I was trying to implement an std::unique_ptr factory that I could use like this:
auto fd = my_make_unique<fclose>(fopen("filename", "r"));
I.e., pass the deleter function as a template argument.
My best attempt in C++11 was:
template<typename D, D deleter, typename P>
struct Deleter {
void operator()(P* ptr) {
deleter(ptr);
}
};
template<typename D, D deleter, typename P>
std::unique_ptr<P, Deleter<D, deleter, P>> my_make_unique(P* ptr)
{
return std::unique_ptr<P, Deleter<D, deleter, P>>(ptr);
}
In C++14 it is much cleaner:
template<typename D, D deleter, typename P>
auto my_make_unique(P* ptr)
{
struct Deleter {
void operator()(P* ptr) {
deleter(ptr);
}
};
return std::unique_ptr<P, Deleter>(ptr);
}
But both solutuions would require me to pass the type of &fclose before fclose itself as template argument:
auto fd = my_make_unique<decltype(&fclose), fclose>(fopen("filename", "r"));
Is it possible to get rid of decltype(&fclose) template argument in C++11? What about in C++14?
EDIT: Why this question is not a duplicate of RAII and smart pointers in C++: referenced question is about general RAII techniques in C++, and one of the answers estates that std::unique_ptr can be used for this purpose. I am already familiar with RAII pattern and how std::unique_ptr is a solution, but I am concerned with present question on how to build an easier to use abstraction to this frequent case I encounter when interacting with C libraries.
Is it possible to get rid of decltype(&fclose) template argument in C++11? What about in C++14?
No, not until C++17 can you get rid of the type of that parameter. Template non-type parameters need a type, which you can't deduce - because it has to be a template non-type parameter. That's one problem.
Additionally, you have the problem that taking the address of functions in the standard library is unspecified. The standard library is always allowed to provide additional overloads, for instance, so &fclose may be invalid. The only truly portable way of doing this is to provide a lambda or write your own wrapper function:
auto my_fclose_lam = [](std::FILE* f) { std::fclose(f); }
void my_fclose_fun(std::FILE* f) { std::fclose(f); }
And with either of those, with C++14 at best you can introduce a macro like:
#define DECL(v) decltype(v), v
auto fd = my_make_unique<DECL(my_fclose_lam)>(fopen("filename", "r"));
C++17 allows you to at least lift your custom function into a template parameter (though not yet a lambda) via template auto:
template <auto deleter, typename P>
auto my_make_unique(P* ptr)
{
struct Deleter {
void operator()(P* ptr) {
deleter(ptr);
}
};
return std::unique_ptr<P, Deleter>(ptr);
}
my_make_unique<my_fclose_fun>(fopen(...));
C++20 will finally allow you to stick a lambda into there:
my_make_unique<[](std::FILE* f){ std::fclose(f); }>(fopen(...));
Old incorrect answer:
So the best you can do is introduce a macro like:
#define DECL(v) decltype(v), v
auto fd = my_make_unique<DECL(&fclose)>(fopen("filename", "r"));
Whether or not you think that's a good idea is probably up to your coworkers.
In C++17, with template auto, you could just be able to write my_make_unique<fclose>, which is great:
template <auto deleter, typename P>
auto my_make_unique(P* ptr)
{
struct Deleter {
void operator()(P* ptr) {
deleter(ptr);
}
};
return std::unique_ptr<P, Deleter>(ptr);
}
Pragmatic approach: Make the deleter a runtime parameter.
template<typename P, typename D>
auto my_make_unique(P* ptr, D deleter)
{
return std::unique_ptr<P, D>(ptr, deleter);
}
int main()
{
auto fd = my_make_unique(fopen("filename", "r"), fclose);
}
Another workaround is use exactly signature of function:
template<typename T, int (*P)(T*)> //for `fclose`
auto my_make_unique(T*) { ... }
template<typename T, void (*P)(T*)> //for other function signatures
auto my_make_unique(T*) { ... }
//etc.
auto uniq = my_make_unique<File, fclose>(fopen("filename", "r"));
This is not universal solution but in 95% cases will work.
The typical way to create a std::unique_ptr for a FILE* pointer is:
auto fd = std::unique_ptr<FILE, decltype(fclose)>(fopen(...), fclose);
You could wrap that in a macro:
#define my_make_unique(ptr, deleter) \
std::unique_ptr<std::remove_pointer<decltype<ptr>>::type, d>(ptr, deleter)
And then use it like this:
auto fd = my_make_unique(fopen(...), fclose);
Related
I am trying to implement a pointer to member function wrapper where the return type and parameter types are known but the class type of the member function is not known. This is for a C++ project targeting resource constrained (No heap allocation/c++ standard library) microprocessor.
