How to write a general function callback test class? - c++

I have several classes which each store and call a callback function. The callback functions' signatures have different parameter and return types, but all of them take just one parameter.
To verify that these classes call their callbacks when they should, I'm trying to write a general test class which (1) provides a callback function which takes one parameter, (2) lets the user query whether that function has been called, and (3) lets the user examine the parameter which was passed to that function. So far, I have something like this:
template<class ReturnType, class ParameterType> class Callable
{
public:
Callable() : m_called(false), m_param() {}
ReturnType operator()(ParameterType param)
{
m_called = true;
m_param = param;
return Returntype();
}
bool Called() { return m_called; }
ParameterType Param() { return m_param; }
private:
bool m_called;
ParameterType m_param;
};
Here's a class which might be tested using class Callable:
#include <boost/function.hpp>
class ToBeTested
{
ToBeTested(boost::function<bool (int)> callback) : m_callback(callback) {};
boost::function<bool (int)> m_callback;
// (methods which should cause the callback to be called here)
};
Here's some test code:
#include <boost/bind.hpp>
int main(int, char**)
{
Callable<bool, int> callable;
ToBeTested tbt(boost::bind(&Callable<bool, int>::operator());
// (tell tbt it should call its callback here)
if (callable.Called()
{
if (EXPECTED_VALUE == callable.Param();
return 0;
}
return -1;
}
This gives me (1) and (2), but there's a problem with (3) when the callback takes its parameter by reference: Callable::m_param is a reference type and therefore can't be default initialised. I could fix that by making Callable::operator() take its parameter by reference, like this:
ReturnType operator()(ParameterType & param)
...but then I can't use class Callable when the callback function takes its parameter by value.
Is there a way to make my test class work regardless of whether the callback function takes its parameter by reference, or do I need to write two nearly-identical test classes?

You could try something like this, where references are actually stored as pointers:
template<typename T>
struct ref_to_ptr
{
typedef T type;
static T wrap(T x) { return x; }
static T unwrap(T x) { return x; }
};
template<typename T>
struct ref_to_ptr<T&>
{
typedef T* type;
static T* wrap(T& x) { return &x; }
static T& unwrap(T* x) { return *x; }
};
template<class ReturnType, class ParameterType> class Callable
{
public:
Callable() : m_called(false), m_param() {}
ReturnType operator()(ParameterType param)
{
m_called = true;
m_param = ref_to_ptr<ParameterType>::wrap(param);
return Returntype();
}
bool Called() { return m_called; }
ParameterType Param() { return ref_to_ptr<ParameterType>::unwrap(m_param); }
private:
bool m_called;
typename ref_to_ptr<ParameterType>::type m_param;
};

