How to keep track of Eigen objects through a solve()? - c++

This question is related to cast from Eigen::CwiseBinaryOp to MatrixXd causes segfault .
It will probably have as simple a solution as the former.
In this minimal example, I define Holder, which holds and Eigen matrix, and returns it via its get() member function. Similarly, Decomp is an expression template for the LDLT decomposition of this matrix, and Solve solves for AX=B, yielding X.
#include <Eigen/Dense>
#include <Eigen/Cholesky>
template <class EigenType> class Holder {
public:
typedef EigenType result_type;
private:
result_type in_;
public:
Holder(const EigenType& in) : in_(in) {}
const result_type& get() const { return in_; }
};
template <class Hold> class Decomp {
public:
typedef typename Eigen::LDLT
<typename Eigen::MatrixBase<typename Hold::result_type>::PlainObject>
result_type;
private:
Hold mat_;
public:
Decomp(const Hold& mat) : mat_(mat) {}
result_type get() const { return mat_.get().ldlt(); }
};
template <class Derived, class OtherDerived> class Solve {
public:
typedef typename Eigen::internal::solve_retval
<typename Derived::result_type, typename OtherDerived::result_type>
result_type;
private:
Derived decomp_;
// typename Derived::result_type decomp_;
OtherDerived mat_;
public:
Solve(const Derived& decomp, const OtherDerived& mat)
: decomp_(decomp), mat_(mat) {}
//: decomp_(decomp.get()), mat_(mat) {}
result_type get() const { return decomp_.get().solve(mat_.get()); }
// result_type get() const { return decomp_.solve(mat_.get()); }
};
typedef Holder<Eigen::MatrixXd> MatrixHolder;
typedef Decomp<MatrixHolder> MatrixDecomp;
typedef Solve<MatrixDecomp, MatrixHolder> SimpleSolve;
The following test fails on X.get()
#include "Simple.h"
#include <Eigen/Dense>
#include <iostream>
int main(int, char * []) {
MatrixHolder A(Eigen::MatrixXd::Identity(3, 3));
MatrixHolder B(Eigen::MatrixXd::Random(3, 2));
MatrixDecomp ldlt(A);
SimpleSolve X(ldlt, B);
std::cout << X.get() << std::endl;
return 0;
}
but if you use the commented out lines in the header file, everything works fine. Unfortunately, this moves the evaluation of the decomposition to the construction of the solver, which is not suited to my use. Typically, I want to build a complicated expression expr involving this Solve, and call expr.get() later on.
How can I solve this problem? Is there a general rule to follow, to avoid further related questions?

To avoid useless and expensive copies, the internal solve_retval structure stores the decomposition and right-hand-side by const references. However, the LDLT object created in the Decomp::get function is deleted at the same time this function returns, and thus the solve_retval object references dead objects.
One possible workaround, is to add a Decomp::result_type object in Solve, and initialize it in Solve::get. Moreover, to avoid multiple deep copies, I suggest to use const references for a few attributes as follow:
#include <Eigen/Dense>
#include <Eigen/Cholesky>
template <class EigenType> class Holder {
public:
typedef EigenType result_type;
private:
result_type in_;
public:
Holder(const EigenType& in) : in_(in) {}
const result_type& get() const { return in_; }
};
template <class Hold> class Decomp {
public:
typedef typename Eigen::LDLT
<typename Eigen::MatrixBase<typename Hold::result_type>::PlainObject>
result_type;
private:
const Hold& mat_;
mutable result_type result_;
mutable bool init_;
public:
Decomp(const Hold& mat) : mat_(mat), init_(false) {}
const result_type& get() const {
if(!init_) {
init_ = true;
result_.compute(mat_.get());
return result_;
}
}
};
template <class Derived, class OtherDerived> class Solve {
public:
typedef typename Eigen::internal::solve_retval
<typename Derived::result_type, typename OtherDerived::result_type>
result_type;
private:
const Derived& decomp_;
const OtherDerived& mat_;
public:
Solve(const Derived& decomp, const OtherDerived& mat)
: decomp_(decomp), mat_(mat) {}
result_type get() const {
return decomp_.get().solve(mat_.get());
}
};
The general rule is to store heavy objects by const reference (to avoid deep copies) and lightweight expressions by value (to reduce temporary life issues).