The following code seems to work (you can run it HERE) and it fulfills our general requirements, but it uses reinterpret_cast hackery.
//can't use any c++ std libs or heap allocation
#include <cstdio>
template<typename Ret,typename... Args>
struct member_cb
{
template<typename T>
void set_cb(T *obj_,Ret (T::*func)(Args...))
{
obj = reinterpret_cast<member_cb*>(obj_);
cb = reinterpret_cast<Ret (member_cb::*)(Args...)>(func);
}
Ret operator()(Args... num)
{
return (obj->*cb)(num...);
}
member_cb* obj;
Ret (member_cb::*cb)(Args...);
};
struct Bar{
void someFunc(int n2){
printf("someFunc called with parameter: %d\n",n2);
}
};
int main()
{
//class type of Bar is not known at wrapper creation
member_cb<void,int> foo;
Bar bar;
foo.set_cb(&bar, &Bar::someFunc);
foo(42);
}
I want to do this without the reinterpret_cast. I believe there is a solution using a lambda that is created by the set_cb() function something like this:
template<typename Ret,typename... Args>
struct member_cb
{
template<typename T>
void set_cb(T *obj_,Ret (T::*func)(Args...))
{
cb_func = [=](Args...num)
{
(obj_->*func)(num...);
};
}
Ret operator()(Args... args)
{
return cb_func(args...);
}
Ret (*cb_func)(Args...);
}
The above code will not compile (you can try it HERE). Problem is that the regular function pointers like cb_func can only point to lambdas that have no captures, and the lambda has to capture the obj_ and the func parameters. Any suggestions on making this work would be appreciated.
A simple, compiler-independent solution:
template <typename Signature>
struct member_cb;
template <typename Ret, typename... Args>
struct member_cb<Ret(Args...)>
{
template <typename T, Ret (T::*func)(Args...)>
static Ret wrapper(void *object, Args&&... args) {
T *o = reinterpret_cast<T *>(object);
return (o->*func)(std::forward<Args>(args)...);
}
template <typename T, Ret (T::*func)(Args...)>
void set_cb(T *obj_)
{
obj = obj_;
cb_func = wrapper<T, func>;
}
// since C++17:
template <auto func, typename T>
void set_cb(T *obj_)
{
obj = obj_;
cb_func = wrapper<T, func>;
}
Ret operator() (Args&& ...args)
{
return cb_func(obj, std::forward<Args>(args)...);
}
void *obj;
Ret (*cb_func)(void *, Args&&...);
};
...
member_cb<void(int)> foo;
foo.set_cb<Bar, &Bar::someFunc>(&bar);
foo.set_cb<&Bar::someFunc>(&bar); // since C++17
The key limitation is that someFunc must be known at compile-time at the place set_cb is called, so you can’t take arbitrary method pointer and convert it to a member_cb. It is possible to change the object keeping the method pointer intact, though (but that’s not type-safe).
Another possible problem is that the compiler may need to emit a wrapper instance for each and every method that is ever used as a callback.
The upside is that it should work on any C++11-compliant compiler (C++17 for nicer syntax), and member_cb itself is small.
As I understand, typedef cannot be used for overloading but what if I need to use some different types as arguments to the function pointer?
How can I make it work with the following functionality?
{
public:
typedef void (*InitFunc)(float x);
typedef void (*InitFunc)(int a, char b); //Needs to be added
virtual void initialize(InitFunc init) = 0;
};
Edit:
I cannot use C++17, so can't use variant
As commented, the easiest way is a union, although not very type safe and C++-y. Here is an example with inheritance, since you commented that you want inheritance.
typedef void (*FloatInit)(float x);
typedef void (*IntCharInit)(int a, char b);
union InitFn {
FloatInit fi;
IntCharInit ici;
};
struct Foo {
void initialize(InitFn) = 0;
};
struct FloatFoo: public Foo {
void initialize(InitFn f) override {
f.fi(42.0f);
}
};
void test(float) {}
// ...
auto x = FloatFoo{};
x.initialize(InitFn{test});
As mentioned by other commenters, you can use std::variant to enhance type safety and get rid of the manual union definition:
typedef void (*FloatInit)(float x);
typedef void (*IntCharInit)(int a, char b);
typedef std::variant<FloatInit, IntCharInit> InitFn;
struct Foo {
void initialize(InitFn) = 0;
};
struct FloatFoo: public Foo {
void initialize(InitFn f) override {
std::get<FloatInit>(f)(42.0f);
}
};
void test(float) {}
// ...
auto x = FloatFoo{};
x.initialize(InitFn{test});
One solution is to create a simple wrapper class template instead, to allow the compiler to automatically generate instantiations as necessary. This is relatively simple if init is always guaranteed to be a non-member function (and by extension, an actual function and not a functor/lambda).
