I'm making use of a generic class member to forward a variable list of arguments to an external object, using forwarding arguments:
#include <utility>
#include <iostream>
#include <string>
class MyClass {
public:
MyClass() = default;
virtual ~MyClass() = default;
// ...
template < typename... Args >
void func(Args&&... args);
};
// generic definition
template < typename... Args >
void MyClass::func(Args&&... args) {
std::cout << "MyClass::" << __func__ << std::endl;
// ...
// e.g. Other(std::forward< Args >(args)...);
// ...
}
However, I'd like to account for a specific case associated with a specific signature for func e.g. func(float, const std::string&). However, I'm not entirely sure how to specialize the template to account for arguments to be passed by-value and by-reference.
I'm only able to compile a specialized definition for func that uses rvalue references:
// specialization
template < >
void MyClass::func< float&&, std::string&& >(float&& first, std::string&& rest) {
std::cout << "MyClass::" << __func__ << "(float,std::string)" << std::endl;
}
However, whereas the generic forwarding definition would be called for any combination of argument types, this specialization will only fit temporary objects. I assume the issue is related to the use of && on the template declaration, but how would one specialize for different combinations of (float, const std::string&) arguments?
Live demo
I can only think of this solution that uses a helper template class. The class method punts the job to a helper template class, using std::remove_cvref_t to trim off the fat:
template < typename... Args >
void func(Args&&... args)
{
helper<std::remove_cvref_t<Args>...>::func(std::forward<Args>(args)...);
}
And this now becomes mostly bog-standard specialization:
template<typename ...Args2> struct helper {
template<typename ...Args>
static void func(Args && ...args)
{
std::cout << "Generic\n";
}
};
template<> struct helper<double, std::string> {
template<typename ...Args>
static void func(Args && ...args)
{
std::cout << "Specialized\n";
}
};
With the following test code:
double v=0;
std::string s;
func(v, v);
func(v, s);
func(v, std::move(s));
The result is:
Generic
Specialized
Specialized
Now, if all the real work needs to be done by a class method, what can be done is have func() pass along *this as the first parameter to the static func()s, which then use it to forwards their remaining argument to two real class methods -- generic and specialized.
It is a reasonable bet that modern compilers will be able to optimize away the extra function calls, this is par for the course in modern C++.
Related
Suppose we have a function that looks like:
template <typename F, typename... A>
inline void execute(F&& functor, A&& ... args) {
std::forward<decltype(functor)>(functor)(std::forward<decltype(args)>(args)...);
}
This works for simple non-templated functions. However, I am trying to perfect-forward a templated function(a quite contrived one):
namespace detail {
template <typename CodecImpl>
class codec
{
public:
//
// Encoding
// Convenient version, returns an std::string.
static std::string encode(const uint8_t* binary, size_t binary_size);
static std::string encode(const char* binary, size_t binary_size);
...
};
class base64_rfc4648
{
public:
template <typename Codec> using codec_impl = stream_codec<Codec, base64_rfc4648>;
static CPPCODEC_ALWAYS_INLINE constexpr size_t alphabet_size() {
static_assert(sizeof(base64_rfc4648_alphabet) == 64, "base64 alphabet must have 64 values");
return sizeof(base64_rfc4648_alphabet);
}
static CPPCODEC_ALWAYS_INLINE constexpr char symbol(alphabet_index_t idx)
{
return base64_rfc4648_alphabet[idx];
}
...
};
} // namespace detail
using base64_rfc4648 = detail::codec<detail::base64<detail::base64_rfc4648>>;
Trying to forward the above:
std::string buf("hello world");
execute(base64_rfc4648::encode, buf.c_str(), buf.size());
Does not work. Template deduction fails:
note: couldn't deduce template parameter 'F'
and it also notes:
No matching function for call to 'execute(<unresolved overloaded function type>, const char*, std::__cxx11::basic_string<char>::size_type)'
How can I fix this?
NOTE: I kept the information short above for readability, but if more info is needed I can add.
I have created a MCVE to work on the problem, rather than the code.
