Using concepts for function overload resolution (instead of SFINAE) - c++

Trying to say goodbye to SFINAE.
Is it possible to use concepts to distinguish between functions, so the compiler can match the correct function based on whether or not a sent parameter meets concept constraints?
For example, overloading these two:
// (a)
void doSomething(auto t) { /* */ }
// (b)
void doSomething(ConceptA auto t) { /* */ }
So when called the compiler would match the correct function per each call:
doSomething(param_doesnt_adhere_to_ConceptA); // calls (a)
doSomething(param_adheres_to_ConceptA); // calls (b)
Related question: Will Concepts replace SFINAE?

Yes concepts are designed for this purpose. If a sent parameter doesn't meet the required concept argument the function would not be considered in the overload resolution list, thus avoiding ambiguity.
Moreover, if a sent parameter meets several functions, the more specific one would be selected.
Simple example:
void print(auto t) {
std::cout << t << std::endl;
}
void print(std::integral auto i) {
std::cout << "integral: " << i << std::endl;
}
Above print functions are a valid overloading that can live together.
If we send a non integral type it will pick the first
If we send an integral type it will prefer the second
e.g., calling the functions:
print("hello"); // calls print(auto)
print(7); // calls print(std::integral auto)
No ambiguity -- the two functions can perfectly live together, side-by-side.
No need for any SFINAE code, such as enable_if -- it is applied already (hidden very nicely).
Picking between two concepts
The example above presents how the compiler prefers constrained type (std::integral auto) over an unconstrained type (just auto). But the rules also apply to two competing concepts. The compiler should pick the more specific one, if one is more specific. Of course if both concepts are met and none of them is more specific this will result with ambiguity.
Well, what makes a concept be more specific? if it is based on the other one1.
The generic concept - GenericTwople:
template<class P>
concept GenericTwople = requires(P p) {
requires std::tuple_size<P>::value == 2;
std::get<0>(p);
std::get<1>(p);
};
The more specific concept - Twople:
class Any;
template<class Me, class TestAgainst>
concept type_matches =
std::same_as<TestAgainst, Any> ||
std::same_as<Me, TestAgainst> ||
std::derived_from<Me, TestAgainst>;
template<class P, class First, class Second>
concept Twople =
GenericTwople<P> && // <= note this line
type_matches<std::tuple_element_t<0, P>, First> &&
type_matches<std::tuple_element_t<1, P>, Second>;
Note that Twople is required to meet GenericTwople requirements, thus it is more specific.
If you replace in our Twople the line:
GenericTwople<P> && // <= note this line
with the actual requirements that this line brings, Twople would still have the same requirements but it will no longer be more specific than GenericTwople. This, along with code reuse of course, is why we prefer to define Twople based on GenericTwople.
Now we can play with all sort of overloads:
void print(auto t) {
cout << t << endl;
}
void print(const GenericTwople auto& p) {
cout << "GenericTwople: " << std::get<0>(p) << ", " << std::get<1>(p) << endl;
}
void print(const Twople<int, int> auto& p) {
cout << "{int, int}: " << std::get<0>(p) << ", " << std::get<1>(p) << endl;
}
And call it with:
print(std::tuple{1, 2}); // goes to print(Twople<int, int>)
print(std::tuple{1, "two"}); // goes to print(GenericTwople)
print(std::pair{"three", 4}); // goes to print(GenericTwople)
print(std::array{5, 6}); // goes to print(Twople<int, int>)
print("hello"); // goes to print(auto)
We can go further, as the Twople concept presented above works also with polymorphism:
struct A{
virtual ~A() = default;
virtual std::ostream& print(std::ostream& out = std::cout) const {
return out << "A";
}
friend std::ostream& operator<<(std::ostream& out, const A& a) {
return a.print(out);
}
};
struct B: A{
std::ostream& print(std::ostream& out = std::cout) const override {
return out << "B";
}
};
add the following overload:
void print(const Twople<A, A> auto& p) {
cout << "{A, A}: " << std::get<0>(p) << ", " << std::get<1>(p) << endl;
}
and call it (while all the other overloads are still present) with:
print(std::pair{B{}, A{}}); // calls the specific print(Twople<A, A>)
Code: https://godbolt.org/z/3-O1Gz
Unfortunately C++20 doesn't allow concept specialization, otherwise we would go even further, with:
template<class P>
concept Twople<P, Any, Any> = GenericTwople<P>;
Which could add a nice possible answer to this SO question, however concept specialization is not allowed.
1 The actual rules for Partial Ordering of Constraints are more complicated, see: cppreference / C++20 spec.