// Quick-and-dirty transparent callable wrapper, to serve as overloadable "type alias".
template<typename>
class InitFunc;
template<typename Ret, typename... Params>
class InitFunc<Ret(*)(Params...)> {
public:
// Supply component types if needed.
// Tuple used for params, for convenience.
using return_type = Ret;
using param_types = std::tuple<Params...>;
using func_type = Ret(Params...);
using func_ptr_type = func_type*;
using func_ref_type = func_type&;
// Create from pointer or reference.
constexpr InitFunc(func_ptr_type p = nullptr) : ptr(p) {}
constexpr InitFunc(func_ref_type r) : ptr(&r) {}
// Transparent invocation.
// Deduces argument types instead of relying on Params, to allow for perfect forwarding.
template<typename... Ts>
constexpr return_type operator()(Ts&&... ts) { return ptr(std::forward<Ts>(ts)...); }
// Convert back to original type if necessary.
operator func_ptr_type() { return ptr; }
operator func_ref_type() { return *ptr; }
private:
// Actual function pointer.
func_ptr_type ptr;
};
// And a nice, clean creator, which can be renamed as necessary.
template<typename Init>
constexpr auto make(Init func) { return InitFunc<Init>(func); }
This creates a nice little wrapper that can easily be optimised out entirely, and will compile as long as C++14 support is available.
Note that you require a C++11 compiler (or variadic templates, rvalue references, perfect forwarding, and constexpr support) at the absolute minimum, and will need to modify make() to have a trailing return type for pre-C++14 compilers. I believe this is compatible with C++11 constexpr, but I'm not 100% sure.
If you want InitFunc to be able to accept pointers/references-to-member-function (including functors and lambdas), you'll need to provide an additional version to isolate it into a non-member "function", and likely bind it to a class instance. It may be worth looking into std::bind() in this case, although I'm not sure if it has any overhead.
In this case, I would suggest splitting the member types off into a base class, to reduce the amount of code you'll need to duplicate.
// Quick-and-dirty transparent callable wrapper, to serve as overloadable "type alias".
template<typename>
class InitFunc;
// Supply component types if needed.
// Tuple used for params, for convenience.
// Using actual function type as a base, similar to std::function.
template<typename Ret, typename... Params>
class InitFunc<Ret(Params...)> {
public:
using return_type = Ret;
using param_types = std::tuple<Params...>;
using func_type = Ret(Params...);
using func_ptr_type = func_type*;
using func_ref_type = func_type&;
};
// Non-member functions.
// As member types are now dependent types, we qualify them and use `typename`.
// Yes, it looks just as silly as you think it does.
template<typename Ret, typename... Params>
class InitFunc<Ret(*)(Params...)> : public InitFunc<Ret(Params...)> {
// Actual function pointer.
typename InitFunc::func_ptr_type ptr;
public:
// Create from pointer or reference.
constexpr InitFunc(typename InitFunc::func_ptr_type p = nullptr) : ptr(p) {}
constexpr InitFunc(typename InitFunc::func_ref_type r) : ptr(&r) {}
// Transparent invocation.
// Deduces argument types instead of relying on Params, to allow for perfect forwarding.
template<typename... Ts>
constexpr typename InitFunc::return_type operator()(Ts&&... ts) { return ptr(std::forward<Ts>(ts)...); }
// Convert back to original type if necessary.
operator typename InitFunc::func_ptr_type() { return ptr; }
operator typename InitFunc::func_ref_type() { return *ptr; }
};
// See ecatmur's http://stackoverflow.com/a/13359520/5386374 for how to accomodate member functions.
// ...
// Non-member function make() is unaffected.
// An overload will likely be needed for member functions.
template<typename Init>
auto make(Init func) { return InitFunc<Init>(func); }
Despite the awkwardness inside our derived specialisation, any code that relies on InitFunc shouldn't (to my knowledge) see any changes to its API; the previous example will work just fine if we swap to this new InitFunc, and be none the wiser after recompilation.
Note that it will change the ABI, though, and thus any code compiled for the simpler InitFunc will need to be recompiled for this version.
Is there a convenient way to re-assign the value of a unique_ptr with a new owned object, without re-specifying the type?