Related

Using templates to set whether class members are const

I have a templated class and I'd like to almost toggle whether things within it are const based on the template type.
pseudocode:
template<bool isConst>
Class ConstableClass
{
public:
// if isConst == true make this method const
void DoSomething() "const";
// if isConst == true make the returned smartpointer type const method const
std::unique_ptr<"const" int> operator->() "const";
private:
// if isConst == true add const at the start of the next line
"const" int foo;
}
Is this sort of thing possible?
With type-traits and SFINAE:
template <bool isConst>
class ConstableClass
{
public:
template <bool C = isConst, typename = std::enable_if_t<C>>
void DoSomething() const;
template <bool C = isConst, typename = std::enable_if_t<!C>>
void DoSomething();
template <bool C = isConst, typename = std::enable_if_t<C>>
std::unique_ptr<const int> operator->();
template <bool C = isConst, typename = std::enable_if_t<!C>>
std::unique_ptr<int> operator->();
private:
std::conditional_t<isConst, const int, int> foo{};
};
With a specialization:
template <bool isConst>
class ConstableClass
{
public:
void DoSomething() const;
std::unique_ptr<const int> operator->() const;
protected:
const int foo{};
};
template <>
class ConstableClass<false>
{
public:
void DoSomething();
std::unique_ptr<int> operator->();
protected:
int foo{};
};
With constraints in c++20:
template <bool isConst>
class ConstableClass
{
public:
void DoSomething() const requires isConst;
void DoSomething() requires not isConst;
std::unique_ptr<const int> operator->() const requires isConst;
std::unique_ptr<int> operator->() requires not isConst;
private:
std::conditional_t<isConst, const int, int> foo{};
};
The first use case is not possible. You can't make function modifiers conditional and based on some template. However, I imagine you are trying to do this because you don't want to copy&paste code between the const and regular version of a function. In that case, just write a private impl method which does the actual work and use it in the cost and non-const version of the class.
private:
int& get_impl() const {...}
public:
const int& get() const {return get_impl();}
int& get() {return get_impl();}
The rest is possible and quite simple:
std::unique_ptr<"const" int>
// we can do this by:
std::unique_ptr<std::conditional_t<isConst, const int, int>> ...
// this can be written more elegantly as
template <typename T, bool isConst>
using const_if_t = std::conditional_t<isConst, const T, T>;
std::unique_ptr<const_if_t<int, isConst>>;
Making member variables const conditionally would be pretty pointless because this already happens when you make a variable const anyways.
No, it is not possible the way you want to do it. While there is a way to manipulate member variable types and function return types, there is no way to add/remove const specifier to a method. This means you have to write 2 separate specializations of the class definition for the true and false values of the template parameter.
I have a templated class and I'd like to almost toggle whether things within it are const based on the template type. Is this sort of thing possible?
Yes, and I find specializations to be more readable (unless you have access to concepts). The example below shows, rather explicitly, the two cases.
template<bool isConst>
class ConstableClass;
template<>
class ConstableClass<false>
{
int foo;
public:
void DoSomething();
std::unique_ptr<int> operator->();
};
template<>
class ConstableClass<true>
{
int const foo;
public:
ConstableClass()
: foo{0} {} // example
void DoSomething() const;
std::unique_ptr<int const> operator->() const;
};
Keep in mind that you do need to add a constructor for the true case, as adding const to foo implicitly deletes it.

