Composing boost::variant visitors for recursive variants - boost

I have an application with several boost::variants which share many of the fields. I would like to be able to compose these visitors into visitors for "larger" variants without copying and pasting a bunch of code. It seems straightforward to do this for non-recursive variants, but once you have a recursive one, the self-references within the visitor (of course) point to the wrong class. To make this concrete (and cribbing from the boost::variant docs):
#include "boost/variant.hpp"
#include <iostream>
struct add;
struct sub;
template <typename OpTag> struct binop;
typedef boost::variant<
int
, boost::recursive_wrapper< binop<add> >
, boost::recursive_wrapper< binop<sub> >
> expression;
template <typename OpTag>
struct binop
{
expression left;
expression right;
binop( const expression & lhs, const expression & rhs )
: left(lhs), right(rhs)
{
}
};
// Add multiplication
struct mult;
typedef boost::variant<
int
, boost::recursive_wrapper< binop<add> >
, boost::recursive_wrapper< binop<sub> >
, boost::recursive_wrapper< binop<mult> >
> mult_expression;
class calculator : public boost::static_visitor<int>
{
public:
int operator()(int value) const
{
return value;
}
int operator()(const binop<add> & binary) const
{
return boost::apply_visitor( *this, binary.left )
+ boost::apply_visitor( *this, binary.right );
}
int operator()(const binop<sub> & binary) const
{
return boost::apply_visitor( *this, binary.left )
- boost::apply_visitor( *this, binary.right );
}
};
class mult_calculator : public boost::static_visitor<int>
{
public:
int operator()(int value) const
{
return value;
}
int operator()(const binop<add> & binary) const
{
return boost::apply_visitor( *this, binary.left )
+ boost::apply_visitor( *this, binary.right );
}
int operator()(const binop<sub> & binary) const
{
return boost::apply_visitor( *this, binary.left )
- boost::apply_visitor( *this, binary.right );
}
int operator()(const binop<mult> & binary) const
{
return boost::apply_visitor( *this, binary.left )
* boost::apply_visitor( *this, binary.right );
}
};
// I'd like something like this to compile
// class better_mult_calculator : public calculator
// {
// public:
// int operator()(const binop<mult> & binary) const
// {
// return boost::apply_visitor( *this, binary.left )
// * boost::apply_visitor( *this, binary.right );
// }
// };
int main(int argc, char **argv)
{
// result = ((7-3)+8) = 12
expression result(binop<add>(binop<sub>(7,3), 8));
assert( boost::apply_visitor(calculator(),result) == 12 );
std::cout << "Success add" << std::endl;
// result2 = ((7-3)+8)*2 = 12
mult_expression result2(binop<mult>(binop<add>(binop<sub>(7,3), 8),2));
assert( boost::apply_visitor(mult_calculator(),result2) == 24 );
std::cout << "Success mult" << std::endl;
}
I would really like something like that commented out better_mult_expression to compile (and work) but it doesn't -- because the this pointers within the base calculator visitor don't reference mult_expression, but expression.
Does anyone have suggestions for overcoming this or am I just barking down the wrong tree?

