first post, so hopefully not violating any etiquette. Feel free to give suggestions for making the question better.
I've seen a few posts similar to this one: Check if a class has a member function of a given signature, but none do quite what I want. Sure it "works with polymorphism" in the sense that it can properly check subclass types for the function that comes from a superclass, but what I'd like to do is check the object itself and not the class. Using some (slightly tweaked) code from that post:
// Somewhere in back-end
#include <type_traits>
template<typename, typename T>
struct HasFunction {
static_assert(integral_constant<T, false>::value,
"Second template parameter needs to be of function type."
);
};
template<typename C, typename Ret, typename... Args>
class HasFunction<C, Ret(Args...)> {
template<typename T>
static constexpr auto check(T*) -> typename is_same<
decltype(declval<T>().myfunc(declval<Args>()...)), Ret>::type;
template<typename>
static constexpr false_type check(...);
typedef decltype(check<C>(0)) type;
public:
static constexpr bool value = type::value;
};
struct W {};
struct X : W { int myfunc(double) { return 42; } };
struct Y : X {};
I'd like to have something like the following:
// somewhere else in back-end. Called by client code and doesn't know
// what it's been passed!
template <class T>
void DoSomething(T& obj) {
if (HasFunction<T, int(double)>::value)
cout << "Found it!" << endl;
// Do something with obj.myfunc
else cout << "Nothin to see here" << endl;
}
int main()
{
Y y;
W* w = &y; // same object
DoSomething(y); // Found it!
DoSomething(*w); // Nothin to see here?
}
The problem is that the same object being viewed polymorphically causes different results (because the deduced type is what is being checked and not the object). So for example, if I was iterating over a collection of W*'s and calling DoSomething I would want it to no-op on W's but it should do something for X's and Y's. Is this achievable? I'm still digging into templates so I'm still not quite sure what's possible but it seems like it isn't. Is there a different way of doing it altogether?
Also, slightly less related to that specific problem: Is there a way to make HasFunction more like an interface so I could arbitrarily check for different functions? i.e. not have ".myfunc" concrete within it? (seems like it's only possible with macros?) e.g.
template<typename T>
struct HasFoo<T> : HasFunction<T, int foo(void)> {};
int main() {
Bar b;
if(HasFoo<b>::value) b.foo();
}
Obviously that's invalid syntax but hopefully it gets the point across.
It's just not possible to perform deep inspection on a base class pointer in order to check for possible member functions on the pointed-to type (for derived types that are not known ahead of time). Even if we get reflection.
The C++ standard provides us no way to perform this kind of inspection, because the kind of run time type information that is guaranteed to be available is very limited, basically relegated to the type_info structure.
Your compiler/platform may provide additional run-time type information that you can hook into, although the exact types and machinery used to provide RTTI are generally undocumented and difficult to examine (This article by Quarkslab attempts to inspect MSVC's RTTI hierarchy)
Related
For a given usage
template<typename T1>
class node{
public:
using sp2node = shared_ptr<node<T1>>;
using r_sp2node = shared_ptr<node<T1>>&;
public:
r_sp2Node getN();
private:
sp2node N;
};
(1)
template<typename T1> decltype(node<T1>::r_sp2node) node<T1>::getN(){
return N;
}
(2)
template<typename T1> typename node<T1>::r_sp2node node<T1>::getN(){
return N;
}
(1) generates the compiler error:
error: missing 'typename' prior to dependent type name
'node<T1>::r_sp2node'
whereas (2) compiles
Can someonene explain what is the difference between the above two?
There are two things wrong with the first example.
In decltype(node<T1>::r_sp2node), reading inside out, the compiler first needs to know what node<T1>::r_sp2node. Is it a type, or something else? This is why the typename disambiguator exists, and that is what the error message is all about.
The second problem is that decltype expects some expression, not a type. So even if you used typename, it still wouldn't compile. (As a simple example, decltype(int) won't even compile.)
To answer the question specifically, the difference between the two is that the first is not valid C++ and the second is the way to go.
When initializing an atomic class member it requires a 'deleted' function, but adding it would make it no longer trivially copyable which is a requirement for an object/struct to be atomic. Am I just not understanding how to do this correctly, or is this a problem in the c++ standard?
Take the example below:
#include <atomic>
#include <cstdint>
template<typename T>
struct A
{
T * data;
std::atomic<uintptr_t> next;
};
template<typename T>
class B
{
std::atomic<A<T>> myA;
public:
B ( A<T> & a ) noexcept
{
myA.store(a, std::memory_order_relaxed );
}
};
int main ()
{
A<int> a;
B<int> b(a);
return 0;
}
Trying to compile this with g++ gives error: use of deleted function 'A<int>::A(const A<int>&)' myA.store(a, std::memory_order_relaxed);. My understanding of this error is that the atomic::store method is looking for that constructor in my struct A but not finding it.