So let us have a generic function called print() which we want to send to your execute()
template <typename Arg>
inline void print(Arg const & a) {
std::cout << a << std::endl;
}
where execute() is:
template <typename F, typename... A>
inline void execute(F&& functor, A&& ... args) {
std::forward<decltype(functor)>(functor)(std::forward<decltype(args)>(args)...);
}
When we try to call execute(print, 10) it fails.
The problem is that the execute() function doesn't understand which overload of print are we trying to call.
Now this problem can be solved in 2 ways :
First approach, specify the complete type of the templated function :
execute(print<int>, 10);
Second approach, create a helper function.
Every problem can be solved by adding one more layer.
This helper function will help us deduce the types, before we pass it to execute()
template <typename Arg>
inline void execute_print(Arg const & a) {
execute(print<Arg>, a); // we need to specify which overload to be invoked
}
And then you can call : execute_print(20);
Here is a full working code for your reference(compiled using C++11):
#include <string>
#include <iostream>
template <typename Arg>
inline void print(Arg const & a) {
std::cout << a << std::endl;
}
template <typename F, typename... A>
inline void execute(F&& functor, A&& ... args) {
std::forward<decltype(functor)>(functor)(std::forward<decltype(args)>(args)...);
}
template <typename Arg>
inline void execute_print(Arg const & a) {
execute(print<Arg>, a); // we need to specify which overload to be invoked
}
int main() {
// execute(print, 5); // wont compile
execute(print<int>, 10);
execute_print(20);
return 0;
}
You can simply use boost::hof to wrap base64_rfc4648::encode in a function object.
execute(BOOST_HOF_LIFT(base64_rfc4648::encode), buf.c_str(), buf.size());
Here is the doc of BOOST_HOF_LIFT
I'm working on a class that schedules functions by binding them in a queue like this:
std::queue <void()> q;
template<typename R,typename... ArgsT>
void
schedule(R& fn, ArgsT&... args)
{
q.push(std::bind(fn, std::forward<ArgsT>(args)...) );
};
template<typename R,typename... ArgsT>
void
schedule(R&& fn, ArgsT&&... args)
{
q.push(std::bind(fn, std::forward<ArgsT>(args)...) );
};
As you see I made the type in the queue void() to make it hold any type of function objects but now I can't get the return when I execute it. What should I do to solve this?
Note: I don't want to use an external library like boost and I don't know what kind of function the user will pass it.
Note: I don't want to use an external library like boost and I don't
know what's the kind of function the user will pass it.
What I usually do in this case is I use a base class (from Command pattern) in my queue, and then have two implementations, the one wrapping the bind, and the other (also wrapping the bind) exposing a function that allows getting the return value.
Here is an example of the returning specialization (at last):
#include <iostream>
#include <functional>
#include <memory>
struct ACmd
{
virtual void exec() = 0;
virtual ~ACmd(){}
};
template <class F>
struct Cmd;
template <class R, class ... Args>
struct Cmd<R(Args...)> : ACmd
{
R result_;
std::function<R()> func_;
template <class F>
Cmd(F&& func, Args&&... args): result_(), func_()
{
auto f = std::bind(std::forward<F>(func), std::forward<Args>(args)...);
func_ = [f](){
return f();
};
}
virtual void exec(){
result_ = func_();
}
const R& getResult() const {return result_;}
};
// Make function for convenience, could return by value or ptr -
// - your choice
template <class R, class F, class ...Args>
Cmd<R(Args...)>* cmd(F&& func, Args&&... args)
{
return new Cmd<R(Args...)>(func, std::forward<Args>(args)...);
}
//... And overload for void...
int foo(int arg) {
return arg;
}
int main() {
auto x = cmd<int>(foo, 10);
x->exec();
std::cout << x->getResult() << std::endl;
return 0;
}
The result of the execution of each element in the queue, it is void, you have already defined it as such. If the functions passed in are required to return a value, then you would need to limit the type(s) returned to a fixed type, use utilities such as std::any, std::variant or some covariant types (possible with a std::unique_ptr or std::shared_ptr).
The simplest is to fix the return type (at compile time);
template <typename R>
using MQ = std::queue<std::function<R()>>;
MQ<int> q;
See the sample below.