Use reference template arguments in inner class

This is my code:
template<class V, class Ref = V&>
class Util {
public:
typedef Ref reference;
};
template<class D, class V, class Ref = V&>
class Base {
public:
class Inner1: public Util<const V, const Ref> {
public:
Inner1(const D &d) :
d(d) {
}
typename Inner1::reference foo() const {
static V v;
return v;
}
private:
const D &d;
};
class Inner2: public Util<V, Ref> {
public:
Inner2(D &d) :
d(d) {
}
typename Inner2::reference foo() const {
static V v;
return v;
}
private:
D &d;
};
};
class Child: public Base<Child, float> {
public:
Inner1 getInner1() const {
return Base<Child, float>::Inner1(*this);
}
Inner2 getInner2() {
return Base<Child, float>::Inner2(*this);
}
};
void print(float & ff) {
}
int main() {
Child c;
Child::Inner1 inner = c.getInner1();
print(inner.foo());
return 0;
}
This code compile without problems but I guess I should receive a compilation error. The method foo of class Inner1 should return a const reference, but for unknown reason Ref is defined without const. Why?
but for unknown reason Ref is defined without const. Why?
Leading const is misleading. In the type Util<const V, const Ref> you const qualify the Ref type. I.e., you supposedly pass V & const as the type, and not V const &. While normally V & const would be ill-formed, when dealing with the type indirectly, such as when it's aliased or a template argument, the extra cv-qualifier is simply ignored.
Personally, I would use the extra template parameter for an entire trait.
template<class V>
class VTrait {
public:
typedef V & reference;
typedef V const & const_reference;
};
template<class D, class V, class Trait = VTrait<V>>
class Base {
public:
class Inner1: public Util<const V, typename Trait::const_reference> {
// ...
};
// ...
};
Client code that wants something different from your framework will need only customize the Trait.
const Ref, given Ref is V&, i.e. float&, note that const is qualified on the reference itself, not the type being referenced. The const qualifier is just omitted, you're still getting V&, but not const V& as you expected. (To be precise there's no const references in C++ but only references to const or to non-const.)
If you change const Ref to const V& (i.e. reference to const), you'll get the error you expected. e.g.
class Inner1: public Util<const V, const V&> {
...

C++ template class operator overloading with the same signatures

A simple C++ OO question regrading templates and operator overloading: In the following class, I have overloaded the index operator twice:
template<class A, class B>
class test
{
A a1;
B a2;
public:
A& operator[](const B&);
B& operator[](const A&);
};
Now, if I instantiate an object of this template class with the same typenames:
test<int, int> obj;
calling the index operator will result in an error, because the two overloaded functions will have the same signatures.
Is there any way to resolve this issue?
Sorry, if this is a basic question. I am still learning!
You can add a partial specialization:
template<class A>
class test<A, A>
{
A a1, a2;
public:
A& operator[](const A&);
};
You can avoid this issue and make the code more robust and expressive by converting the index to some other type that clarifies what the user wants. Usage would be like this:
bidirectional_map<int, int> myTest;
int& valueFor1 = myTest[Key{1}];
int& key1 = myTest[Value{valueFor1}];
Implemented like this:
template<class TKey>
struct Key { const TKey& key; };
template<class TValue>
struct Value { const TValue& value; };
// Deduction guides (C++17), or use helper functions.
template<class TValue>
Value(const TValue&) -> Value<TValue>;
template<class TKey>
Key(const TKey&) -> Key<TKey>;
template<class TKey, class TValue>
class bidirectional_map
{
TKey a1; // Probably arrays
TValue a2; // or so?
public:
TValue & operator[](Key<TKey> keyTag) { const TKey & key = keyTag.key; /* ... */ }
TKey & operator[](Value<TValue> valueTag) { const TValue& value = valueTag.value; /* ... */ }
};
Now, Key and Value are popular names so having them "taken up" by these auxiliary functions is not the best. Also, this is all just a pretty theoretical exercise, because member functions are of course a much better fit for this task:
template<class TKey, class TValue>
class bidirectional_map
{
TKey a1; // Probably arrays
TValue a2; // or so?
public:
TValue& getValueForKey(const TKey& key) { /* ... */ }
TKey& getKeyForValue(const TValue& value) { /* ... */ }
};
In C++2a, you might use requires to "discard" the function in some case:
template<class A, class B>
class test
{
A a1;
B a2;
public:
A& operator[](const B&);
B& operator[](const A&) requires (!std::is_same<A, B>::value);
};
Demo
Here is an example solution using if constexpr that requires C++17:
#include <type_traits>
#include <cassert>
#include <string>
template <class A, class B>
class test
{
A a1_;
B b1_;
public:
template<typename T>
T& operator[](const T& t)
{
constexpr bool AequalsB = std::is_same<A,B>();
constexpr bool TequalsA = std::is_same<T,A>();
if constexpr (AequalsB)
{
if constexpr (TequalsA)
return a1_; // Can also be b1_, same types;
static_assert(TequalsA, "If A=B, then T=A=B, otherwise type T is not available.");
}
if constexpr (! AequalsB)
{
constexpr bool TequalsB = std::is_same<T,B>();
if constexpr (TequalsA)
return a1_;
if constexpr (TequalsB)
return b1_;
static_assert((TequalsA || TequalsB), "If A!=B, then T=A || T=B, otherwise type T is not available.");
}
}
};
using namespace std;
int main()
{
int x = 0;
double y = 3.14;
string s = "whatever";
test<int, int> o;
o[x];
//o[y]; // Fails, as expected.
//o[s]; // Fails, as expected
test<double, int> t;
t[x];
t[y];
//t[s]; // Fails, as expected.
return 0;
};

