How do I implement "perfect forwarding" in a class template? - c++

I have a class template with template parameter T and a member of type T. I want to initialize that member with a parameter passed to the ctor and I also want the passed parameter to be moved if it's a rvalue reference and if T supports move semantic:
template <typename T>
class C {
public:
explicit C(T t) : t_(t)
{
}
explicit C(T&& t) : t_(std::move(t))
{
}
...
private:
T t_;
};
g++ 4.8 gives the following error if I try to pass rvalue reference to the ctor:
int main()
{
int x = 0;
C<int> p1{x}; // OK
C<int> p2{1}; // error g++-4.8: call of overloaded ‘C(<brace-enclosed initializer list>)’ is ambiguous
return 0;
}
The full error text:
g++-4.8 -std=c++11 -O2 -Wall -pedantic -pthread main.cpp
main.cpp: In function ‘int main()’:
main.cpp:23:16: error: call of overloaded ‘C()’ is ambiguous
C p2{1}; // error g++-4.8: call of overloaded ‘C()’ is ambiguous
^
main.cpp:23:16: note: candidates are:
main.cpp:12:11: note: C::C(T&&) [with T = int]
explicit C(T&& t) : t_(std::move(t))
^
main.cpp:8:14: note: C::C(T) [with T = int]
explicit C(T t) : t_(t)
^
main.cpp:6:7: note: constexpr C::C(const C&)
class C {
^
main.cpp:6:7: note: constexpr C::C(C&&)
Could someone help me, please?
Thanks!

I solved the problem by making the parameter t in C(T t) const reference.

Related

No matching function for call std::forward(const std::string &) with variadic arguments

I'm trying to make a movable wrapper to non-copyable, non-movable class, however I have a problem passing a const std::string variable to the constructor. The minimal example below produces following error:
#include <iostream>
#include <memory>
#include <string>
#include <utility>
struct X {
std::string x;
X(const std::string &x) : x(x) {}
X(const X &x) = delete;
X(X &&x) = delete;
};
struct Wrapper {
std::unique_ptr<X> x;
Wrapper(const Wrapper & wrapper) = delete;
Wrapper(Wrapper && wrapper) = default;
template<typename... Args>
Wrapper(Args&&... args) : x(std::make_unique<X>(std::forward(args)...)) {}
};
int main() {
const std::string XXX = "XXX";
Wrapper w{XXX};
std::cout << w.x->x << std::endl;
}
Error message here:
forwarding.cc:21:53: error: no matching function for call to 'forward'
Wrapper(Args&&... args) : x(std::make_unique<X>(std::forward(args)...)) {}
^~~~~~~~~~~~
forwarding.cc:26:13: note: in instantiation of function template specialization 'Wrapper::Wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > &>' requested here
Wrapper w{XXX};
^
/usr/bin/../lib/gcc/x86_64-linux-gnu/7.2.0/../../../../include/c++/7.2.0/bits/move.h:73:5: note: candidate template ignored: couldn't infer template argument '_Tp'
forward(typename std::remove_reference<_Tp>::type& __t) noexcept
^
/usr/bin/../lib/gcc/x86_64-linux-gnu/7.2.0/../../../../include/c++/7.2.0/bits/move.h:84:5: note: candidate template ignored: couldn't infer template argument '_Tp'
forward(typename std::remove_reference<_Tp>::type&& __t) noexcept
^
1 error generated.
You need to explicitly pass template parameters to std::forward:
std::forward<Args>(args)...
This is because std::forward needs some way of knowing the "original value category" of args..., which is impossible through template argument deduction alone as args is always an lvalue.
Lvalues will deduced as lvalue references in the context of template argument deduction for forwarding references (as a special rule), so std::forward can do its job by looking at the types inside Args....

Why don't the any_cast function overloads cause ambiguity?

Boost's <boost/any.hpp> has:
template<typename ValueType>
ValueType any_cast(any & operand);
template<typename ValueType>
inline ValueType any_cast(const any & operand);
(among other variants.) Shouldn't this combination cause ambiguity in calls such as boost::any_cast<int>(my_any); ?
I'm asking because if I write this program:
#include <boost/any.hpp>
#include <iostream>
template<typename ValueType>
ValueType any_cast(boost::any & operand)
{
return boost::any_cast<ValueType>(operand);
}
int main()
{
int x = 123;
boost::any my_any(x);
std::cout << "my_any = " << any_cast<int>(my_any) << "\n";
return 0;
}
I do get a complaint about ambiguity:
g++ -std=c++14 -O3 -Wall -pedantic -pthread main.cpp && ./a.out
main.cpp: In function 'int main()':
main.cpp:14:57: error: call of overloaded 'any_cast(boost::any&)' is ambiguous
std::cout << "my_any = " << any_cast<int>(my_any) << "\n";
^
main.cpp:5:11: note: candidate: ValueType any_cast(boost::any&) [with ValueType = int]
ValueType any_cast(boost::any & operand)
^~~~~~~~
In file included from main.cpp:1:0:
/usr/local/include/boost/any.hpp:281:22: note: candidate: ValueType boost::any_cast(const boost::any&) [with ValueType = int]
inline ValueType any_cast(const any & operand)
^~~~~~~~
/usr/local/include/boost/any.hpp:258:15: note: candidate: ValueType boost::any_cast(boost::any&) [with ValueType = int]
ValueType any_cast(any & operand)
^~~~~~~~
Why would the calls be ambiguous? The way you call the function the any argument is an lvalue. Thus, the any argument will either be const-qualified in which case the second overload is the only potential match or it is not const-qualified in which case the first overload is the better match (there is no conversion needed while the second overload would need a conversion from any& to any const&). If you call the function with a temporary any, it could bind to an rvalue overload (i.e., taking any&&) or, if that doesn't exist, it can bind to the const-qualified overload but not the non-const-qualified overload, again, not causing any ambiguity.
Actually, there is something interesting happening here: without the overload in the global namespace the function using the explicit template argument cannot be used! However, as soon as any function template is present, even a non-matching one, it can be used! Here is an example:
namespace foo {
struct bar {};
template <typename T> void bar_cast(bar&) {}
template <typename T> void bar_cast(bar const&) {}
template <typename T> void bar_cast(bar&&) {}
}
struct whatever;
template <typename T> void bar_cast(whatever);
int main()
{
foo::bar b;
bar_cast<int>(b);
}

deduce of argument of type class method (overloads by const qualifier) fails with trailing return type in gcc, but not in clang