The queue declaration needs to be a queue of objects, such as std::function objects. The return value from a bind can be assigned to a function and then used as expected.
std::function is a polymorphic function wrapper, it implements type erasure patterns akin to any, but is specifically designed for functions and other callable objects.
By way of example;
template <typename R>
using MQ = std::queue<std::function<R()>>;
MQ<int> q;
template<typename R,typename... ArgsT>
void
schedule(R&& fn, ArgsT&&... args)
{
q.push(std::bind(std::forward<R>(fn), std::forward<ArgsT>(args)...) );
};
int main()
{
schedule([](int a) { std::cout << "function called" << std::endl; return a; }, 42);
std::cout << q.front()() << std::endl;
}
I've found this interesting code here on stackoverflow from:
Using a STL map of function pointers
template<typename T,typename... Args>
T searchAndCall(std::string s1, Args&&... args){
// ....
// auto typeCastedFun = reinterpret_cast<T(*)(Args ...)>(mapVal.first);
auto typeCastedFun = (T(*)(Args ...))(mapVal.first);
//compare the types is equal or not
assert(mapVal.second == std::type_index(typeid(typeCastedFun)));
return typeCastedFun(std::forward<Args>(args)...);
}
};
Basically, mapVal is a map of function pointers casted to void(*)(void) that will be casted back to their original type with this function. What I would like to do know is how typeCastedFun will be deduced when you don't specify the template parameters.
For instance, let's suppose that you had:
int f(const MyClass& a, MyClass b) {...}
... if you have:
MyClass first, second;
searchAndCall<int>(first, second);
What Args... parameter will be deduced? if I recall correctly, using the function casted back to a function with a different signature compared to the original one, should yield undefined behavior. Is there any other alternative?
What I would like to do is a way to store the type of the function somewhere and use this information to do the correct cast. Everything in the most efficient way.
Thanks
[edit1]
More specifically, I'm trying to build a kind of generic function dispatcher, able to call functions (templated with an enum class value) with different signatures using a lookup table for efficiency reasons. No boost::any as it internally uses a new
[edit2] Use of macros is not allowed
The key problem is that by taking the calling argument types directly, and attempting to cast the function pointer, you are losing all implicit conversions.
Your function signature has to match exactly, or you will get UB if you try to call it. And there is generally no way to get the signature from the args without manually specifying it at the call site.
One workaround to try would be to add a wrapper lambda which takes standardized args with pre-specified implicit coversions applied, e.g. T -> const T&, and possibly numeric types -> double.
Then, when you look up the function, you can cast it to use these standardized args, and the calling args will be implicitly converted.
This would rule out functions taking rvalue refs and non-const references, but I don't thing this is unreasonable for a function that you don't know the signature of, unless you want to disregard const-correctness completely.
Also, other implicit conversions wouldn't happen, e.g. Derived& -> Base&, or char* -> std::string, and I don't think there would be an easy way to make that happen without creating extra limitations.
Overall, it's definitely a tricky thing to do in c++, and anything you try will be hacky. This way should be decent enough. The performance overhead of one extra function call (which can be inlined), and possibly some extraneous argument conversions will be overshadowed by the unavoidable RTTI checking.
Here is a sample implementation (also here on ideone):
#include <unordered_map>
#include <typeinfo>
#include <typeindex>
#include <string>
#include <type_traits>
#include <iostream>
#include <assert.h>
#include <cxxabi.h>
#include <sstream>
#include <stdexcept>
template <typename Func, Func f>
struct store_func_helper;
// unix-specific
std::string demangle(const std::string& val) {
int status;
char *realname;
std::string strname = realname = abi::__cxa_demangle(val.c_str(), 0, 0, &status);
free(realname);
return strname;
}
// args will be implicitly converted to arg<T>::type before calling function
// default: convert to const Arg&
template <typename Arg, typename snifae=void>
struct arg {
using type = const Arg&;
};
// numeric types: convert to double.
template <typename Arg>
struct arg <Arg, typename std::enable_if<std::is_arithmetic<Arg>::value, void>::type> {
using type = double;
};
// set more special arg types here.