C++ template class overload [] operator

I would like to overload the [] operator for a template class in respect with the template parameters. Like so:
template<
typename T,
template<typename> class Property,
template<typename> class Key1,
template<typename> class Key2>
class a_map
{
public:
const Property<T>& operator[](const Key1<T>& k) const
{ return _values[k.index()]; }
const Property<T>& operator[](const Key2<T>& k) const
{ return _values[k.index()]; }
protected:
std::vector<Property<T> > _values;
};
I would use this class like so:
int main()
{
a_map<float, prop, key_a, key_b> pm;
}
Basically I want to be able to access the elements inside the _values vector without having to worry to much about the Key types. All that matters is that they have an index() member.
However I get the following error
error C2535: 'const Property &a_map::operator
[](const Key1 &) const' : member function already defined or
declared
even though key_a and key_b are two totaly different types class templates.
Am I missing something? Is the compiler afraid that in certain situation Key1<T> and Key2<T> might actually be the same type?
Edit
These are the class templates used in main
template<typename T>
struct prop
{
T weight;
T height;
};
template<typename T>
class key_a
{
public:
int index() { return _i; }
private:
int _i;
};
template<typename T>
class key_b
{
public:
int index() { return 3; } // Always return 3
Edit
I'm using MVC++ 2008 compiler.
Since both of your operator[] are the same except for the argument type, why not template them?
template <typename TT>
const Property<T>& operator[](const TT& k) const
{
return _values[k.index()];
}
You need to declare your index() functions as const because inside the a_map template you are invoking them through const objects
template<typename T> class key_a {
public:
int index() const // <- `const` is necessary
{ return _i; }
private:
int _i;
};
template<typename T> class key_b {
public:
int index() const // <- `const` is necessary
{ return 3; }
};
But otherwise, everything compiles and work fine for me.
Tried it in VS2010 compiler and get the same error as yours. This is obviously a compiler bug in MSVC++ compilers. Handling of template-template arguments is implemented incorrectly.
To work around the problem I was able to use this technique
template<
typename T,
template<typename> class Property,
template<typename> class Key1,
template<typename> class Key2>
class a_map
{
public:
const Property<T>& operator[](const typename Key1<T>::self& k) const
{ return _values[k.index()]; }
const Property<T>& operator[](const typename Key2<T>::self& k) const
{ return _values[k.index()]; }
protected:
std::vector<Property<T> > _values;
};
and define the key templates as
template<typename T> class key_a {
public:
typedef key_a self;
int index() const { return _i; }
private:
int _i;
};
template<typename T> class key_b {
public:
typedef key_b self;
int index() const { return 3; }
};
This is inelegant, but it makes it work correctly with MSVC++ compilers.
Compiles fine like this... note how Prop/K1/K2 should be defined.
#include <vector>
template<
typename T,
template<typename> class Property,
template<typename> class Key1,
template<typename> class Key2>
class a_map
{
public:
const Property<T>& operator[](const Key1<T>& k) const
{ return _values[k.index()]; }
const Property<T>& operator[](const Key2<T>& k) const
{ return _values[k.index()]; }
protected:
std::vector<Property<T> > _values;
};
template <typename T> struct K1 { int index() const { return 0; } };
template <typename T> struct K2 { int index() const { return 0; } };
template <typename T> struct Prop { };
int main()
{
a_map<float, Prop, K1, K2> m;
}