Firstly, I'd suggest the variant to include all possible node types, not distinguishing between mult and expression. This distinction makes no sense at the AST level, only at a parser stage (if you implement operator precedence in recursive/PEG fashion).
Other than that, here's a few observations:
if you encapsulate the apply_visitor dispatch into your evaluation functor you can reduce the code duplication by a big factor
your real question seems not to be about composing variants, but composing visitors, more specifically, by inheritance.
You can use using to pull inherited overloads into scope for overload resolution, so this might be the most direct answer:
Live On Coliru
struct better_mult_calculator : calculator {
using calculator::operator();
auto operator()(const binop<mult>& binary) const
{
return boost::apply_visitor(*this, binary.left) *
boost::apply_visitor(*this, binary.right);
}
};
IMPROVING!
Starting from that listing let's shave off some noise!
remove unncessary AST distinction (-40 lines, down to 55 lines of code)
generalize the operations; the <functional> header comes standard with these:
namespace AST {
template <typename> struct binop;
using add = binop<std::plus<>>;
using sub = binop<std::minus<>>;
using mult = binop<std::multiplies<>>;
using expr = boost::variant<int,
recursive_wrapper<add>,
recursive_wrapper<sub>,
recursive_wrapper<mult>>;
template <typename> struct binop { expr left, right; };
} // namespace AST
Now the entire calculator can be:
struct calculator : boost::static_visitor<int> {
int operator()(int value) const { return value; }
template <typename Op>
int operator()(AST::binop<Op> const& binary) const {
return Op{}(boost::apply_visitor(*this, binary.left),
boost::apply_visitor(*this, binary.right));
}
};
Here your variant can add arbitrary operations without even needing to touch the calculator.
Live Demo, 43 Lines Of Code
Like I mentioned starting off, encapsulate visitation!
struct Calculator {
template <typename... T> int operator()(boost::variant<T...> const& v) const {
return boost::apply_visitor(*this, v);
}
template <typename T>
int operator()(T const& lit) const { return lit; }
template <typename Op>
int operator()(AST::binop<Op> const& bin) const {
return Op{}(operator()(bin.left), operator()(bin.right));
}
};
Now you can just call your calculator, like intended:
Calculator calc;
auto result1 = calc(e1);
It will work when you extend the variant with operatios or even other literal types (like e.g. double). It will even work, regardless of whether you pass it an incompatible variant type that holds a subset of the node types.
To finish that off for maintainability/readability, I'd suggest making operator() only a dispatch function:
Full Demo
Live On Coliru
#include <boost/variant.hpp>
#include <iostream>
namespace AST {
using boost::recursive_wrapper;
template <typename> struct binop;
using add = binop<std::plus<>>;
using sub = binop<std::minus<>>;
using mult = binop<std::multiplies<>>;
using expr = boost::variant<int,
recursive_wrapper<add>,
recursive_wrapper<sub>,
recursive_wrapper<mult>>;
template <typename> struct binop { expr left, right; };
} // namespace AST
struct Calculator {
auto operator()(auto const& v) const { return call(v); }
private:
template <typename... T> int call(boost::variant<T...> const& v) const {
return boost::apply_visitor(*this, v);
}
template <typename T>
int call(T const& lit) const { return lit; }
template <typename Op>
int call(AST::binop<Op> const& bin) const {
return Op{}(call(bin.left), call(bin.right));
}
};
int main()
{
using namespace AST;
std::cout << std::boolalpha;
auto sub_expr = add{sub{7, 3}, 8};
expr e1 = sub_expr;
expr e2 = mult{sub_expr, 2};
Calculator calc;
auto result1 = calc(e1);
std::cout << "result1: " << result1 << " Success? " << (12 == result1) << "\n";
// result2 = ((7-3)+8)*2 = 12
auto result2 = calc(e2);
std::cout << "result2: " << result2 << " Success? " << (24 == result2) << "\n";
}
Still prints
result1: 12 Success? true
result2: 24 Success? true

Related

clang - how to declare a static const int in header file?