// Functions stored in the map are first wrapped in a lambda with this signature.
template <typename Ret, typename... Arg>
using func_type = Ret(*)(typename arg<Arg>::type...);
class func_map {
template <typename Func, Func f>
friend class store_func_helper;
public:
template <typename Func, Func f>
void store(const std::string& name){
store_func_helper<Func, f>::call(this, name );
}
template<typename Ret, typename... Args>
Ret call(std::string func, Args... args){
using new_func_type = func_type<Ret, Args...>;
auto& mapVal = m_func_map.at(func);
if (mapVal.second != std::type_index(typeid(new_func_type))){
std::ostringstream ss;
ss << "Error calling function " << func << ", function type: "
<< demangle(mapVal.second.name())
<< ", attempted to call with " << demangle(typeid(new_func_type).name());
throw std::runtime_error(ss.str());
}
auto typeCastedFun = (new_func_type)(mapVal.first);
//args will be implicitly converted to match standardized args
return typeCastedFun(std::forward<Args>(args)...);
};
private:
std::unordered_map<std::string, std::pair<void(*)(),std::type_index> > m_func_map;
};
#define FUNC_MAP_STORE(map, func) (map).store<decltype(&func),&func>(#func);
template <typename Ret, typename... Args, Ret(*f)(Args...)>
struct store_func_helper<Ret(*)(Args...), f> {
static void call (func_map* map, const std::string& name) {
using new_func_type = func_type<Ret, Args...>;
// add a wrapper function, which takes standardized args.
new_func_type lambda = [](typename arg<Args>::type... args) -> Ret {
return (*f)(args...);
};
map->m_func_map.insert(std::make_pair(
name,
std::make_pair((void(*)()) lambda, std::type_index(typeid(lambda)))
));
}
};
//examples
long add (int i, long j){
return i + j;
}
int total_size(std::string arg1, const std::string& arg2) {
return arg1.size() + arg2.size();
}
int main() {
func_map map;
FUNC_MAP_STORE(map, total_size);
FUNC_MAP_STORE(map, add);
std::string arg1="hello", arg2="world";
std::cout << "total_size: " << map.call<int>("total_size", arg1, arg2) << std::endl;
std::cout << "add: " << map.call<long>("add", 3, 4) << std::endl;
}
I have a class foo with a method bar which takes something callable (function-pointer/ functor). this callable something should be passed to another method doit as an binded element with a third method bar_cb method.
#include <functional>
#include <iostream>
class foo {
public:
template<typename T>
void bar(T&& t) {
std::cout << "bar\n";
doit(std::bind(&foo::template bar_cb<T>, this, std::forward<T>(t)));
}
template<typename T>
void doit(T&& t) {
std::cout << "doit\n";
t();
}
template<typename T>
void bar_cb(T&& t) {
std::cout << "bar_cb\n";
t();
}
};
void lala() {
std::cout << "lala\n";
}
class functor {
public:
void operator()() {
std::cout << "functor::operator()\n";
}
};
int main() {
foo f;
functor fn;
f.bar(fn);
f.bar(std::bind(lala)); // error
return 0;
}
This works fine for functors but not for binded functions as argument for foo::bar (lala in my example). Is it possible to pass an unknowable type to a method and bind it in this method as an argument to another (and if so how)?
I know I could wrap a functor (std::function for example) around the function but since I can call an unknowable type I think there is a way to also bind it (I think I'm just missing something simple).
Here a link to an example.
The primary problem is that your bar_cb(T&&) doesn't deduce the template argument because the template argument is actually specified when using &foo::template bar_cb<X> with some template argument X. The bind() expression will, however, copy the bound function, i.e., it may or may not have the type which would be deduced. Also, std::bind() will not pass bind()-expression through but will rather call them!
The easiest work around is to not use std::bind() to bind the function but rather to use a lambda function:
template<typename T>
void bar(T&& t) {
std::cout << "bar\n";
doit([=](){ this->bar_cb(t); });
}
Doing so let's the compiler deduce the correction argument type for bar_cb() (with C++14 you may want to use the capture [this,t = std::forward<T>(t)] although your bar_cb() still won't see an rvalue).