Using STL algorithms (specifically std::sort) from within a templated class

I've declared a template class MyContainer as bellow, then created an instance of it of type DataType1. The DataType1 class provides a friend function "DataSpecificComparison" which is used by std::sort to compare DataType1 objects. The program compiled and sorted correctly.
I then defined a class called DataType2, gave it a friend implementation of "DataSpecificComparison" and used it to create another instance of MyContainer.
I am now unable to compile the program as a "C2914: 'std::sort' : cannot deduce template argument as function argument is ambiguous" compile time error is reported.
How can a developer specify that the DataSpecificComparison binary predicate is to take arguments of template type T*? Or is there another way around this issue?
template <class T>
class MyContainer
{
private:
vector<T*> m_vMyContainerObjects;
....
public:
....
void SortMyContainerObjects()
{
std::sort(m_vMyContainerObjects.begin(), m_vMyContainerObjects.end(), DataSpecificComparison)
}
}
class DataType1
{
....
friend bool DataSpecificComparison(const DataType1 * lhs, const DataType1 * rhs)
}
class DataType2
{
....
friend bool DataSpecificComparison(const DataType2* lhs, const DataType2* rhs)
}
You can use a temporary local function pointer variable of the required type to select the correct overload of DataSpecificComparison:
void SortMyContainerObjects()
{
typedef bool (*comparer_t)(const T*, const T*);
comparer_t cmp = &DataSpecificComparison;
std::sort(m_vMyContainerObjects.begin(), m_vMyContainerObjects.end(), cmp);
}
Here the compiler can deduce that you want to use the DataSpecificComparison overload that matches the comparer_t type, which resolves the ambiguity.
sth already gave a correct answer, but there's also a direct alternative based on the same principle:
void SortMyContainerObjects()
{
std::sort(m_vMyContainerObjects.begin(), m_vMyContainerObjects.end(),
static_cast<bool (*comparer_t)(const T*, const T*)>(&DataSpecificComparison));
}
This uses essentially the same mechanism. The cast forces overload resolution to happen before the Template Argument Deduction for std::sort.
template<typename T>
struct DataSpecificComp : public binary_function<T, T, bool>
{
public:
bool operator()(const T* lhs, const T* rhs)
{
return *lhs < *rhs;
}
};
call the sort function as shown below:
sort(vi.begin(), vi.end(), DataSpecificComp<int>());
I'd prefer something along the following lines: by default it compares objects with less_than (so you wouldn't have to remember to provide a function with a funny name), and there's an overload that allows giving your own comparison functor (again, value-based):
#include <vector>
#include <algorithm>
#include <functional>
template <class T, class Func>
struct indirect_binary_call_type: public std::binary_function<const T*, const T*, bool>
{
Func f;
indirect_binary_call_type(Func f): f(f) {}
bool operator()(const T* a, const T* b) const
{
return f(*a, *b);
}
};
template <class T, class Func>
indirect_binary_call_type<T, Func> indirect_binary_call(Func f)
{
return indirect_binary_call_type<T, Func>(f);
}
template <class T>
class MyContainer
{
private:
std::vector<T*> m_vMyContainerObjects;
public:
void Sort()
{
Sort(std::less<T>());
}
template <class Func>
void Sort(Func f )
{
std::sort(m_vMyContainerObjects.begin(), m_vMyContainerObjects.end(), indirect_binary_call<T>(f));
}
};
int main()
{
MyContainer<int> m;
m.Sort();
m.Sort(std::greater<int>());
}
Did you try defining DataSpecificComparison as template with bunch of specializations and giving it the type?
template<T>
bool DataSpecificComparison(const T* t1, const T* t2)
{
// something non compilable here
}
template<> bool DataSpecificComparison<Data1>(const Data1* t1, const Data1* t2)
{
// return *t1 < *t2;
}
....
void SortMyContainerObjects()
{
std::sort(m_vMyContainerObjects.begin(), m_vMyContainerObjects.end(), DataSpecificComparison<T>)
}
....
Templating DataSpecificComparison should work. You can also specifically call the proper std::sort template, but it's a bit cumbersome:
template <class T>
class MyContainer
{
private:
vector<T*> m_vMyContainerObjects;
typedef bool (*compsT)(T, T);
public:
....
void SortMyContainerObjects()
{
std::sort<std::vector<T*>::iterator, compsT>(m_vMyContainerObjects.begin(), m_vMyContainerObjects.end(), DataSpecificComparison);
}
}