Given the following code here in IDEOne:
#include <iostream>
#include <vector>
#include <list>
template<typename T>
class MyVectorCollection
{
using collection = std::vector<T>;
};
template<typename C, typename T>
class MyGenericCollection
{
using collection = C;
};
template<typename C, typename T>
class MyMoreGenericCollection
{
using collection = C<T>;
};
int main() {
// your code goes here
MyVectorCollection<int> a;
MyGenericCollection<std::list<int>, int> b;
MyMoreGenericCollection<std::list, int> c; // How to do this?
return 0;
}
I get the error:
prog.cpp:20:24: error: ‘C’ is not a template
using collection = C<T>;
^
prog.cpp: In function ‘int main()’:
prog.cpp:27:43: error: type/value mismatch at argument 1 in template parameter list for ‘template<class C, class T> class MyMoreGenericCollection’
MyMoreGenericCollection<std::list, int> c;
^
prog.cpp:27:43: note: expected a type, got ‘list’
How can I write my code such that I can use a C<T> at compile time without having an explicit list of potential specialisations, and avoiding macros, if possible? I realise std::list isn't a typename, but I don't know how to progress, and I have been unable to find a similar question here.
(Note that this is just an MCVE, my actual usage is much more involved.)
Just to tidy up, here is the solution. The search term I was looking for is template template, and whilst I did find a possible duplicate, I think this question and answer is much simpler to follow.
So, thanks to the hint from #Some programmer dude, I looked up template template and updated the code to this, which does compile:
template<template<typename, typename> class C, typename T>
class MyMoreGenericCollection
{
using collection = C<T, std::allocator<T>>;
};
We declare the first template parameter as a template itself, and remembering that the standard library constructors take two parameters, we need to make the inner template take two parameters. There is (as far as I am aware) no way to automatically use the default second parameter, so for the sake of this example I explicitly state the default.
I could of course add a third parameter to the master template that could be used to specify the allocator, which itself would also be a template, but I leave that as an exercise for th reader.
Related
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)
I learned that from C++17, with the deduction guides, template arguments of std::vector can be deduced e.g. from the initialization:
std::vector vec = { function_that_calculate_and_return_a_specifically_templated_type() }
However I do not have the luxury of C++17 in the machine where I want to compile and run the code now.
Is there any possible workaround for C++11? If more solutions exist, the best would be the one that keep the readability of the code.
At the moment the only idea that I have is to track the various cases along the code (luckily they should not be too many) and make some explicit typedef/using.
Any suggestion is very welcome
The usual way to use type deduction for class template when CTAD is not available is providing a make_* function template, e.g. for your case (trailing return type is necessary for C++11):
#include <vector>
#include <type_traits>
#include <tuple>
template <class ...Args>
auto make_vec(Args&&... args) ->
std::vector<typename std::decay<typename std::tuple_element<0, std::tuple<Args...>>::type>::type>
{
using First = typename std::decay<typename std::tuple_element<0, std::tuple<Args...>>::type>::type;
return std::vector<First>{std::forward<Args>(args)...};
}
You can invoke the above with
const auto v = make_vec(1, 2, 3);
which gets at least kind of close to CTAD in the sense that you don't have to explicitly specify the vector instantiation.
While the answer by lubgr is a correct way, the following template is simpler and seems to work as well:
#include <vector>
#include <string>
template <typename T>
std::vector<T> make_vec(const std::initializer_list<T> &list)
{
return std::vector<T>(list);
}
int main()
{
auto v = make_vec({1,2,3});
auto v2 = make_vec({std::string("s")});
std::string s("t");
auto v3 = make_vec({s});
return v.size() + v2.size() + v3.size();
}
One advantage of using the initializer_list template directly are more clear error messages if you pass mixed types like in make_vec({1,2,"x"});, because the construction of the invalid initializer list now happens in non-templated code.
Using C++11, g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-18).
Lets pretend I have a templated function (pardon my terminology if it isn't quite right).
I want to perform a "general" algorithm based on what was supposed to be compile-time instances of "field". Where the only things that really changed are these constants which I moved into trait classes (only added one here but imagine there are more). Originally I was declaring it as
constexpr field FIELD1{1};
However in C++11, non-type template params need to have external linkage (unlike C++14 which can have internal and external linkage?). So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr and it seems that losing that constexpr constructor this field is no longer a valid constant expression to qualify as a non-type template param.
Any suggestions if there is some way I can get around this? Open to a new method of doing things. Below is a simplified (incomplete, and non-compiling version to get the gist of the organization).
So the error I am seeing is along the lines of
error: the value of ‘FIELD1’ is not usable in a constant expression
note: ‘FIELD1’ was not declared ‘constexpr’
extern const field FIELD1;
Not quite sure what could be a best alternative.
I can get rid of the second error by removing the constexpr from the constructor. But then I don't know how to approach the constant expression issue.
field.H
struct field
{
int thingone;
constexpr field(int i):thingone(i){}
};
extern const field FIELD1;
field.C
#include "field.H"
const field FIELD1{0};
field_traits.H
#include "field.H"
template< const field& T >
class fieldTraits;
template< >
class fieldTraits<FIELD1>
{
public:
// Let's say I have common field names
// with different constants that I want to plug
// into the "function_name" algorithm
static constexpr size_t field_val = 1;
};
function.H
#include "field.H"
template< const field& T, typename TT = fieldTraits<T> >
void function_name()
{
// Let's pretend I'm doing something useful with that data
std::cout << T.thingone << std::endl;
std::cout << TT::field_val << std::endl;
}
So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr [...]
Per my comment, you can. It wouldn't work for you, but it's a step that helps in coming up with something that would work:
extern constexpr int i = 10;
This is perfectly valid, gives i external linkage, and makes i usable in constant expressions.
But it doesn't allow multiple definitions, so it can't work in a header file which is included in multiple translation units.
Ordinarily, the way around that is with inline:
extern inline constexpr int i = 10;
But variables cannot be declared inline in C++11.
Except... when they don't need to be declared inline because the effect has already been achieved implicitly:
struct S {
static constexpr int i = 10;
};
Now, S::i has external linkage and is usable in constant expressions!
You may not even need to define your own class for this, depending on the constant's type: consider std::integral_constant. You can write
using i = std::integral_constant<int, 10>;
and now i::value will do exactly what you want.
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?