Related

Callback using lambda with closures

I'm trying to implement a callback which passes control from an Interrupt Service Routine to a member function on a c++ class. I thought lambdas and closures would be a convenient means of doing this, but I'm having trouble implementing it. Below is a simplified version of my code.
The issue I'm stuck on is how to store the "function pointer" to the "lambda".
class Gpio
{
public:
typedef void (*ExtiHandler)();
private:
ExtiHandler handler;
public:
void enable_irq(ExtiHandler handler_in)
{
// enable interrupt
// ...
// save handler so callback can be issued later
handler = handler_in;
}
};
class Button
{
private:
Gpio& pin;
public:
Button(Gpio& pin_in) : pin(pin_in)
{
};
void button_pressed()
{
// do something
}
void init()
{
pin.enable_irq([this]() { this->button_pressed(); });
}
};
Compiling fails with the following error message;
no matching function for call to 'Gpio::enable_irq(Button::init()::<lambda()>)'candidate: void Gpio::enable_irq(Gpio::ExtiHandler) no known conversion for argument 1 from 'Button::init()::<lambda()>' to 'Gpio::ExtiHandler {aka void (*)()}' Build failed
How can I modify this code to resolve the compile error?
The problem is, that enable_irq function expects a typed function pointer of type void (*ExtiHandler)() not a lambda function.
That means, here
pin.enable_irq([this]() { this->button_pressed(); });
you are trying to store a lambda function(with capturing the instance) to a typed function pointer. You could have converted the lambda to a function pointer(easily) if it would have been a capture-less lambda.
See [expr.prim.lambda.closure] (sec 7)
The closure type for a non-generic lambda-expression with no
lambda-capture whose constraints (if any) are satisfied has a
conversion function to pointer to function with C++ language linkage
having the same parameter and return types as the closure type's
function call operator.
Since lambdas are not just ordinary functions and capturing it need to preserve a state,
you can not find any simple or conventional solution to make them assign to function pointers.
Solution - 1
The simplest solution is to use std::function instead, by paying some type erasure overhead. That means, in your code, just need to change the
typedef void(*ExtiHandler)();
to
typedef std::function<void()> ExtiHandler;
// or
// using ExtiHandler = std::function<void()>;
Solution - 2
Can this be accomplished without using the STL?
Yes. After making a small research on this topic, I came up with a type traits solution to store the lambdas with closure to the equivalent typed function pointer.
#include <iostream>
template<typename Lambda> struct convert_lambda : convert_lambda<decltype(&Lambda::operator())> {};
template<typename Lambda, typename ReType, typename... Args>
struct convert_lambda<ReType(Lambda::*)(Args...) const>
{
using funPtr = ReType(*)(Args...);
static funPtr make_function_ptr(const Lambda& t)
{
static const Lambda& lmda = t;
return [](Args... args) { return lmda(args...); };
}
};
template<typename Lambda> using convert_lambda_t = typename convert_lambda<Lambda>::funPtr;
template<typename Lambda> constexpr convert_lambda_t<Lambda> make_function_ptr(const Lambda& t)
{
return convert_lambda<Lambda>::make_function_ptr(t);
}
Usage: SEE LIVE EXAMPLE
You can now simply continue with your Gpio and Button classes, without
changing anything.:
pin.enable_irq(make_function_ptr([this]() { this->button_pressed(); }));
// or
// pin.enable_irq(make_function_ptr([&]() { this->button_pressed();}));
Or with arguments. For example
int aa = 4;
auto lmda = [&aa](const int a, const float f) { std::cout << a * aa * f << std::endl; };
void(*fPtrTest)(const int, const float) = make_function_ptr(lmda);
fPtrTest(1, 2.0f);
Drawbacks: The solution - 2:
is not capable of recognizing the optional sequence of specifiers.(i.e, mutable, constexpr)
is not capable of forwarding parameter pack to the traits. i.e,
the following is not possible:
return [](Args&&... args) { return lmda(std::forward<Args>(args)...); };
Closure object can be assigned to function pointer only if capture list of lambda is empty, in your case this condition is not met - [this].
You can use std::function as wrapper to store your closures:
#include <functional>
class Gpio
{
public:
using ExtiHandler = std::function<void()>;
private:
std::function<void()> handler;
public:
void enable_irq(const ExtiHandler& handler_in)
{
handler = handler_in;
}
};
If you don't have std library then you could implement the type erasure yourself.
Something like this ...
#include <iostream>
#include <memory>
struct function
{
struct base
{
virtual void call() = 0;
virtual base* clone() = 0;
};
template <typename Fn>
struct impl : base
{
Fn fn_;
impl(Fn&& fn) : fn_(std::forward<Fn>(fn)){}
impl(Fn& fn) : fn_(fn){}
virtual void call()
{
fn_();
}
virtual base* clone() { return new impl<Fn>(fn_); }
};
base* holder_;
function() : holder_(nullptr)
{};
template <typename Fn>
function(Fn&& fn) : holder_(nullptr)
{
holder_ = new impl<Fn>(std::forward<Fn>(fn));
}
function( function&& other)
{
holder_ = other.holder_;
other.holder_ = nullptr;
}
function(const function& other)
{
holder_ = other.holder_->clone();
}
~function()
{
if (holder_) delete holder_;
}
function& operator=(function&& other)
{
if (holder_) delete holder_;
holder_ = other.holder_;
other.holder_ = nullptr;
return *this;
}
function& operator=(const function& other)
{
if (holder_) delete holder_;
holder_ = other.holder_->clone();
return *this;
}
void operator()()
{
holder_->call();
}
};
class Gpio
{
public:
using ExtiHandler = function;
//private:
ExtiHandler handler;
//public:
void enable_irq(ExtiHandler handler_in)
{
// enable interrupt
// ...
// save handler so callback can be issued later
handler = handler_in;
}
};
class Button
{
private:
Gpio& pin;
public:
Button(Gpio& pin_in) : pin(pin_in)
{
};
void button_pressed()
{
std::cout << "Button pressed" << std::endl;
}
void init()
{
pin.enable_irq([this]() { this->button_pressed(); });
}
};
int main() {
Gpio some_pin;
Button b(some_pin);
b.init();
some_pin.handler();
return 0;
}
Demo