For instance:
std::unique_ptr<int> foo;
// .... Later, once we actually have a value to store...
foo = std::make_unique<int>(my_cool_value);
Of course int is not too much of an eyesore, but foo::element_type could be long or subject to change after a refactoring.
So, to use type inference, we could do:
foo = std::make_unique<decltype(foo)::element_type>(value);
...But that's pretty hideous (foo::element_type doesn't work because foo can't be used in a constant expression).
Ideally, std::unique_ptr would support a forwarding emplace-like method:
foo.reassign(value);
This would release the old value and, just like std::vector::emplace, construct the new owned object in-place.
....But as far as I can tell, there's nothing more concise than make_unique<decltype(foo)::element_type>.
EDIT: The most concise way to reassign the value for a type that supports operator= is, of course, to use operator=:
*foo = value;`
...But I do not want to rely on the copyability of element_type (for instance, I initially ran into this issue when trying to work with input-file streams).
Stash the arguments (or references thereto) into a proxy object with a templated conversion operator that deduces the target type. Then construct the new object once you have that deduced.
template<class... Args>
struct maker {
template<class T>
operator std::unique_ptr<T>() && {
return make<T>(std::index_sequence_for<Args...>());
}
std::tuple<Args...> args;
private:
template<class T, size_t ... Is>
std::unique_ptr<T> make(std::index_sequence<Is...>) {
return std::make_unique<T>(std::get<Is>(std::move(args))...);
}
};
template<class... Args>
auto maybe_make_unique_eventually(Args&&... args){
return maker<Args&&...>{std::forward_as_tuple(std::forward<Args>(args)...)};
}
It won't be a member function, but a free function could essentially achieve this:
template<typename T, typename D, typename...Args>
void TakeNew(std::unique_ptr<T,D>& up, Args&&... args)
{
up.reset(new T{std::forward<Args>(args)...});
// or use parentheses for consistency with `make_unique`; see comments
}
// usage...
auto foo = std::make_unique<int>(3);
// .... Later...
TakeNew(foo, 5);
(I do not consider this solution ideal.)
#include <memory>
// a class with a long and unweildy name
namespace mary {
namespace poppins {
struct supercalafragalisticexpialadocious
{
};
}
}
int main()
{
// what we don't want to have to do:
auto a = std::make_unique<mary::poppins::supercalafragalisticexpialadocious>();
// so alias the typename
using atrocious = mary::poppins::supercalafragalisticexpialadocious;
// same type with a shorter name
a = std::make_unique<atrocious>();
}
As you have unique ownership, unless the type is not copyable, you may simply do
*foo = value;
I am looking for a way to convert function object to function pointer.
Captureless lambda has implicit conversion that allows to:
using fptr_t = int (*)(int);
fptr_t ptr = nullptr;
ptr = [](int) { return 2; };
ptr = [](auto) { return 3; };
(*ptr)(42);
I try to do the same with old-fashioned, empty class function objects like:
struct Foo {
int operator()(int) const { return 5; }
} foo;
Or std predicates like std::less<int>.
One way I found is to wrap call of foo with lambda.
If I can assure that foo is stateless and const
I dont really need this ptr and lambda-capture:
template <typename R, typename... Args>
struct to_function_pointer<R(Args...)> {
private:
template <typename T, REQUIRES(std::is_empty<T>::value)>
static T const& stateless_const() {
return (*static_cast<T const*>(nullptr));
}
public:
using pointer = R (*)(Args...);
template <typename U>
pointer operator()(U) const {
return [](Args... args) {
return stateless_const<std::decay_t<U>>()(args...);
};
}
};
But here I do not know how to provide perfect forwarding,
cause [](Args&&...) or [](auto&&...) cannot convert to R(*)(Args...).
Such trick fails when args is noncopyable like std::unique_ptr<int>.
I know that I could use std::function, but it's kind of heavy-weight, while I am trying to get a light-weight solution.
Live example.
Any advice appreciated.
I believe you can simplify your to_function_pointer with just:
template <typename R, typename... Args>
struct to_function_pointer<R(Args...)> {
using pointer = R(*)(Args...);
template <typename U, REQUIRES(std::is_empty<U>::value && std::is_trivially_constructible<U>::value)>
pointer operator()(U ) const
{
return [](Args... args) {
return U{}(std::forward<Args>(args)...);
}
}
};
Few things to note. Args... will already be references or not, you're providing that signature. So forward<> will still do the same thing - the function just happens to not be a template here. Also, scrap the weird nullptr cast. That just looks bad - if we just require trivial constructibility we can just write U{} which seems way cleaner to me.
