I'm a little lost in how to cast templates. I have a function foo which takes a parameter of type ParamVector<double>*. I would like to pass in a ParamVector<float>*, and I can't figure out how to overload the casting operator for my ParamVector class, and Google isn't helping me that much. Does anyone have an example of how to do this? Thanks.
EDIT: Adding some code, sorry I'm an idiot and didn't phrase the original question well at all;
template<class T> class ParamVector
{
public:
vector <T> gnome;
vector <T> data_params;
}
template<class T> class ParamVectorConsumer
{
public:
ParamVector<T> test;
}
ParamVector<float> tester;
ParamVectorConsumer<double> cons;
cons.ParamVector = tester
will fail to compile. I would like to know how to write it so that I can cast the float version of tester to a ParamVector double. Thanks
EDIT2: Casting was the wrong word. I don't mind writing extra code, I just need to know how to get this to be accepted by the compiler so that I can write some sort of conversion code.
I'm not sure but maybe you need some like this:
template< typename TypeT >
struct ParamVector
{
template < typename NewTypeT >
operator ParamVector< NewTypeT >()
{
ParamVector< NewTypeT > result;
// do some converion things
return result;
}
template< typename NewTypeT >
ParamVector( const ParamVector< NewTypeT > &rhs )
{
// convert
}
template < typename NewTypeT >
ParamVector& operator=( const ParamVector< NewTypeT > &rhs )
{
// do some conversion thigns
return *this;
}
};
ParamVector< double > d1;
ParamVector< float > f1;
f1 = d1;
You can choose use conversion operator or operator= - I've provided both in my example.
Well, you can't. Each different actual template parameter, makes an entirely new class, which has no* relation inheritance relation with any any other class, with a diffent actual argument, made from that template.
No relationship. Well, except that each provides the same interface, so that inside a template you can handle then the same.
But neither the static types or the dynamic types have any relation.
Let me drop back here, and explain.
When I declare a pointer to classtype, like
Foo fp*;
fp has what we call a static type, of pointer-to Foo. If class Bar is a subclass of Foo, and I point fp at new Bar:
fp = new Bar1();
then we say that the object pointed to by fp has the dynamic type of Bar.
if Bar2 also publicly derives from Foo, I can do this:
fp = new Bar2();
and without ever even knowing what fp points to, I can call virtual methods declared in Foo, and have the compiler make sure that the method defined in he dynamic type pointed to is what's called.
For a template< typename T > struct Baz { void doSomething(); };
Baz<int> and Baz<float> are two entirely different class types, with no relationship.
The only "relationship" is that I can call doSomething() on both, but since the static types have no relationship, if I have a Baz<int> bi*, I can't point it to a Baz<float>. Not even with a cast. The compiler has no way to "translate" a call to the Baz doSotheing method into a call to a Baz::doSomething() method. That's because there is no "Baz method", there is no Baz, there are ony Baz<int>s and Baz<float>s, and Baz<whatevers>, but there's no common parent. Baz is not a class, Baz is a template, a set of instructions about how to make a class if and only if we have a T parameter that's bound to an actual type (or to a constant).
Now there is one way I can treat those Bazes alike: in a template, they present the same interface, and the compiler, if it knows what kind of Baz we're really dealing with, can make a static call to that method (or a static access of a member variable).
But a template is not code, a template is meta-code, the instructions of how to synthesize a class. A "call" in a template is not a call,it's an instruction of how to write the code to make a call.
So. That was long winded and confusing. Outside of a template definition, there is no relationship between a ParamVector and aParamVector. So your assignment can't work.
Well. Almost.
Actually, with partial application of templates, you can write a template function which gives a "recipe" of how to transform a Paramvector<T> to a ParamVector<U>. Notice the T and the U. If you can write code to turn any kind of ParamVector, regardless of actual template parameter into any other kind of ParamVector, you can package that up as a partially applied template, and the compiler will add that function to, for example, ParamVector.