Related

Deducing type from template class conversion operator

I'm debugging an issue in a large C++ codebase where an attribute of a struct is occasionally being changed to a bad value. Unfortunately, the attribute is public and is accessed or changed in hundreds of places, so simply adding a breakpoint on a mutator is not possible. Also, I don't know the instance of the struct, so adding an address watchpoint wouldn't help.
Instrumenting the code would be a major job. However, a colleague helpfully suggested creating a proxy class which could wrap the existing type in the struct declaration. For example, instead of using MyType _type I would replace this with ChangeProxy<MyType> _type in the struct and the application should compile and work with the proxy taking the place of the direct type in the same manner as, for example, a smart pointer.
However, when I build an example, the implicit conversion operation in the template class doesn't appear to get invoked in type deduction. Here's the code:
#include <iostream>
class MyType {
long _n = 0;
public:
MyType() {}
MyType(const long n) : _n{n} {}
MyType& operator=(const long n) { _n = n; return *this; }
bool isZero() const { return _n != 0; }
};
template <class T>
class ChangeProxy {
public:
ChangeProxy() {}
ChangeProxy(const T& t) : _t{t} {}
ChangeProxy(const T&& t) : _t{std::move(t)} {}
ChangeProxy& operator=(const T& t) {onChange(t); _t = t; return *this;}
ChangeProxy& operator=(const T&& t) {onChange(t); _t = std::move(t); return *this;}
operator T() {return _t;}
private:
T _t;
void onChange(const T& newVal) { /* something here to notify me of changes */ };
};
struct MyStruct {
// MyType _type; // this works ...
ChangeProxy<MyType> _type; // .. but this doesn't
};
int main() {
MyStruct i;
std::cout << "i._type.isZero() : " << std::boolalpha << i._type.isZero() << std::endl;
i._type = 1;
std::cout << "i._type.isZero() : " << std::boolalpha << i._type.isZero() << std::endl;
return 0;
}
Unfortunately, when I build this I get the following errors:
proxy-variable~test.cpp:35:73: error: ‘class ChangeProxy<MyType>’ has no member named ‘isZero’
35 | std::cout << "i._type.isZero() : " << std::boolalpha << i._type.isZero() << std::endl;
| ^~~~~~
proxy-variable~test.cpp:37:73: error: ‘class ChangeProxy<MyType>’ has no member named ‘isZero’
37 | std::cout << "i._type.isZero() : " << std::boolalpha << i._type.isZero() << std::endl;
| ^~~~~~
So it seems that the compiler isn't deducing that it can cast a ChangeProxy<MyType> to a MyType. What have I done wrong here?
The context here doesn't let the compiler try out implicit conversions. Calling a member function on some object never does. You can force this by e.g.
std::cout << "i._type.isZero() : " << std::boolalpha <<
static_cast<MyType>(i._type).isZero() << '\n';
// ^^^^^^^^^^^^^^^^^^^ Here, enforce conversion
Another option would be:
MyStruct i;
const MyType& underlying = i._type; // Again, request conversion manually
std::cout << underlying.isZero() << '\n';
What you are doing is invoking a method on the class ChangeProxy<MyType> which indeed doesn't have any method isZero() defined on it, hence the compilation error. You could probably add something like
T const& operator()() const {return _t;}
And then call it using
i._type().isZero()
The reason that the wrapped i._type.isZero() can never work is that implicit conversions of the i._type object aren't considered for direct method calls, and you can't overload operator. like you can operator->.
It's nothing to do with type deduction, there's simply no mechanism in the language to do what you want.
Luckily, you're solving the wrong problem anyway.
... a colleague helpfully suggested creating a proxy class which could wrap the existing type in the struct declaration
Hmm, you didn't mention that here - or am I a colleague now?
an attribute of a struct is occasionally being changed to a bad value
Which attribute? Be specific!
In your code, you're treating the MyType instance as the problematic attribute. However, the only state in MyType is its long _n member.
Writing
class MyType {
ChangeProxy<long> _n = 0;
which is what I actually suggested when I referred to wrapping built-in types, avoids this problem entirely. You may of course need operator!= to make isZero work, but that's a normally overloadable operator.
Oddly the code in your question doesn't permit any mutation of _n anyway, so it's unclear how it can be getting a bad value. However, I assume this is just an artefact of a simplified example.