To pass an already bind()-expression through another bind()-expression, without having bind() consider the inner bind()-expression a bind()-expression you need to make it look as if it is not a bind()-expression. You could do so with a thin function wrapper:
template <typename Fun>
class unbinder {
Fun fun;
public:
template <typename F>
unbinder(F&& fun): fun(std::forward<F>(fun)) {}
template <typename... Args>
auto operator()(Args&&... args) const
-> decltype(fun(std::forward<Args>(args)...)) {
return fun(std::forward<Args>(args)...);
}
};
template <typename Fun>
auto unbind(Fun&& fun)
-> unbinder<Fun> {
return unbinder<Fun>(std::forward<Fun>(fun));
}
Since the function stored in the bind() expression will be passed by lvalue, you'll need a different declaration for your bar_cb(), however:
template<typename T>
void bar_cb(T& t) {
...
}
With that, you can register the bind()-expression using
f.bar(unbind(std::bind(lala)));
If you want to use f.bar(std::bind(lala)) you'll need a conditional definition of bar(): if it receives a bind()-expression it needs to automatically hide the fact that it is a bind()-expression by applying unbind() or something similar:
template<typename T>
typename std::enable_if<!std::is_bind_expression<typename std::decay<T>::type>::value>::type
bar(T&& t) {
std::cout << "bar (non-bind)\n";
doit(std::bind(&foo::template bar_cb<T>, this, std::forward<T>(t)));
}
template<typename T>
typename std::enable_if<std::is_bind_expression<typename std::decay<T>::type>::value>::type
bar(T&& t) {
std::cout << "bar (bind)\n";
doit(std::bind(&foo::template bar_cb<unbinder<T>>, this, unbind(std::forward<T>(t))));
}
In the code I register one or multiple function pointer in a manager class.
In this class I have a map that maps the argument types of the function to said function. It may look like so: std::map< std::vector<std::type_index> , void*>
template<typename Ret, typename... Args>
void Register(Ret(*function)(Args...)) {
void* v = (void*)function;
// recursively build type vector and add to the map
}
At runtime the code gets calls (from an external script) with an arbitrary number of arguments. These arguments can be read as primitive data types or as custom types that will be specified at compile time.
With every call from the script, I have to find out which function to call, and then call it. The former is easy and already solved (filling a vector with type_index in a loop), but I can't think of a solution for the latter.
My first approach was using variadic templates in recursion with an added template argument for each read type - but this turned out to be impossible since templates are constructed at compile time, and the arbitrary number of arguments is read at runtime.
Without variadic templates however, I don't see any possibility to achieve this. I considered boost::any instead of void*, but I didn't see how that would solve the need to cast back to the original type. I also thought of using std::function but that would be a templated type, so it could not be stored in a map for functions with different arguments.
(If it's unclear what I'm asking, think of LuaBinds possibility to register overloaded functions. I tried to understand how it's implemented there (without variadic templates, pre-C++11), but to no avail.)
Suppose you had the arguments in a vector of some kind, and a known function (fully).
You can call this. Call the function that does this invoke.
Next, work out how to do this for template<class... Args>. Augment invoke.
So you have written:
typedef std::vector<run_time_stuff> run_time_args;
template<class... Args>
void invoke( void(*func)(Args...), run_time_args rta )
at this point. Note that we know the types of the argument. I do not claim the above is easy to write, but I have faith you can figure it out.
Now we wrap things up:
template<class...Args>
std::function<void(run_time_args)> make_invoker(void(*func)(Args...)){
return [func](run_time_args rta){
invoke(func, rta);
};
}
and now instead of void* you store std::function<void(run_time_args)> -- invokers. When you add the function pointers to the mechanism you use make_invoker instead of casting to void*.
Basically, at the point where we have the type info, we store how to use it. Then where we want to use it, we use the stored code!
Writing invoke is another problem. It will probably involve the indexes trick.
Suppose we support two kinds of arguments -- double and int. The arguments at run time are then loaded into a std::vector< boost::variant<double, int> > as our run_time_args.