That probably involves making a ParamVector<U>, and transforming each T in the ParamVector<T> into a U to put in the ParamVector<U>. Which still won't let you asign to a ParamConsumer<T>.
So maybe you want to have both templates and inheritance. In that case, you can same that all ParamVectors regardless of type inherit from some non-template class. And then there would be a relationship between ParamVectors, they'd all be sibling subclasses of that base class.
Notice that when you do an implicit cast, what the compiler can do without your help (I mean, without additional code) is just reference-upcast. That means that, seeing the object as a reference (for cast purposes only, the nature of the object doesn't change of course), it can look at it as one of its ancestors. When you have two template instances, none of them is an ancestor of the other (neither they are necessarily in the same hierarchy).
After trying that, the compiler looks for cast operators, constructors, etc. At this stage, probably a temporary object needs to be created, except when you're doing attribution and there's an attribution operator that fits.
One solution to your problem would be to use a conversion constructor:
template<class T> class ParamVector
{
public:
vector <T> gnome;
vector <T> data_params;
ParamVector()
{
}
template <class T2> ParamVector(const ParamVector<T2> &source)
{
gnome.reserve(source.gnome.size());
copy(source.gnome.begin(), source.gnome.end(), gnome.begin());
data_params.reserve(source.data_params.size());
copy(source.data_params.begin(), source.data_params.end(), data_params.begin());
}
};
This would create a temporary object whenever you use an instance of the template and other is required. Not a good solution if you're dealing with large containers, the overhead isn't acceptable. Also, if you pass a template instance to a function that requires not an object but a reference, the compiler won't call the conversion constructor automatically (you have to do an explicit call).
You are lost because you can't do it - the two types are completely different. Whenever you come across the need for a cast in your code, you should examine both your code and your design very closely - one or both is probably wrong.
You can't do this with a direct cast because double and float are completly different sizes. Doubles are going to be 64 bits while floats are 32. A pointer forced to cast from a
ParamVector<float>
to
ParamVector<double>
is going to misinterpret the data and give you garbage. You may want to google "pointer aliasing" or just learn more about pointers in general to see how this isn't going to work.
Think about it for a second you have one array that is a bunch of 64 bit values with fields layed out like this
0 => abcdabcd12341234
1 => abcdabcd12341234
If you force this to be interpreted as an array of 32 bit values, its going to not be interpreted correctly. You may or may not get something like
0 => abcdabcd
1 => 12341234
2 => abcdabcd
3 => abcdabcd
or it could be switched so that the 12341234's come first, or something stranger due to how the word ordering works out.
You mentioned "template casting" in your headline, so I'll presume that ParamVector is a templated type. That means that foo could be templated as well, and that would solve your problem.
template <typename T>
void foo(ParamVector<T> const& data)
{
}
You can't cast templates like this because the types are unrelated.
However, you can add a conversion function, such as:
(Your code wasn't really complete, so I can post complete code either. Hopefully you will get the idea.)
template<class T> class ParamVectorConsumer
{
public:
ParamVector<T> test;
template<T2> ParamVectorConsumer<T2> convert()
{
ParamVectorConsumer<T2> ret;
ret = this->...
}
Related
I'm working on a program where some data is statically allocated and some is dynamically allocated. Now I want to have another type that can be called with any template of the type as its argument.
#include <array>
#include <vector>
template <int size> class Foo {
std::array<int, size> data;
public:
int& operator[](std::size_t idx) {return data[idx];}
};
template <> class Foo<-1> {
std::vector<int> data;
public:
int& operator[](std::size_t idx) {return data[idx];}
};
// option 1- polymorphism
struct FooCaller {
virtual void operator()(Foo data) = 0; // how would I make this work with both forms of Foo?
};
// option 2- generic programming
template <class T> concept CanCallFoo = requires (const T& t) {
t(std::declval<Foo&>()); // how do I ensure that this can call any overload of Foo?
};
Both methods would be fine, but I'm not sure how to go about this. Because the full code is more complex, I'd rather not have both Foos inherit from a base.
