_t value) {
return value * 2;
}
// Because the free function still needs a "this" reference (unlike Javascript which has a
// floating `this`), we can't bind as `std::uint64_t do_method() = free_do_method`
std::uint64_t do_method() {
return free_do_method(this);
}
};
```
# Require concept methods to take `const this`?
# Move/consume `self` as opposed to `&self`?
Is there a way to force `std::move(object).method()`?
# Require static methods on a class?
# `override`, or other means of verifying a function implements a requirement?
# Local trait implementation of remote types?
AKA "extension methods". UFCS can accomplish this, and could use free functions to handle instead,
but having the IDE auto-complete `.<the next thing>` is exceedingly useful, as opposed to memorizing
what functions are necessary for conversion. We're not changing what's possible, just making it
easier for humans.
Likely requires sub-classing the remote class. Implicit conversions don't _really_ work because they
must be defined on the remote type.
Rust makes this weird because you have to `use ClientExt` to bring the methods in scope, but the
trait name might not show up because `impl ClientExt for RemoteStruct` is defined elsewhere.
Alternately, `ClientExt: AnotherTrait` implementations where the default `ClientExt` implementation
is used.
# Automatic markers?
Alternately, conditional inheritance based on templates?
# Trait objects as arguments
```rust
trait MyTrait {
fn some_method(&self);
}
fn my_function(value: &dyn MyTrait) {
}
```
C++ can't explicitly use vtable as part of concepts:
```c++
template<typename T, typename = std::enable_if_t<...>>
void my_function(T& value) {}
```
...is equivalent to:
```rust
fn my_function<T: MyTrait>(value: &T) {}
```
vtable is automatically used if declared virtual.
`dyn Trait` seems to be used in Rust mostly for type erasure - `Box<Pin<dyn Future>>` for example,
but is generally fairly rare, and C++ probably doesn't suffer for not having it. Can use inheritance
to force virtual if truly necessary, but not sure why you'd need that.