Is there anyway that I can determine whether the construct function is exist?

#include <iostream>
template<typename _OutType, typename _InType>
struct ConvertClass
{
_OutType operator()(_InType src)
{
return _OutType(src);
}
};
class OutClass
{
public:
OutClass(std::string str)
{
std::cout << "construct function works well!" << std::endl;
}
};
int main()
{
ConvertClass<OutClass, int>()(20); // this is wrong, because the OutClass only have one construct which takes the std::string type parameter.
// ConvertClass<OutClass, std::string>()(std::string("Hello!"));
/*
if (...) // So I wonder if there is any way that we can know whether the construct function is exists or not before we call the OutClass(int i) function
{
std::cout << "there is no such construct function of OutClass to take that parameter type" << std::endl;
return -1;
}
else
{
std::cout << "construct function works well!" << std::endl;
return 0;
}
*/
}
My Problem:
I know the main function is definitely wrong for the OutClass don't have the construct function OutClass(string str).
I wonder if there is a way only to change the Comment 1 section, the template class to make this file be compiled and linked successfully.
My English is not good, hoping you guys don't mind!
Thank you !
to my knowledge there is not runtime checking if given class is constructible using argument of given type
as said in my previous answer you can resort to Concepts and check the types at compiletime, but if clause does not work at compile time
To me it looks like the best solution would be indeed making a template class out of OutClass, then you have a single class with serves diverse purposes, dependent on you needs
one more edit to your code, I see that you pass the _OutType and _InType to your template.
In the setting where we have the following class template
template<class srcType>
class OutType:{
srcType src;
public:
OutType(srcType src) : src(src) {std::cout << "constructor works well!" << std::endl;}
}
then while invoking the class ConvertClass:
auto val = ConvertClass<OutClass<std::string>, std::string>()(std::string("Hello!"));
and also this will work:
auto val = ConvertClass<OutClass<int>, int>(20);
however, since operator() is not a static method you need first to construct object of class ConvertClass
In c++20 or even in c++17 you can in fact check if OutClass is constructible from int:
so your if clause should look like this
if(std::is_constructible<OutClass, int>::value) {
std::cout << "all is well" << std::endl;
}else{
std::cout << "you can't construct OutClass from int" << std::endl;
}
you can make the the following class template from the OutClass
template<class SrcType>
class OutClass {
SrcType src;
public:
OutClass(SrcType src) : src(src) {}
}
then in your code
return OutType<InType>(src);
if you need to check what the classes passed as template arguments actually can do (if they are arithmetic or additive or copy constructible e.g.) use Concepts from the C++20 standard

class parametrized with classes and confusion with helper functions

I'm learning some new concepts about c++ and I'm playing with them.
I wrote some piece of code that really confuses me in terms of how it works.
#include <iostream>
class aid {
public:
using aid_t = std::string;
void setaid(const std::string& s) {
aid_ = s;
}
const aid_t& getaid() const {
return aid_;
}
private:
aid_t aid_;
};
class c {
public:
using c_t = std::string;
void setc(const aid::aid_t& aid_val) {
if (aid_val.size() < 4)
c_ = "yeah";
else
c_ = aid_val + aid_val;
}
const c_t& getc() {
return c_;
}
private:
c_t c_;
};
template<typename ...Columns>
class table : public Columns... {
};
template <typename... Columns>
void f(table<Columns...>& t) {
t.setaid("second");
std::cout << t.getaid() << "\n";
}
void f2(table<aid>& t) {
t.setaid("third");
std::cout << t.getaid() << "\n";
}
int main() {
table<aid, c> tb;
tb.setaid("first");
std::cout << tb.getaid() << " " << "\n";
// f<c>(tb); // (1) doesnt compile, that seem obvious
f<aid>(tb); // (2) works?
f(tb); // (3) works too -- template parameter deduction
// f2(tb); // (4) doesnt work? worked with (2)...
}
The idea here is simple, I have some table with columns. And then I would like to create some functions that require only some set of columns and doesn't care if passed argument has some extra columns.
My confusion is mostly about points (2) and (4) in code... My intuition says it should be the same, why it isn't and (2) compiles and (4) doesn't? Is there any major topic I'm missing and should read up?
Is there a way to achieve this particular functionality?
In the second case, the compiler still deduces the rest of the template parameter pack, so that you get table<aid, c> & as the function parameter. This is different from (4) (table<aid> &).
[temp.arg.explicit]/9:
Template argument deduction can extend the sequence of template arguments corresponding to a template parameter pack, even when the sequence contains explicitly specified template arguments.