Now here is what happens when I add that constructor:
#include <atomic>
#include <cstdint>
template<typename T>
struct A
{
T * data;
std::atomic<uintptr_t> next;
A(const A<T>& obj) { }
A( ) { }
};
template<typename T>
class B
{
std::atomic<A<T>> myA;
public:
B ( A<T> & a ) noexcept
{
myA.store(a, std::memory_order_relaxed );
}
};
int main ()
{
A<int> a;
B<int> b(a);
return 0;
}
I no longer receive the above compiler error but a new one coming from the requirements of the atomic class required from 'class B<int>' .... error: static assertion failed: std::atomic requires a trivially copyable type ... In other words by adding the used-defined constructors I have made my struct A a non-trivially copyable object which cannot be initialized in class B. However, without the user-defined constructors I cannot use the store method in myA.store(a, std::memory_order_relaxed).
This seems like a flaw in the design of the std::atomic class. Now maybe I am just doing something wrong because I don't have a lot of experience using C++11 and up (I'm old school). Since 11 there have been a lot of changes and the requirements seem to be a lot stricter. I'm hoping someone can tell me how to achieve what I want to achieve.
Also I cannot change std::atomic<A<T>> myA; to std::atomic<A<T>> * myA; (changed to pointer) or std::atomic<A<T>*> myA;. I realize this will compile but it will destroy the fundamental design of a class I am trying to build.
The problem here resides in the fact that std::atomic requires a trivially copiable type. This because trivially copyable types are the only sure types in C++ which can be directly copied by copying their memory contents directly (eg. through std::memcpy). Also non-formerly trivially copyable types could be safe to raw copy but no assumption can be made on this.
This is indeed important for std::atomic since copy on temporary values is made through std::memcpy, see some implementation details for Clang for example.
Now at the same time std::atomic is not copy constructible, and this is for reasonable reasons, check this answer for example, so it's implicitly not trivially copyable (nor any type which contains them).
If, absurdly, you would allow a std::atomic to contain another std::atomic, and the implementation of std::atomic contains a lock, how would you manage copying it atomically? How should it work?
Ok, muddling though Stack on the particulars about void*, books like The C Programming Language (K&R) and The C++ Programming Language (Stroustrup). What have I learned? That void* is a generic pointer with no type inferred. It requires a cast to any defined type and printing void* just yields the address.
What else do I know? void* can't be dereferenced and thus far remains the one item in C/C++ from which I have discovered much written about but little understanding imparted.
I understand that it must be cast such as *(char*)void* but what makes no sense to me for a generic pointer is that I must somehow already know what type I need in order to grab a value. I'm a Java programmer; I understand generic types but this is something I struggle with.
So I wrote some code
typedef struct node
{
void* data;
node* link;
}Node;
typedef struct list
{
Node* head;
}List;
Node* add_new(void* data, Node* link);
void show(Node* head);
Node* add_new(void* data, Node* link)
{
Node* newNode = new Node();
newNode->data = data;
newNode->link = link;
return newNode;
}
void show(Node* head)
{
while (head != nullptr)
{
std::cout << head->data;
head = head->link;
}
}
int main()
{
List list;
list.head = nullptr;
list.head = add_new("My Name", list.head);
list.head = add_new("Your Name", list.head);
list.head = add_new("Our Name", list.head);
show(list.head);
fgetc(stdin);
return 0;
}
I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out? This implies I already need to know the type, and this reveals nothing about the generic nature of void* while I follow what is here although still no understanding.
Why am I expecting void* to cooperate and the compiler to automatically cast out the type that is hidden internally in some register on the heap or stack?
I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out?
You can't. You must know the valid types that the pointer can be cast to before you can dereference it.
Here are couple of options for using a generic type:
If you are able to use a C++17 compiler, you may use std::any.
If you are able to use the boost libraries, you may use boost::any.
Unlike Java, you are working with memory pointers in C/C++. There is no encapsulation whatsoever. The void * type means the variable is an address in memory. Anything can be stored there. With a type like int * you tell the compiler what you are referring to. Besides the compiler knows the size of the type (say 4 bytes for int) and the address will be a multiple of 4 in that case (granularity/memory alignment). On top, if you give the compiler the type it will perform consistency checks at compilation time. Not after. This is not happening with void *.
In a nutshell, you are working bare metal. The types are compiler directives and do not hold runtime information. Nor does it track the objects you are dynamically creating. It is merely a segment in memory that is allocated where you can eventually store anything.
The main reason to use void* is that different things may be pointed at. Thus, I may pass in an int* or Node* or anything else. But unless you know either the type or the length, you can't do anything with it.
But if you know the length, you can handle the memory pointed at without knowing the type. Casting it as a char* is used because it is a single byte, so if I have a void* and a number of bytes, I can copy the memory somewhere else, or zero it out.