I'd like to create a function that takes a weak pointer and any kind of functor (lambda, std::function, whatever) and returns a new functor that only executes the original functor when the pointer was not removed in the meantime (so let's assume there is a WeakPointer type with such semantics). This should all work for any functor without having to specify explicitly the functor signature through template parameters or a cast.
EDIT:
Some commenters have pointed out that std::function - which I used in my approach - might not be needed at all and neither might the lambda (though in my original question I also forgot to mention that I need to capture the weak pointer parameter), so any alternative solution that solves the general problem is of course is also highly appreciated, maybe I didn't think enough outside the box and was to focused on using a lambda + std::function. In any case, here goes what I tried so far:
template<typename... ArgumentTypes>
inline std::function<void(ArgumentTypes...)> wrap(WeakPointer pWeakPointer, const std::function<void(ArgumentTypes...)>&& fun)
{
return [=] (ArgumentTypes... args)
{
if(pWeakPointer)
{
fun(args...);
}
};
}
This works well without having to explicitly specify the argument types if I pass an std::function, but fails if I pass a lambda expression. I guess this because the std::function constructor ambiguity as asked in this question. In any case, I tried the following helper to be able to capture any kind of function:
template<typename F, typename... ArgumentTypes>
inline function<void(ArgumentTypes...)> wrap(WeakPointer pWeakPointer, const F&& fun)
{
return wrap(pWeakPointer, std::function<void(ArgumentTypes...)>(fun));
}
This now works for lambdas that don't have parameters but fails for other ones, since it always instantiates ArgumentTypes... with an empty set.
I can think of two solution to the problem, but didn't manage to implement either of them:
Make sure that the correct std::function (or another Functor helper type) is created for a lambda, i.e. that a lambda with signature R(T1) results in a std::function(R(T1)) so that the ArgumentTypes... will be correctly deduced
Do not put the ArgumentTypes... as a template parameter instead have some other way (boost?) to get the argument pack from the lambda/functor, so I could do something like this:
-
template<typename F>
inline auto wrap(WeakPointer pWeakPointer, const F&& fun) -> std::function<void(arg_pack_from_functor(fun))>
{
return wrap(pWeakPointer, std::function<void(arg_pack_from_functor(fun))(fun));
}
You don't have to use a lambda.
#include <iostream>
#include <type_traits>
template <typename F>
struct Wrapper {
F f;
template <typename... T>
auto operator()(T&&... args) -> typename std::result_of<F(T...)>::type {
std::cout << "calling f with " << sizeof...(args) << " arguments.\n";
return f(std::forward<T>(args)...);
}
};
template <typename F>
Wrapper<F> wrap(F&& f) {
return {std::forward<F>(f)};
}
int main() {
auto f = wrap([](int x, int y) { return x + y; });
std::cout << f(2, 3) << std::endl;
return 0;
}
Assuming the weak pointer takes the place of the first argument, here's how I would do it with a generic lambda (with move captures) and if C++ would allow me to return such a lambda:
template<typename Functor, typename Arg, typename... Args>
auto wrap(Functor&& functor, Arg&& arg)
{
return [functor = std::forward<Functor>(functor)
, arg = std::forward<Arg>(arg)]<typename... Rest>(Rest&&... rest)
{
if(auto e = arg.lock()) {
return functor(*e, std::forward<Rest>(rest)...);
} else {
// Let's handwave this for the time being
}
};
}
It is possible to translate this hypothetical code into actual C++11 code if we manually 'unroll' the generic lambda into a polymorphic functor:
template<typename F, typename Pointer>
struct wrap_type {
F f;
Pointer pointer;
template<typename... Rest>
auto operator()(Rest&&... rest)
-> decltype( f(*pointer.lock(), std::forward<Rest>(rest)...) )
{
if(auto p = lock()) {
return f(*p, std::forward<Rest>(rest)...);
} else {
// Handle
}
}
};
template<typename F, typename Pointer>
wrap_type<typename std::decay<F>::type, typename std::decay<Pointer>::type>
wrap(F&& f, Pointer&& pointer)
{ return { std::forward<F>(f), std::forward<Pointer>(pointer) }; }
There are two straightforward options for handling the case where the pointer has expired: either propagate an exception, or return an out-of-band value. In the latter case the return type would become e.g. optional<decltype( f(*pointer.lock(), std::forward<Rest>(rest)...) )> and // Handle would become return {};.
Example code to see everything in action.
[ Exercise for the ambitious: improve the code so that it's possible to use auto g = wrap(f, w, 4); auto r = g();. Then, if it's not already the case, improve it further so that auto g = wrap(f, w1, 4, w5); is also possible and 'does the right thing'. ]