Static templated member function as function pointer

I'm attempting to replace a series of static to non static member function pairs throughout my code, I can achieve this with a macro but I was hoping I could do so with a static function which takes the non static member function as a template argument, which then gets stored as a function pointer. See the following code:
struct widget_param
{
void* ptr;
};
struct widget_data
{
bool(*callback)(widget_param*);
};
template <class CLASSNAME>
class widget_base
{
protected:
static CLASSNAME* get_instance(widget_param* param)
{
return static_cast<CLASSNAME*>(param->ptr);
}
public:
template <bool(CLASSNAME::*f)(widget_param*)>
static bool static_to_nonstatic(widget_param* param)
{
return get_instance(param)->*f(param);
}
};
class widget_derived : public widget_base<widget_derived>
{
public:
// Attempting to replace this function
static bool static_do_stuff(widget_param* param)
{
return get_instance(param)->do_stuff(param);
}
bool do_stuff(widget_param* param)
{
param;
cout << "Success!";
return true;
}
};
int main() {
widget_derived derived;
//widget_data data{ widget_derived::static_do_stuff}; // Current approach
widget_data data{ widget_derived::static_to_nonstatic<widget_derived::do_stuff> };
widget_param param{ &derived };
data.callback(&param);
return 0;
}
I was expecting the template to evaluate to:
static bool static_to_nonstatic(widget_param* param);
Is it possible to do what I'm trying to achieve, without resorting to using a macro?
Resolved, since the call is to a member function pointer - it needed to be surrounded by parenthesis, changed the template to the following and it worked:
template <bool(CLASSNAME::*f)(widget_param*)>
static bool static_to_nonstatic(widget_param* param)
{
return (get_instance(param)->*f)(param);
}

C++ auto deduction of return type

I want to write a function that return different types based on different input as below.
enum MyType
{
A,
B
};
template<MyType T> struct MyStruct
{
};
static auto createMyStruct(MyType t)
{
if(t==A)
return MyStruct<A>();
else
return MyStruct<B>();
}
It didn't work out because there are two return types for one auto. Is there any other way to do this?
There is absolutely no way of having a (single) function that returns different types based on a runtime decision. The return type has to be known at compile time. However, you can use a template function, like this (thanks to #dyp for making me simplify the code):
#include <iostream>
#include <typeinfo>
enum MyType
{
A,
B
};
template<MyType>
struct MyStruct {};
template<MyType type>
MyStruct<type> createMyStruct()
{
return {};
}
int main()
{
auto structA = createMyStruct<A>();
auto structB = createMyStruct<B>();
std::cout << typeid(structA).name() << std::endl;
std::cout << typeid(structB).name() << std::endl;
}
I am assuming you want to write code like this:
void foo (MyType t) {
auto x = createMyStruct(t);
//... do something with x
}
You are attempting to derive the right type for x at runtime. However, the return type of a function must be known at compile time, and the type resolution for auto is also determined at compile time.
You could instead restructure your code to be like this:
template<MyType T> struct MyStruct
{
//...
static void foo () {
MyStruct x;
//... do something with x
}
};
The idea is to write a single foo() function whose only difference is the type of thing it is manipulating. This function is encapsulated within the type itself. You can now make a runtime decision if you have a mapping between MyType and MyStruct<MyType>::foo.
typedef std::map<MyType, void(*)()> MyMap;
template <MyType...> struct PopulateMyMap;
template <MyType T> struct PopulateMyMap<T> {
void operator () (MyMap &m) {
m[T] = MyStruct<T>::foo;
}
};
template <MyType T, MyType... Rest> struct PopulateMyMap<T, Rest...> {
void operator () (MyMap &m) {
m[T] = MyStruct<T>::foo;
PopulateMyMap<Rest...>()(m);
}
};
template<MyType... Types> void populateMyMap (MyMap &m) {
PopulateMyMap<Types...>()(m);
}
//...
populateMyMap<A, B>(myMapInstance);
Then, to make a runtime decision:
void foo (MyType t) {
myMapInstance.at(t)();
}
I think you should learn abstract factory design pattern.
For use objects of type MyStruct<A> or MyStruct<B> you need common interface.
Common interface provided in abstract base class.
struct MyStruct
{
virtual ~MyStruct() {}
virtual void StructMethod() = 0;
};
struct MyStructA: public MyStruct
{
void StructMethod() override {}
};
struct MyStructB: public MyStruct
{
void StructMethod() override {}
};
std::unique_ptr<MyStruct> createMyStruct(MyType t)
{
if (t==A)
return std::make_unique<MyStructA>();
else
return std::make_unique<MyStructB>();
}