Additionally, if it is a pointer to a class, but you don't know if it is a parent or inherited class, you may be able to assume one and find out a flag inside the data which tells you which one. But no matter what, when you want to do much beyond passing it to another function, you need to cast it as something. char* is just the easiest single byte value to use.
Your confusion derived from habit to deal with Java programs. Java code is set of instruction for a virtual machine, where function of RAM is given to a sort of database, which stores name, type, size and data of each object. Programming language you're learning now is meant to be compiled into instruction for CPU, with same organization of memory as underlying OS have. Existing model used by C and C++ languages is some abstract built on top of most of popular OSes in way that code would work effectively after being compiled for that platform and OS. Naturally that organization doesn't involve string data about type, except for famous RTTI in C++.
For your case RTTI cannot be used directly, unless you would create a wrapper around your naked pointer, which would store the data.
In fact C++ library contains a vast collection of container class templates that are useable and portable, if they are defined by ISO standard. 3/4 of standard is just description of library often referred as STL. Use of them is preferable over working with naked pointers, unless you mean to create own container for some reason. For particular task only C++17 standard offered std::any class, previously present in boost library. Naturally, it is possible to reimplement it, or, in some cases, to replace by std::variant.
Assuming I have no understanding of the type stored in void*, how do I get the value out
You don't.
What you can do is record the type stored in the void*.
In c, void* is used to pass around a binary chunk of data that points at something through one layer of abstraction, and recieve it at the other end, casting it back to the type that the code knows it will be passed.
void do_callback( void(*pfun)(void*), void* pdata ) {
pfun(pdata);
}
void print_int( void* pint ) {
printf( "%d", *(int*)pint );
}
int main() {
int x = 7;
do_callback( print_int, &x );
}
here, we forget thet ype of &x, pass it through do_callback.
It is later passed to code inside do_callback or elsewhere that knows that the void* is actually an int*. So it casts it back and uses it as an int.
The void* and the consumer void(*)(void*) are coupled. The above code is "provably correct", but the proof does not lie in the type system; instead, it depends on the fact we only use that void* in a context that knows it is an int*.
In C++ you can use void* similarly. But you can also get fancy.
Suppose you want a pointer to anything printable. Something is printable if it can be << to a std::ostream.
struct printable {
void const* ptr = 0;
void(*print_f)(std::ostream&, void const*) = 0;
printable() {}
printable(printable&&)=default;
printable(printable const&)=default;
printable& operator=(printable&&)=default;
printable& operator=(printable const&)=default;
template<class T,std::size_t N>
printable( T(&t)[N] ):
ptr( t ),
print_f( []( std::ostream& os, void const* pt) {
T* ptr = (T*)pt;
for (std::size_t i = 0; i < N; ++i)
os << ptr[i];
})
{}
template<std::size_t N>
printable( char(&t)[N] ):
ptr( t ),
print_f( []( std::ostream& os, void const* pt) {
os << (char const*)pt;
})
{}
template<class T,
std::enable_if_t<!std::is_same<std::decay_t<T>, printable>{}, int> =0
>
printable( T&& t ):
ptr( std::addressof(t) ),
print_f( []( std::ostream& os, void const* pt) {
os << *(std::remove_reference_t<T>*)pt;
})
{}
friend
std::ostream& operator<<( std::ostream& os, printable self ) {
self.print_f( os, self.ptr );
return os;
}
explicit operator bool()const{ return print_f; }
};
what I just did is a technique called "type erasure" in C++ (vaguely similar to Java type erasure).
void send_to_log( printable p ) {
std::cerr << p;
}
Live example.
Here we created an ad-hoc "virtual" interface to the concept of printing on a type.
The type need not support any actual interface (no binary layout requirements), it just has to support a certain syntax.
We create our own virtual dispatch table system for an arbitrary type.
This is used in the C++ standard library. In c++11 there is std::function<Signature>, and in c++17 there is std::any.
std::any is void* that knows how to destroy and copy its contents, and if you know the type you can cast it back to the original type. You can also query it and ask it if it a specific type.
Mixing std::any with the above type-erasure techinque lets you create regular types (that behave like values, not references) with arbitrary duck-typed interfaces.
I wanted to test a simple thing like the following:
#include <iostream>
#include <boost/variant.hpp>
template<typename T1,typename T2>
std::ostream& operator<<(std::ostream& os, const std::pair<T1,T2>& dt){
os << dt.first << dt.second;
return os;
}
int main(){
boost::variant<int, std::pair<int,int>, bool> v;
v = std::pair<int,int>(3,3);
std::cout << v << std::endl;
}
This should actually work, because for normal types, like int, double and so on, it compiles.
boost::variant has a printer vistor which it uses internally to output the content to the stream.