A callable F could write a restriction that it can be called by Foo<x> such that an arbitrary function of x must be true to be valid.
In order for your "can be called with any Foo" test to work, you would have to invert an arbitrary function at compile time.
There is no practical way to do this short of examinjng all 2^32 possible values of x. No, really. The problem you have is that the type F is possibly too powerful for you to determine its properties. This is related to Halt and Rice's theorem and the fact that template metaprogramming and C++ overload resolution is Turing complete (only related, because 2^32 is finite).
In the other case, you could type erase. Write a type RefAnyFoo/AnyFooValue that can be constructed from any Foo type and "type erases" the operations you want, like how std::function<void(int)> type erases. It could either duck type the Foo or actually restrict to instsnces of types made from the template.
Then your interface is one that takes a AnyFooValue/RefAnyFoo (depending if you are talking about copies or references).
Now, flipping this over, you can write a concept that accepts F that take RefFooAny. This isn't quite what you asked for, but plausibly you have an X/Y problem; you had a real problem, came up with incomplete solutions, then asked about how to get your solutions working, instead of the original problem.
Similarly, it is possible you only care about certain ducktype properties of Foo<?> and not the specific type. Then a concept that checks those ducktypes could be used by your callable; this again deosn't solve your problem, as much as it flips it upside down, because you cannot verify a callable accepts an entire concept from outside the callable.
Say an Object has a templated TYPE (for a very good reason)
template <typename T> class Object {
};
I want an Object to be able to contain an array of subobjects.
template <typename T> class Object {
vector< Object<T>* > subobjects;
};
However, the subobjects CAN BE OF ANY TYPE. That is, a subobject can be an Object of a different T than it’s parent.
The language is enforcing a restriction that doesn’t need to be there. How do I fix this?
A template is a type-factory. Object<int> and Object<double> are unrelated types.
Object<int> generates one type, and Object<double> generates a different type.
If you want these two types to be related, you can make them related.
struct IObject {
virtual ~IObject() {}
// common interface
};
then
template<class T>
struct Object<T>:IObject {
std::vector<IObject*> subObjects;
};
now your Object<T> has subobjects. All the code knows (statically) is that they are IObjects.
Now, if you want the type T to be runtime determined, you could use std::any (this has type safety issues, sort of; it is "brittle" in that the reader needs to know the exact type that went into it) or std::variant<double, int, std::string> (but here you have to name all of the supported types).
Basically, you are probably using a round peg and a square hole.
You have a problem. You thought templates where the solution. You found a problem with your solution. You looked for a way to make templates work.
The language is enforcing a restriction that doesn’t need to be there. How do I fix this?
The language isn't enforcing anything. You wrote a template with one type parameter, so that parameter can only take one value at a time.
However, the subobjects CAN BE OF ANY TYPE. That is, a subobject can be an Object of a different T than it’s parent.
If you mean each Object<T> can have children of a single different type U != T, just write
template <typename T, typename U> class Object {
vector< Object<U>* > subobjects;
};
If you mean each object can have children of a variety of instantiations, you either need some form of type erasure (eg. std::any), or polymorphism.
In either case, you need to give some thought to what you're going to do with these things you don't know the concrete type of. How will you use them? This is the information needed to choose the appropriate mechanism.
Using #WayneVanderLinde's comment:
struct Object{
vector<Object*> subobjects;
};
template<typename T> struct TObject : public Object {};
you can now create
TObject<int> i;
TObject<double>* d = new TObject<double>();
i.subobjects.push_back( d );
The example below is a minimal, maybe not so good example of a well known idiom.
It compiles and it is so ugly in order to be able to maintain it minimal, because the question is not about the idiom itself.
struct Foo {
virtual void fn() = 0;
};
template<class T>
struct Bar: public Foo {
void fn() override {
T{}.fn();
}
};
struct S {
void fn() { }
};
int main() {
Foo *foo = new Bar<S>{};
foo->fn();
}
What I'm struggling with since an hour ago is how to change it (or even, if there exists an alternative idiom) to introduce a variadic template member method.