How to implement template function like add(T a, K b) with std::any elegant and without template? [duplicate]

I have a "dictionary" std::map<std::string, boost::any> (or std::any, if you want) that can possibly be nested. Now, I would like to display the map. Since boost::any obviously doesn't play nicely with <<, things are getting a little nasty. So far, I'm checking the type, cast it, and pipe the cast to cout:
for (const auto &p: map) {
std::cout << std::string(indent + 2, ' ') << p.first << ": ";
if (p.second.type() == typeid(int)) {
std::cout << boost::any_cast<int>(p.second);
} else if (p.second.type() == typeid(double)) {
std::cout << boost::any_cast<double>(p.second);
} else if (p.second.type() == typeid(std::string)) {
std::cout << boost::any_cast<std::string>(p.second);
} else if (p.second.type() == typeid(const char*)) {
std::cout << boost::any_cast<const char*>(p.second);
} else if (p.second.type() == typeid(std::map<std::string, boost::any>)) {
show_map(
boost::any_cast<std::map<std::string, boost::any>>(p.second),
indent + 2
);
} else {
std::cout << "[unhandled type]";
}
std::cout << std::endl;
}
std::cout << std::string(indent, ' ') << "}";
This prints, for example
{
fruit: banana
taste: {
sweet: 1.0
bitter: 0.1
}
}
Unfortunately, this is hardly scalable. I'd have to add another else if clause for every type (e.g., float, size_t,...), which is why I'm not particularly happy with the solution.
Is there a way to generalize the above to more types?
One thing you can do to lessen (but not remove) the pain is to factor the type determination logic into one support function, while using static polymorphism (specifically templates) for the action to be applied to the values...
#include <iostream>
#include <boost/any.hpp>
#include <string>
struct Printer
{
std::ostream& os_;
template <typename T>
void operator()(const T& t)
{
os_ << t;
}
};
template <typename F>
void f_any(F& f, const boost::any& a)
{
if (auto p = boost::any_cast<std::string>(&a)) f(*p);
if (auto p = boost::any_cast<double>(&a)) f(*p);
if (auto p = boost::any_cast<int>(&a)) f(*p);
// whatever handling for unknown types...
}
int main()
{
boost::any anys[] = { std::string("hi"), 3.14159, 27 };
Printer printer{std::cout};
for (const auto& a : anys)
{
f_any(printer, a);
std::cout << '\n';
}
}
(With only a smidge more effort, you could have the type-specific test and dispatch done for each type in a variadic template parameter pack, simplifying that code and the hassle of maintaining the list. Or, you could just use a preprocessor macro to churn out the if-cast/dispatch statements....)
Still - if you know the set of types, a boost::variant is more appropriate and already supports similar operations (see here).
Yet another option is to "memorise" how to do specific operations - such as printing - when you create your types:
#include <iostream>
#include <boost/any.hpp>
#include <string>
#include <functional>
struct Super_Any : boost::any
{
template <typename T>
Super_Any(const T& t)
: boost::any(t),
printer_([](std::ostream& os, const boost::any& a) { os << boost::any_cast<const T&>(a); })
{ }
std::function<void(std::ostream&, const boost::any&)> printer_;
};
int main()
{
Super_Any anys[] = { std::string("hi"), 3.14159, 27 };
for (const auto& a : anys)
{
a.printer_(std::cout, a);
std::cout << '\n';
}
}
If you have many operations and want to reduce memory usage, you can have the templated constructor create and store a (abstract-base-class) pointer to a static-type-specific class deriving from an abstract interface with the operations you want to support: that way you're only adding one pointer per Super_Any object.
Since you're already using Boost you could consider boost::spirit::hold_any.
It already has pre-defined streaming operators (both operator<<() and operator>>()).
Just the embedded type must have the corresponding operator defined, but in your use context this seems to be completely safe.
Despite being in the detail namespace, hold_any is quite widespread and almost a ready-to-use boost:any substitute (e.g. Type Erasure - Part IV, Why you shouldn’t use boost::any)
A recent version of Boost is required (old versions had a broken copy assignment operator).