Actually this fails to compile, but I do not really know the problem:
The codes fails here: in variant_io.hpp
template <typename OStream>
class printer
: public boost::static_visitor<>
{
private: // representation
OStream& out_;
public: // structors
explicit printer(OStream& out)
: out_( out )
{
}
public: // visitor interface
template <typename T>
void operator()(const T& operand) const
{
out_ << operand; // HEEEEEEERRRRREE!!!!!!!!!!!!
}
private:
printer& operator=(const printer&);
};
With the message:
/usr/local/include/boost/variant/detail/variant_io.hpp|64|error: cannot bind 'std::basic_ostream<char>' lvalue to 'std::basic_ostream<char>&&'
Does someone know what I did wrong, and why?
Thanks a lot!
Most likely it's not finding your overload of operator <<, and then gets confused trying to match some other overload, leading to whatever message you're getting.
What you did wrong: You overloaded the stream operator in the global namespace instead of the namespace the right-hand-side class is defined in, so it's not found by ADL.
Trying to overload the stream operator for a standard class is a doomed exercise in the first place, unfortunately. You can't actually do that. I'm not sure if there is an explicit rule against it. However, if you place the operator in namespace std as you have to in order to make it properly findable by ADL, you violate the rule that you can't add your own stuff to namespace std except in very specific cases, this not being one of them.
The bottom line is that std::pair doesn't have a stream operator, and it's not possible to legally add a generic one that is useful. You can add one for a specific instantiation, if one of the parameters is a class you defined yourself; in this case the operator needs to be placed next to your own class.
Overloaded operator<< must be findable by argument dependent lookup. That means you have to put it in associated namespace of one of the arguments.
The first argument has only one associated namespace, std. The second also has only one associated namespace, std. However it is only permitted to overload symbols in std for user-defined types. Since std::pair<int, int> is not user-defined type, this is not allowed. However it is allowed for a structure or class you define yourself. Obviously in that case it is easier to place the overload to your namespace, not std.
That said if you put that overload in namespace std, it will actually work.
Also note, that boost::tuple does have operator<< (in separate header that you have to include, but it does), so you can use that instead.
I am trying to learn rvalue references, as an exercise I tried to do answer the following.
Is it possible to write a function that can tell (at least at runtime, better if at compile time) if the passed value is a value (non reference), a rvalue or an lvalue? for a generic type? I want to extract as much information about the type as possible.
An alternative statement of the problem could be:
Can I have a typeid-like function that can tell as much as possible about the calling expression?, for example (and ideally) if the type is T, T&, T const&, or T&&.
Currently, for example, typeid drops some information about the type and one can do better (as in the example the const and non-const reference are distiguished). But how much better than typeid can one possibly do?
This is my best attempt so far. It can't distinguish between a rvalue and a "constant". First and second case in the example).
Maybe distiguishing case 1 and 2 is not possible in any circumstance? since both are ultimately rvalue? the the question is Even if both are rvalues can the two cases trigger different behavior?
In any case, it seems I overcomplicated the solution as I needed to resort to rvalue conditional casts, and ended up with this nasty code and not even 100% there.
#include<iostream>
#include<typeinfo>
template<class T>
void qualified_generic(T&& t){
std::clog << __PRETTY_FUNCTION__ << std::endl;
std::clog
<< typeid(t).name() // ok, it drops any qualification
<< (std::is_const<typename std::remove_reference<decltype(std::forward<T>(t))>::type>::value?" const":"") // seems to detect constness rigth
<< (std::is_lvalue_reference<decltype(std::forward<T>(t))>::value?"&":"")
<< (std::is_rvalue_reference<decltype(std::forward<T>(t))>::value?"&&":"") // cannot distiguish between passing a constant and an rvalue expression
<< std::endl
;
}
using namespace std;
int main(){
int a = 5;
int const b = 5;
qualified_generic(5); // prints "int&&", would plain "int" be more appropriate?
qualified_generic(a+1); // prints "int&&" ok
qualified_generic(a); // print "int&", ok
qualified_generic(b); // print "int const&", ok
}
Maybe the ultimate solution to distiguish between the cases will involve detecting a constexpr.
UPDATE: I found this talk by Scott Meyers where he claims that "The Standard sometimes requires typeid to give the 'wrong' answer". http://vimeo.com/97344493 about minute 44. I wonder if this is one of the cases.
UPDATE 2015: I revisited the problem using Boost TypeIndex and the result is still the same. For example using:
template<class T>
std::string qualified_generic(T&& t){
return boost::typeindex::type_id_with_cvr<decltype(t)>().pretty_name();
// or return boost::typeindex::type_id_with_cvr<T>().pretty_name();
// or return boost::typeindex::type_id_with_cvr<T&&>().pretty_name();
// or return boost::typeindex::type_id_with_cvr<T&>().pretty_name();
}
Still it is not possible to distinguish the type of 5 and a+1 in the above example.