Given the following template in a header file, and a couple of specializations:
template<typename> class A {
static const int value;
};
template<> const int A<int>::value = 1;
template<> const int A<long>::value = 2;
and building with clang-5, it results in errors for each source unit that included the file, all complaining about multiple definitions for A<int>::value and A<long>::value.
At first, I thought that maybe the template specializations needed to be put in a specific translation unit, but on checking the spec, this apparently should be allowed, because the value is a constant integer.
Am I doing something else wrong?
EDIT: if I move the definition into a single translation unit, then I can no longer use the value of A<T>::value in the context of a const int (eg, where its value is being used to calculate the value of another const assignment) , so the value really needs to be in a header.
In c++11 you maybe can go that way:
template<typename> class B {
public:
static const int value = 1;
};
template<> class B<long> {
public:
static const int value = 2;
};
template<typename T> const int B<T>::value;
If you only want to specialize the value var, you can use CRTP for that.
From C++17 you can make your definition inline:
template<> inline const int A<int>::value = 1;
template<> inline const int A<long>::value = 2;
Also from c++17 you can remove the 'template const int B::value;' for constexpr:
template<typename> class C {
public:
static constexpr int value = 1;
};
template<> class C<long> {
public:
static constexpr int value = 2;
};
// no need anymore for: template<typename T> const int C<T>::value;
And another solution for c++11 can be to use a inline method instead of inline vars which are allowed from c++17:
template<typename T> class D {
public:
static constexpr int GetVal() { return 0; }
static const int value = GetVal();
};
template <> inline constexpr int D<int>::GetVal() { return 1; }
template <> inline constexpr int D<long>::GetVal() { return 2; }
template< typename T>
const int D<T>::value;
In addition to your last edit:
To use your values also in other dependent definitions it seems to be the most readable version if you use the inline constexpr methods.
Edit: "Special" version for clang, because as OP tells us, clang complains with "specialization happening after instantiation". I don't know if clang or gcc is wrong in that place...
template<typename T> class D {
public:
static constexpr int GetVal();
static const int value;
};
template <> inline constexpr int D<int>::GetVal() { return 1; }
template <> inline constexpr int D<long>::GetVal() { return 2; }
template <typename T> const int D<T>::value = D<T>::GetVal();
int main()
{
std::cout << D<int>::value << std::endl;
std::cout << D<long>::value << std::endl;
}
I told already that CRTP is possible if not the complete class should be redefined. I checked the code on clang and it compiles without any warning or error, because OP comments that he did not understand how to use it:
template<typename> class E_Impl {
public:
static const int value = 1;
};
template<> class E_Impl<long> {
public:
static const int value = 2;
};
template<typename T> const int E_Impl<T>::value;
template < typename T>
class E : public E_Impl<T>
{
// rest of class definition goes here and must not specialized
// and the values can be used here!
public:
void Check()
{
std::cout << this->value << std::endl;
}
};
int main()
{
E<long>().Check();
std::cout << E<long>::value << std::endl;
E<int>().Check();
std::cout << E<int>::value << std::endl;
}

Pass f(a) and f(a,b) as same slot of template parameter : without require explicit signature passing

I want to pass two functions with different signature into a same slot of class template parameter (one each time).
Ignore the strict syntax, this is what I want :-
void hf1 (int a) { std::cout<< "hf1" <<std::endl; }
void hf2 (int a, int i){ std::cout<< "hf2 "<<i<<std::endl; }
template<hfX> class Collection{
int i_=56;
public: test(){
if( "hfX has 1 param" ){hfX(0);} //call hf1
else {hfX(0,i_);} //call hf2
}
};
int main(){
Collection<&hf1> test1; test1.test(); // print "hf1"
Collection<&hf2> test2; test2.test(); // print "hf2 56"
}
Here is the code that works OK, but its usage is not so convenient :-
template<typename ... AA> using hfX = void(*)(AA ... );
void hf1 (int a) { std::cout<< "hf1" <<std::endl; }
void hf2 (int a, int i) { std::cout<< "hf2 "<<i <<std::endl; }
template <typename Tf, Tf F>
class Collection;
template <typename ... I, hfX<I...> F>
class Collection<hfX<I...>, F>{
public:
int i_=56;
template <std::size_t N = sizeof...(I)>
typename std::enable_if<N == 1U, void>::type test (){
F(0);
}
template <std::size_t N = sizeof...(I)>
typename std::enable_if<N == 2U, void>::type test (){
F(0,i_);
}
};
The usage:-
int main () {
Collection<hfX<int>, hf1> test1; //<--- #A dirty signature
Collection<hfX<int,int>, hf2> test2; //<--- #B dirty signature
test1.test(); // print "hf1"
test2.test(); // print "hf2 56"
}
Live version : https://ideone.com/f20BEk
Question
It would be nice if I can call it without explicit redundant signature.
Collection<hf1> test1; //or &hf
Collection<hf2> test2;
How to improve code (especially around hfX and Collection) to make its usage easier?
I don't know how to make what do you want with functions.
But if you can, instead of functions, accept to use static method in classes or structs (and pass that classes/structs as template argument)...
#include <iostream>
struct sf1
{ static void hf (int a) { std::cout << "hf1" << std::endl; } };
struct sf2
{ static void hf (int a, int i) { std::cout << "hf2 " << i << std::endl; } };
template <typename S>
class Collection
{
private:
int i_ = 56;
public:
template <typename T = S>
decltype(T::hf(0)) test() { S::hf(0); /*call sf1::hf */ }
template <typename T = S>
decltype(T::hf(0, 0)) test() { S::hf(0,i_); /*call sf2::hf */ }
};
int main ()
{
Collection<sf1> test1; test1.test(); // print "hf1"
Collection<sf2> test2; test2.test(); // print "hf2 56"
}