cout map with boost::any

I have a "dictionary" std::map<std::string, boost::any> (or std::any, if you want) that can possibly be nested. Now, I would like to display the map. Since boost::any obviously doesn't play nicely with <<, things are getting a little nasty. So far, I'm checking the type, cast it, and pipe the cast to cout:
for (const auto &p: map) {
std::cout << std::string(indent + 2, ' ') << p.first << ": ";
if (p.second.type() == typeid(int)) {
std::cout << boost::any_cast<int>(p.second);
} else if (p.second.type() == typeid(double)) {
std::cout << boost::any_cast<double>(p.second);
} else if (p.second.type() == typeid(std::string)) {
std::cout << boost::any_cast<std::string>(p.second);
} else if (p.second.type() == typeid(const char*)) {
std::cout << boost::any_cast<const char*>(p.second);
} else if (p.second.type() == typeid(std::map<std::string, boost::any>)) {
show_map(
boost::any_cast<std::map<std::string, boost::any>>(p.second),
indent + 2
);
} else {
std::cout << "[unhandled type]";
}
std::cout << std::endl;
}
std::cout << std::string(indent, ' ') << "}";
This prints, for example
{
fruit: banana
taste: {
sweet: 1.0
bitter: 0.1
}
}
Unfortunately, this is hardly scalable. I'd have to add another else if clause for every type (e.g., float, size_t,...), which is why I'm not particularly happy with the solution.
Is there a way to generalize the above to more types?
One thing you can do to lessen (but not remove) the pain is to factor the type determination logic into one support function, while using static polymorphism (specifically templates) for the action to be applied to the values...
#include <iostream>
#include <boost/any.hpp>
#include <string>
struct Printer
{
std::ostream& os_;
template <typename T>
void operator()(const T& t)
{
os_ << t;
}
};
template <typename F>
void f_any(F& f, const boost::any& a)
{
if (auto p = boost::any_cast<std::string>(&a)) f(*p);
if (auto p = boost::any_cast<double>(&a)) f(*p);
if (auto p = boost::any_cast<int>(&a)) f(*p);
// whatever handling for unknown types...
}
int main()
{
boost::any anys[] = { std::string("hi"), 3.14159, 27 };
Printer printer{std::cout};
for (const auto& a : anys)
{
f_any(printer, a);
std::cout << '\n';
}
}
(With only a smidge more effort, you could have the type-specific test and dispatch done for each type in a variadic template parameter pack, simplifying that code and the hassle of maintaining the list. Or, you could just use a preprocessor macro to churn out the if-cast/dispatch statements....)
Still - if you know the set of types, a boost::variant is more appropriate and already supports similar operations (see here).
Yet another option is to "memorise" how to do specific operations - such as printing - when you create your types:
#include <iostream>
#include <boost/any.hpp>
#include <string>
#include <functional>
struct Super_Any : boost::any
{
template <typename T>
Super_Any(const T& t)
: boost::any(t),
printer_([](std::ostream& os, const boost::any& a) { os << boost::any_cast<const T&>(a); })
{ }
std::function<void(std::ostream&, const boost::any&)> printer_;
};
int main()
{
Super_Any anys[] = { std::string("hi"), 3.14159, 27 };
for (const auto& a : anys)
{
a.printer_(std::cout, a);
std::cout << '\n';
}
}
If you have many operations and want to reduce memory usage, you can have the templated constructor create and store a (abstract-base-class) pointer to a static-type-specific class deriving from an abstract interface with the operations you want to support: that way you're only adding one pointer per Super_Any object.
Since you're already using Boost you could consider boost::spirit::hold_any.
It already has pre-defined streaming operators (both operator<<() and operator>>()).
Just the embedded type must have the corresponding operator defined, but in your use context this seems to be completely safe.
Despite being in the detail namespace, hold_any is quite widespread and almost a ready-to-use boost:any substitute (e.g. Type Erasure - Part IV, Why you shouldn’t use boost::any)
A recent version of Boost is required (old versions had a broken copy assignment operator).