Obviously, I cannot modify the fn function of the Foo class, because it's a virtual one and virtual specifier doesn't goes along with templates. The same is valid for the fn specification of Bar, because it has to override somehow the one in the base class.
Note.
For I strongly suspect that this question could be one of the greatest XYProblem ever seen, I'd like also to give a brief description of the actual problem.
I have a class that exposes two templated member methods:
the first one accepts a template class T that is not used immediately, instead it should be stored somehow in order to be used later.
the second one accepts a variadic number of arguments (it is actually a variadic templated member function) and those arguments should be perfectly forwarded to a newly created instance of T.
Well, the problem is far more complex, but this is a good approximation of it and should give you an idea of what's the goal.
Edit
I guess that it is somehow similar to higher order functions.
I mean, what would solve the problem is indeed a templated function to which to bind the first argument, but as far as I know this is impossible as well as any other approach I've explored so far.
Any viable solution that expresses the same concept?
What I mentioned in the comments is the following approach:
template<typename T> class Factory {
public:
template<typename ...Args>
auto construct(Args && ...args)
{
return T(std::forward<Args>(args)...);
}
};
So now, your first exposed class method will be something like this:
template<typename T>
auto getFactory() {
return Factory<T>();
}
So:
auto factory=object.getFactory<someClass>();
// Then later:
factory.construct(std::string("Foo"), bar()); // And so on...
Instead of construct() you could use operator() too, so the second part of this becomes, simply:
factory(std::string("Foo"), bar()); // And so on...
As I mentioned, this is not really type erasure. You can't use type erasure here.
Having given this a few minutes' more thought, the reason that type erasure cannot be used here is because a given instance of type erasure must be "self contained", or atomic, and what you need to do is to break atomic type erasure into two parts, or two class methods, in your case.
That won't work. Type erasure, by definition, takes a type and "erases" it. Once your first function type-erases its class method template parameter, what you end up with is an opaque, type-erased object of some kind. What was type-erased is no longer available, to the outside world. But you still haven't type-erased your constructor parameters, which occurs somewhere else.
You can type-erase the template class, and the constructor parameters together. You can't type-erase the template class, and the constructor parameters, separately and then somehow type-erase the result again.
The simple factory-based approach, like the one I've outlined, would be the closest you can get to results that are similar to type erasure, if both halfs of your desired type-erasure appear in the same scope, so you can actually avoid type-erasure, and instead rely on compiler-generated bloat.
I also agree that you cannot do exactly what you want here. I will post what I think the closest option is (at least a close option that is different from SamVarshavchik's answer).
I don't expect this answer to solve your problem exactly, but hopefully it will give you some ideas.
struct Delay // I have no idea what to call this
{
template <class T>
void SetT()
{
function_ = [](boost::any params){return T(params);}
}
template <class ... Args>
boost::any GetT(Args ... args)
{
return function_(std::make_tuple(args...));
}
private:
std::function<boost::any(boost::any)> function_;
};
The obvious limitation of this is that anyone calling GetT will somehow have to know what T was already, though you can query the boost::any object for the type_info of its class if that helps. The other limitation here is that you have to pass in T's that take a boost::any object and know what to do with it. If you cannot have T do that, then you can change SetT (or create a new member function) like this:
template <class F>
SetTFactory(F f)
{
function_ = f;
}
and then use it like:
Delay d;
d.SetTFactory([](boost::any s){return std::string(boost::any_cast<const char*>(s));});
auto s = d.GetT("Message");
assert(s.type() == typeid(std::string));
This of course introduces a whole new set of difficulties to deal with, so I don't know how viable this solution will be for you. I think regardless of anything else, you're going to have to rethink your design quite a bit.