c++11 segmentation fault while playing with variadic templates

So I was Playing around with c++11 Varidiacs, and I wanted to create a thing called CallClass, basically a class that warps a function, for later call,when all variables are set(truly I have No Idea If It can Be Useful):
#include <tuple>
template <typename OBJ,typename F,typename... VARGS>
class CallClass
{
public:
CallClass(OBJ& object,F callFunction)
:_object(&object),_func(callFunction)
{ }
CallClass(const CallClass& other)
:_func_args(other._func_args)
,_object(other._object)
,_func(other._func)
{ }
template <size_t INDEX>
auto get(){ return std::get<INDEX>(_func_args); }
template <size_t INDEX,typename T>
void set(const T& val){ std::get<INDEX>(_func_args) = val; }
template <size_t INDEX,typename T>
void set(T&& val){ std::get<INDEX>(_func_args) = val; }
auto Call()
{
//throws segmentation Fault Here
return InnerCall<0>(_func_args);
}
virtual ~CallClass() {}
protected:
private:
std::tuple<VARGS...> _func_args;
OBJ* _object;
F _func;
template <size_t INDEX,typename... ARGS>
auto InnerCall(std::tuple<VARGS...>& tup,ARGS... args)
{
auto arg = std::get<INDEX>(tup);
return InnerCall<INDEX + 1>(tup,args...,arg);
}
template <size_t INDEX,VARGS...>
auto InnerCall(std::tuple<VARGS...>& tup,VARGS... args)
{
return (_object->*_func)(args...);
}
};
Now when I try to compile(compiling using IDE:code::blocks, configured to use MINGW On windows ), it prints Compiler:Segmentation Fault, anybody any Ideas?
Usage:
class obj{
public:
obj(int a)
:_a(a)
{ }
virtual ~obj() {}
int add(int b,int c){
return _a + b + c;
}
private:
int _a;
};
int main(){
obj ob(6);
CallClass<obj,decltype(obj::add),int,int> callAdd(ob,obj::add);
callAdd.set<0,int>(5);
callAdd.set<1,int>(7);
cout << "result is " << callAdd.Call() << endl;
return 0;
}
After a Bit of a search i stumbled upon a similar issue, in a way.
apparently the way I'm unpacking the tuple is an issue, so i decided to use a different approach as shown in: enter link description here
had to add a few changes to suit my needs:
changes:
namespace detail
{
template <typename OBJ,typename F, typename Tuple, bool Done, int Total, int... N>
struct call_impl
{
static auto call(OBJ& obj,F f, Tuple && t)
{
return call_impl<OBJ,F, Tuple, Total == 1 + sizeof...(N), Total, N..., sizeof...(N)>::call(obj,f, std::forward<Tuple>(t));
}
};
template <typename OBJ,typename F, typename Tuple, int Total, int... N>
struct call_impl<OBJ,F, Tuple, true, Total, N...>
{
static auto call(OBJ& obj,F f, Tuple && t)
{
return (obj.*f)(std::get<N>(std::forward<Tuple>(t))...);
}
};
}
// user invokes this
template <typename OBJ,typename F, typename Tuple>
auto call(OBJ& obj,F f, Tuple && t)
{
typedef typename std::decay<Tuple>::type ttype;
return detail::call_impl<OBJ,F, Tuple, 0 == std::tuple_size<ttype>::value, std::tuple_size<ttype>::value>::call(obj,f, std::forward<Tuple>(t));
}
and changed Call():
auto Call()
{
std::tuple<VARGS...> func_args = _func_args;
return call(*_object,_func, std::move(func_args));
}
I will probably make a few more changes, like passing the tuple as a reference, and making the structs a part of my class.