Dynamically define a function return type

I have a Message class that is able to pack its payload to binary and unpack it back. Like:
PayloadA p;
msg->Unpack(&p);
where PayloadA is a class.
The problem is that I have a bunch of payloads, so I need giant if or switch statement:
if (msg->PayloadType() == kPayloadTypeA)
{
PayloadA p;
msg->Unpack(&p); // void Unpack(IPayload *);
// do something with payload ...
}
else if ...
I want to write a helper function that unpacks payloads. But what would be the type of this function? Something like:
PayloadType UnpackPayload(IMessage *msg) { ... }
where PayloadType is a typedef of a proper payload class. I know it is impossible but I looking for solutions like this. Any ideas?
Thanks.
I would split one level higher to avoid the problem entirely:
#include <map>
#include <functional>
...
std::map<int, std::function<void()> _actions;
...
// In some init section
_actions[kPayloadA] = [](IMessage* msg) {
PayloadA p;
msg->Unpack(&p);
// do something with payload ...
};
// repeat for all payloads
...
// decoding function
DecodeMsg(IMessage* msg) {
_actions[id](msg);
}
To further reduce the code size, try to make Unpack a function template (possible easily only if it's not virtual, if it is you can try to add one level of indirection so that it isn't ;):
class Message {
template <class Payload>
Payload Unpack() { ... }
};
auto p = msg->Unpack<PayloadA>();
// do something with payload ...
EDIT
Now let's see how we can avoid writing the long list of _actions[kPayloadN]. This is highly non trivial.
First you need a helper to run code during the static initialization (i.e. before main):
template <class T>
class Registrable
{
struct Registrar
{
Registrar()
{
T::Init();
}
};
static Registrar R;
template <Registrar& r>
struct Force{ };
static Force<R> F; // Required to force all compilers to instantiate R
// it won't be done without this
};
template <class T>
typename Registrable<T>::Registrar Registrable<T>::R;
Now we need to define our actual registration logic:
typedef map<int, function<void()> PayloadActionsT;
inline PayloadActionsT& GetActions() // you may move this to a CPP
{
static PayloadActionsT all;
return all;
}
Then we factor in the parsing code:
template <class Payload>
struct BasePayload : Registrable<BasePayload>
{
static void Init()
{
GetActions()[Payload::Id] = [](IMessage* msg) {
auto p = msg->Unpack<Payload>();
p.Action();
}
}
};
Then we define all the payloads one by one
struct PayloadA : BasePayload<PayloadA>
{
static const int Id = /* something unique */;
void Action()
{ /* what to do with this payload */ }
}
Finally we parse the incoming messages:
void DecodeMessage(IMessage* msg)
{
static const auto& actions = GetActions();
actions[msg->GetPayloadType]();
}
How about a Factory Method that creates a payload according to the type, combined with a payload constructor for each payload type, taking a message as a parameter?
There's no avoiding the switch (or some similar construct), but at least it's straightforward and the construction code is separate from the switch.
Example:
class PayloadA : public Payload
{
public:
PayloadA(const &Message m) {...} // unpacks from m
};
class PayloadB : public Payload
{
public:
PayloadB(const &Message m) {...} // as above
};
Payload * UnpackFromMessage(const Message &m)
{
switch (m.PayloadType) :
case TypeA : return new PayloadA(m);
case TypeB : return new PayloadB(m);
... etc...
}
I seen this solved with unions. The first member of the union is the type of packet contained.
Examples here: What is a union?
An important question is how the payloads differ, and how they are the same. A system whereby you produce objects of a type determined by the payload, then interact with them via a virtual interface that is common to all types of payload, is reasonable in some cases.
Another option assuming you have a finite and fixed list of types of payload, returning a boost::variant is relatively easy. Then to process it, call apply_visitor with a functor that accepts every type in the variant.
If you only want to handle one type of payload differently, a "call and run the lambda if and only if the type matches T" function isn't that hard to write this way.
So you can get syntax like this:
struct State;
struct HandlePayload
{
typedef void return_type;
State* s;
HandlePayload(State* s_):s(s_) {}
void operator()( int const& payload ) const {
// handle int here
}
void operator()( std::shared_ptr<bob> const& payload ) const {
// handle bob ptrs here
}
template<typename T>
void operator()( T const& payload ) const {
// other types, maybe ignore them
}
}
which is cute and all, but you'll note it is quite indirect. However, you'll also note that you can write template code with a generic type T above to handle the payload, and use stuff like traits classes for some situations, or explicit specialization for others.
If you expect the payload to be one particular kind, and only want to do some special work in that case, writing a single-type handler on a boost::variant is easy.
template<typename T, typename Func>
struct Helper {
typedef bool return_type;
Func f;
Helper(Func f_):f(f_) {}
bool operator()(T const& t) {f(t); return true; }
template<typename U>
bool operator()(U const& u) { return false; }
};
template<typename T, typename Variant, typename Func>
bool ApplyFunc( Variant const& v, Func f )
{
return boost::apply_visitor( Helper<T, Func>(f), v );
}
which will call f on a variant v but only on the type T in the Variant, returning true iff the type is matched.
Using this, you can do stuff like:
boost::variant<int, double> v = 1.0;
boost::variant<int, double> v2 = int(1);
ApplyFunc<double>( v, [&](double d) { std::cout << "Double is " << d << "\n"; } );
ApplyFunc<double>( v2, [&](double d) { std::cout << "code is not run\n"; } );
ApplyFunc<int>( v2, [&](int i) { std::cout << "code is run\n"; } );
or some such variant.
One good solution is a common base class + all payloads inheriting from that class:
class PayLoadBase {
virtual char get_ch(int i) const=0;
virtual int num_chs() const=0;
};
And then the unpack would look like this:
class Unpacker {
public:
PayLoadBase &Unpack(IMessage *msg) {
switch(msg->PayLoadType()) {
case A: a = *msg; return a;
case B: b = *msg; return b;
...
}
}
private:
PayLoadA a;
PayLoadB b;
PayLoadC c;
};
You can make the function return a void *. A void pointer can be cast to any other type.