I am working on a fairly tightly coupled library which up until now has explicitly assumed all computations are done with doubles. I'm in the process of converting some of the core classes to templates so that we can start computing with std::complex<double>. I've templated about 10 of our classes so far have noticed a tendency toward proliferation of templates. As one class becomes templated, any other class that uses the templated class appears to need templating as well. I think I can avoid some of this proliferation by defining abstract base classes for my templates so that other classes can just use pointers to the abstract class and then refer to either a double or std::complex<double> version of the derived class. This seems to work on at the header level, but when I dive into the source files, the templated class will often have functions which compute a value or container of values of type double or std::complex<double>. It seems like a waste to template a whole class just because a couple of lines in the source file are different because of some other classes return type.
The use of auto seems like a possible way to fix this, but I'm not 100% sure it would work. Suppose I have an abstract base class AbstractFunction from which Function<Scalar> derives, where Scalar can be double or std::complex<double>. Now suppose we have two member functions:
virtual Scalar Function<Scalar>::value(double x);
virtual void Function<Scalar>::values(std::vector<Scalar> &values, std::vector<double> x);
And suppose I have some other class (that I don't want to template) with a member function that calls one of these.
// populate double x and std::vector<double> xs
auto value = functionPtr->value(x);
std::vector<auto> values;
functionPtr->values(values, xs);
// do something with value and values
where functionPtr is of type std::shared_ptr<AbstractFunction>.
I could see auto working for the first case, but I don't believe I could construct a vector of auto to be filled with the second one. Does this necessitate making the calling class a template? Can someone recommend another strategy to cut down on the proliferation of templates?
I think you are already wrong in assuming that the first use-case is going to work. If you have an abstract base class, then either value is a member of it and you can call it through std::shared_ptr<AbstractFunction> or value is not a member of it and only available if you know the derived class' type. In the first case, the AbstractFunction::value method must have a fixed return type, it can not depend on Scalar, which is the template parameter of the derived class.
That said: In my experience the two concept often don't mix well. You either want to create an abstract base class with the full interface or you want a template. In the latter case, there is often no need / no benefit for having an abstract base class. It then follows that also the code using your template works with templates.
What might help you is to "export" the template parameter from Function, i.e.
template<typename T>
class Function
{
public:
using value_type = T;
value_type value() const;
// ...
};
and in other parts of the code, use a template which takes any T which behaves like Function if you don't want to directly write out (and limit yourself) to Function:
template<typename T>
void something( const std::shared_ptr<T>& functionPtr )
{
// ignoring where x comes from...
using V = typename T::value_type;
V value = functionPtr->value(x);
std::vector<V> values;
functionPtr->values(values, xs);
}
Note that this is just one option, I don't know if it is the best option for your use-case.
So, I have a single templated class, B, which inherits from a non-template class A. Now, this templated version B, has a set of four valid and known template classes, call them B<uint8_t>, B<uint16_t>, B<uint32_t>, B<uint64_t>.
Currently, I am storing these in a map, boost::unordered_map<std::string, A>, and I would like to create some way to cast to the different types. I can easily differentiate based on a A::width() method which returns 8, 16, 32, or 64 respectively.
I have been racking my brain for a good design but I can't figure out a way to get a better solution than having every use casting themselves.
Here is a minimum working case:
class A {
std::size_t width() const = 0;
};
template <typename value> class B : public A {
std::size_t width() const {
return sizeof(value) * 8;
}
};
Within usage:
boost::unordered_map<std::string, A *> _bmap;
template <class T> T cast_A(A * const a) {
switch (a->width()) {
case 8: return dynamic_cast<B<uint8_t> *>(a);
// removed for brevity
}
The problem with this is that it forces the user to check the runtime type of the A class after, aka making the cast method absolutely useless.
I welcome any input. :)
You could use, e.g., a templated get() method which converts to the asked for size via a virtual method going through the largest supported type. If the conversion would loose bits (checked at run-time) you could produce an error.
I think the downcasting is not very good idea in this case. It seems to be a design problem.
However, you could try to implement something like visitor pattern for these classes. It may solve your problem, but I'm not sure, because there are no use examples of B objects.