Ambiguous operator<< selection

I have some code which, very much simplified, looks somewhat like this:
#include <iostream>
#include <type_traits>
namespace X {
struct Foo {int x;};
struct Bar {int x;};
template <typename T , typename = typename std::enable_if<
std::is_same<decltype(T::x),int>::value
>::type>
std::ostream & operator<<(std::ostream & os, const T&) {
return os;
}
}
namespace Y {
struct Faa : X::Foo {int y;};
struct Baz {int x; int y;};
template <typename T , typename = typename std::enable_if<
std::is_same<decltype(T::x),int>::value &&
std::is_same<decltype(T::y),int>::value
>::type>
std::ostream & operator<<(std::ostream & os, const T&) {
return os;
}
}
int main() {
// Everything is ok
X::Foo x;
std::cout << x;
Y::Baz k;
std::cout << k;
// Problems..
Y::Faa y;
// std::cout << y; // <--operator is ambiguous
Y::operator<<(std::cout, y);
return 0;
}
Is there any way to avoid the ambiguous operator for Y::Faa and having to manually specify Y::operator<<? If not, why?
Two functions have a conflict because conditions on their arguments have non-empty intersection (actually, 1st supersedes 2nd). Function overloading works only if signatures are different. So, to solve this we have 2 options:
Change conditions so that they have empty intersection (manually forbid having y field by adding && !sfinae_has_member_y<T>::value condition to the 1st enable_if)
template<typename T>
struct sfinae_has_member_y {
static int has(...);
template<typename U = T, typename = decltype(U::y)>
static char has(const U& value);
enum { value = sizeof(char) == sizeof(has(std::declval<T>())) };
};
OR use another C++ feature that supports arguments overlapping, like struct/class template specialization. If you replace bool with int, other fields may be added too:
template<typename T, bool>
struct Outputter {
};
template<typename T>
struct Outputter<T, false> {
static std::ostream & output(std::ostream & os, const T&) {
os << "x";
return os;
}
};
template<typename T>
struct Outputter<T, true> {
static std::ostream & output(std::ostream & os, const T&) {
os << "y";
return os;
}
};
template<typename T, typename = std::enable_if_t<std::is_same<decltype(T::x), int>::value>>
std::ostream & operator<<(std::ostream & os, const T& a) {
return Outputter<T, sfinae_has_member_y<T>::value>::output(os, a);
}