instantiating a free template function within a template class

I need to instantiate a free template function (FTF) within a template class (TC). The FTF takes as a template parameter one of the template parameters of the TC. The TC also holds generic pointers to these FTF's, and these functions are called through the pointers.
The step of taking a pointer to a FTF is not enough to instantiate it, and I receive linker errors from the GCC toolchain. MSDN illustrates FTF specification as so -- however my instantion of the FTF is dependant on a template parameter of my TC, and therefore the FTF instantiation cannot be placed in free scope.
Is this possible ? I am attaching some basic generated code, the issue is in the constructor of the class test_service, where I assign the pointer of a free function into a custom container. I get a linker error telling me the free function cannot be found (uninstantiated). I know that specifying a call to the template function in the class somewhere will produce a instantiation, however I am only going to be making a call via a pointer.
#include "rpc_common.h"
#include <boost/cstdint.hpp>
namespace rubble { namespace rpc {
struct test_service_dummy_tag{};
template<typename T>
class test_service_skel
{
public:
bool Init() {}
bool TearDown() {}
bool test_one(TestRequest,TestResponse){};
private:
};
template<typename T_IMPL>
bool test_service_test_one(T_IMPL & impl,ClientRequest & request)
{
return 0;
}
template<typename T_IMPL=test_service_skel<test_service_dummy_tag> >
class test_service
{
public:
test_service()
{
// uncomment the following two lines and a instantiation will occur.
// ClientRequest cr;
//test_service_test_one<T_IMPL>(m_impl,cr);
m_dispatch_table.SetEntry( Oid("test_one",0),(void *) & test_service_test_one<T_IMPL>);
}
bool Init() { return m_impl.Init(); };
bool TearDown() { return m_impl.TearDown(); };
private:
T_IMPL m_impl;
OidContainer<Oid,void *> m_dispatch_table;
};
} }
EDIT: self-contained minimal version
class test_skel
{
bool test_function()
{
return true;
}
};
template<typename T>
bool test_function()
{
}
template<typename T = test_skel>
class test
{
public:
test()
{
dispatch = (void *) & test_function<T>;
}
void * dispatch;
};
int main()
{
test<> t;
return 0;
}
There is no problem iff you don't use a void*, i.e.: http://www.ideone.com/eRgUG
However, if you insist on storing the pointer in a void*, then you need to take the address using a specific function pointer first and then cast - e.g.
bool (*temp)() = &test_function<T>;
dispatch = reinterpret_cast<void*>(temp); // YUCK
This gives the compiler enough context to generate the address for you.
Ahh - just saw DeadMG's answer, the function to generate the void* is neater...
Your self-contained example wouldn't compile for me with a strange error about overloaded functions, when there is no overloading going on, with MSVC. I did, however, manage to work around it.
class test_skel
{
bool test_function()
{
return true;
}
};
template<typename T> void* to_void_pointer(T t) {
return reinterpret_cast<void*>(t);
}
template<typename T>
bool test_function()
{
return true;
}
template<typename T = test_skel>
class test
{
public:
test()
{
dispatch = to_void_pointer(&test_function<T>);
}
void * dispatch;
};
int main()
{
test<> t;
return 0;
}
This compiles cleanly. I suspect that whatever behaviour you're seeing and I saw is a compiler error.