Next, let us extend the above invoke function to return an error in the case of parameter type mismatch.
enum class invoke_result {
everything_ok,
error_parameter_count_mismatch,
parameter_type_mismatch,
};
typedef boost::variant<int,double> c;
typedef std::vector<run_time_stuff> run_time_args;
template<class... Args>
invoke_result invoke( void(*func)(Args...), run_time_args rta );
now some boilerplate for the indexes trick:
template<unsigned...Is>struct indexes{typedef indexes type;};
template<unsigned Max,unsigned...Is>struct make_indexes:make_indexes<Max-1, Max-1,Is...>{};
template<unsigned...Is>struct make_indexes<0,Is...>:indexes<Is...>{};
template<unsigned Max>using make_indexes_t=typename make_indexes<Max>::type;
With that, we can write an invoker:
namespace helpers{
template<unsigned...Is, class... Args>
invoke_result invoke( indexes<Is...>, void(*func)(Args...), run_time_args rta ) {
typedef void* pvoid;
if (rta.size() < sizeof...(Is))
return invoke_result::error_parameter_count_mismatch;
pvoid check_array[] = { ((void*)boost::get<Args>( rta[Is] ))... };
for( pvoid p : check_array )
if (!p)
return invoke_result::error_parameter_type_mismatch;
func( (*boost::get<Args>(rts[Is]))... );
}
}
template<class... Args>
invoke_result invoke( void(*func)(Args...), run_time_args rta ) {
return helpers::invoke( make_indexes_t< sizeof...(Args) >{}, func, rta );
}
And that should work when func's args exactly match the ones passed in inside run_time_args.
Note that I was fast and loose with failing to std::move that std::vector around. And that the above doesn't support implicit type conversion. And I didn't compile any of the above code, so it is probably littered with typos.
I was messing around with variadic templates a few weeks ago and came up with a solution that might help you.
DELEGATE.H
template <typename ReturnType, typename ...Args>
class BaseDelegate
{
public:
BaseDelegate()
: m_delegate(nullptr)
{
}
virtual ReturnType Call(Args... args) = 0;
BaseDelegate* m_delegate;
};
template <typename ReturnType = void, typename ...Args>
class Delegate : public BaseDelegate<ReturnType, Args...>
{
public:
template <typename ClassType>
class Callee : public BaseDelegate
{
public:
typedef ReturnType (ClassType::*FncPtr)(Args...);
public:
Callee(ClassType* type, FncPtr function)
: m_type(type)
, m_function(function)
{
}
~Callee()
{
}
ReturnType Call(Args... args)
{
return (m_type->*m_function)(args...);
}
protected:
ClassType* m_type;
FncPtr m_function;
};
public:
template<typename T>
void RegisterCallback(T* type, ReturnType (T::*function)(Args...))
{
m_delegate = new Callee<T>(type, function);
}
ReturnType Call(Args... args)
{
return m_delegate->Call(args...);
}
};
MAIN.CPP
class Foo
{
public:
int Method(int iVal)
{
return iVal * 2;
}
};
int main(int argc, const char* args)
{
Foo foo;
typedef Delegate<int, int> MyDelegate;
MyDelegate m_delegate;
m_delegate.RegisterCallback(&foo, &Foo::Method);
int retVal = m_delegate.Call(10);
return 0;
}
Not sure if your requirements will allow this, but you could possibly just use std::function and std::bind.
The below solution makes the following assumptions:
You know the functions you want to call and their arguments
The functions can have any signature, and any number of arguments
You want to use type erasure to be able to store these functions and arguments, and call them all at a later point in time
Here is a working example:
#include <iostream>
#include <functional>
#include <list>
// list of all bound functions
std::list<std::function<void()>> funcs;
// add a function and its arguments to the list
template<typename Ret, typename... Args, typename... UArgs>
void Register(Ret(*Func)(Args...), UArgs... args)
{
funcs.push_back(std::bind(Func, args...));
}
// call all the bound functions
void CallAll()
{
for (auto& f : funcs)
f();
}
////////////////////////////
// some example functions
////////////////////////////
void foo(int i, double d)
{
std::cout << __func__ << "(" << i << ", " << d << ")" << std::endl;
}
void bar(int i, double d, char c, std::string s)
{
std::cout << __func__ << "(" << i << ", " << d << ", " << c << ", " << s << ")" << std::endl;
}
int main()
{
Register(&foo, 1, 2);
Register(&bar, 7, 3.14, 'c', "Hello world");
CallAll();
}