Using find method that require only key in boost splay_set

The find method of boost::splay_set that require only the key accepts an argument of type KeyValueCompare to compare objects with the key. To be able to use this, we need to supply two methods of the form:
struct KeyValCompare {
inline bool operator() (const std::int64_t key, const MyType& val) const {
//TODO:
}
inline bool operator() (const MyType& val, const std::int64_t key) const {
//TODO:
}
};
However there is no mention in the documentation about how to implement these. Any pointers?
Found a solution here:
http://boost.cowic.de/rc/pdf/intrusive.pdf
they should return true if key (or key from the value) of lhs is less than the key (or key from the value) of rhs.
I don't see why the comparator would be so complicated. The set just stores elements of MyType, so you need to define a strict weak total ordering on them:
struct Comparator {
bool operator()(MyType const& a, MyType const& b) const;
};
Indeed, the default comparer is std::less<MyType>
E.g. to sort
class MyType : public splay_set_base_hook<>
{
int int_;
public:
MyType(int i) : int_(i) {}
int getValue() const { return int_; }
};
By the value, after reversing the digits (e.g. "431" before "322" because 134<223):
struct CompareReversed {
bool operator()(MyType const& a, MyType const& b) const {
return reversed(a.getValue()) < reversed(b.getValue());
}
private:
static int reversed(int i)
{
auto s = std::to_string(i);
std::reverse(s.begin(), s.end());
return boost::lexical_cast<int>(s);
}
};
See it Live On Coliru:
#include <boost/intrusive/splay_set.hpp>
#include <boost/lexical_cast.hpp>
#include <vector>
#include <algorithm>
using namespace boost::intrusive;
class MyType : public splay_set_base_hook<>
{
int int_;
public:
MyType(int i) : int_(i)
{}
// default ordering
friend bool operator< (const MyType &a, const MyType &b) { return a.int_ < b.int_; }
friend bool operator> (const MyType &a, const MyType &b) { return a.int_ > b.int_; }
friend bool operator== (const MyType &a, const MyType &b) { return a.int_ == b.int_; }
int getValue() const { return int_; }
};
struct CompareReversed {
bool operator()(MyType const& a, MyType const& b) const {
return reversed(a.getValue()) < reversed(b.getValue());
}
private:
static int reversed(int i)
{
auto s = std::to_string(i);
std::reverse(s.begin(), s.end());
return boost::lexical_cast<int>(s);
}
};
#include <iostream>
int main()
{
//typedef splay_set<MyType, compare<std::less<MyType> > > Set;
typedef splay_set<MyType, compare<CompareReversed> > Set;
std::vector<MyType> v { 24, 42, 123, 321 };
Set set;
set.insert(v[0]);
set.insert(v[1]);
set.insert(v[2]);
set.insert(v[3]);
for (auto& el : set)
{
std::cout << el.getValue() << "\n";
}
std::cout << set.count(24) << "\n"; // 1
std::cout << set.count(25) << "\n"; // 0
std::cout << set.count(42) << "\n"; // 1
}
If you want to suppor mixed type comparisons, just supply the overloads, obviously:
struct CompareReversed {
bool operator()(MyType const& a, MyType const& b) const {
return reversed(a.getValue()) < reversed(b.getValue());
}
bool operator()(MyType const& a, int b) const {
return reversed(a.getValue()) < reversed(b);
}
bool operator()(int a, MyType const& b) const {
return reversed(a) < reversed(b.getValue());
}
// ...
};
Thanks sehe for the support.
That is exactly what I am doing there. But please have a look at following sample code which fails.
#include <boost/intrusive/splay_set.hpp>
#include <algorithm>
using namespace boost::intrusive;
class MyClass {
public:
MyClass(const std::int64_t& k)
: key(k) {
}
std::int64_t key;
splay_set_member_hook<> member_hook_;
friend bool operator <(const MyClass& lhs, const MyClass& rhs) {
return lhs.key < rhs.key;
}
friend bool operator >(const MyClass& lhs, const MyClass& rhs) {
return lhs.key > rhs.key;
}
friend bool operator ==(const MyClass& lhs, const MyClass& rhs) {
return lhs.key == rhs.key;
}
};
struct KeyValCompare {
inline bool operator()(const std::int64_t key, const MyClass& val) const {
return key < val.key;
}
inline bool operator()(const MyClass& val, const std::int64_t key) const {
return val.key < key;
}
};
typedef member_hook<MyClass, splay_set_member_hook<>, &MyClass::member_hook_> MemberOption;
typedef splay_set<MyClass, MemberOption, compare<std::greater<MyClass> > > MyClassObjectsType;
TEST(MyClass, test) {
MyClassObjectsType set;
set.insert(*new MyClass(10));
set.insert(*new MyClass(20));
set.insert(*new MyClass(100));
auto ite = set.find(100, KeyValCompare());
ASSERT_TRUE(ite != set.end()); // Fails here
}
If I use std::less instead of std::greater, it passes.
Figured it out:
The greater than operator must be change from:
friend bool operator >(const MyClass& lhs, const MyClass& rhs) {
return lhs.key > rhs.key;
}
to this:
friend bool operator >(const MyClass& lhs, const MyClass& rhs) {
return lhs.key < rhs.key;
}

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