Nothing clearer than an old good MCVE:
struct X {
auto get(int) const -> int { return {}; }
auto get(int) -> int { return {}; }
};
template <class R> auto f(auto (X::*)(int) const -> R) {}
// ^~~~ ~~~~
// trailing return type
int main() {
f(&X::get);
}
This fails in g++ (4.9.2 & 5.1.0). However if the old return type is used:
template <class R> auto f(R (X::*)(int) const) {}
// ^
// old return type
it works.
On clang (3.5.0) both variants work.
I know that trailing return type changes when the return type is inferred and the scope of it, so I wouldn't be quick to cast it as a gcc bug. So what does the standard says? Which compiler is right?
The most significant message in the error I think is
couldn't deduce template parameter ‘R’`
g++ full message:
main2.cpp: In function ‘int main()’:
main2.cpp:21:12: error: no matching function for call to ‘f(<unresolved overloaded function type>)’
f(&X::get);
^
main2.cpp:18:25: note: candidate: template<class R, class auto:1> auto f(auto:1 (X::*)(int) const)
template <class R> auto f(auto (X::*)(int) const -> R) {}
^
main2.cpp:18:25: note: template argument deduction/substitution failed:
main2.cpp:21:12: note: types ‘auto:1 (X::)(int) const’ and ‘int (X::)(int)’ have incompatible cv-qualifiers
f(&X::get);
^
main2.cpp:21:12: note: couldn't deduce template parameter ‘R’
<builtin>: recipe for target 'main2' failed
make: *** [main2] Error 1
As pointed in the question this is a gcc bug which was beed fixed in version 6
gcc.gnu.org/bugzilla/show_bug.cgi?id=69139

g++ 4.9.2 regression on pass reference to 'this'

This is a minimized part of Pointer to implementation code:
template<typename T>
class PImpl {
private:
T* m;
public:
template<typename A1>
PImpl(A1& a1) : m(new T(a1)) {
}
};
struct A{
struct AImpl;
PImpl<AImpl> me;
A();
};
struct A::AImpl{
const A* ppub;
AImpl(const A* ppub)
:ppub(ppub){}
};
A::A():me(this){}
A a;
int main (int, char**){
return 0;
}
It compilable on G++4.8 and prior and works as well. But G++4.9.2 compiler raise following errors:
prog.cpp: In constructor 'A::A()':
prog.cpp:24:15: error: no matching function for call to 'PImpl<A::AImpl>::PImpl(A*)'
A::A():me(this){}
^
prog.cpp:24:15: note: candidates are:
prog.cpp:9:5: note: PImpl<T>::PImpl(A1&) [with A1 = A*; T = A::AImpl]
> PImpl(A1& a1) : m(new T(a1)) {
^
prog.cpp:9:5: note: no known conversion for argument 1 from 'A*' to 'A*&'
prog.cpp:2:7: note: PImpl<A::AImpl>::PImpl(const PImpl<A::AImpl>&)
class PImpl {
^
prog.cpp:2:7: note: no known conversion for argument 1 from 'A*' to 'const PImpl<A::AImpl>&'
But it can be fixed by small hack. If i pass '&*this' instead of 'this' then it bring to compilable state.
Is it G++ regression or new C++ standards feature which eliminate backward compatibility?
We can make a simpler example that compiles on neither g++ 4.9 nor clang:
template <typename T>
void call(T& ) { }
struct A {
void foo() { call(this); }
};
int main()
{
A().foo();
}
That is because this is, from the standard, [class.this] (§9.3.2):
In the body of a non-static (9.3) member function, the keyword this is a prvalue expression whose value
is the address of the object for which the function is called.
You cannot take an lvalue reference to a prvalue, hence the error - which gcc explains better than clang in this case:
error: invalid initialization of non-const reference of type A*& from an rvalue of type A*
If we rewrite call to either take a const T& or a T&&, both compilers accept the code.
This didn't compile for me with gcc-4.6, so it seems that gcc-4.8 is where the regression occurred. It seems what you want is to take A1 by universal reference, ie: PImpl(A1 && a1). This compiles for me with both gcc-4.6, gcc-4.8 and gcc-4.9.

I am trying to pass a lambda as argument

Trying to pass a lambda to a constructor:
#include <functional>
#include <exception>
template<typename R>
class Nisse
{
private:
Nisse(Nisse const&) = delete;
Nisse(Nisse&&) = delete;
Nisse& operator=(Nisse const&) = delete;
Nisse& operator=(Nisse&&) = delete;
public:
Nisse(std::function<R> const& func) {}
};
int main()
{
Nisse<int> nisse([](){return 5;});
}
When I compile I get an error message:
Test.cpp: In function ‘int main()’:
Test.cpp:19:39: error: no matching function for call to ‘Nisse<int>::Nisse(main()::<lambda()>)’
Test.cpp:19:39: note: candidate is:
Test.cpp:14:9: note: Nisse<R>::Nisse(const std::function<R>&) [with R = int]
Test.cpp:14:9: note: no known conversion for argument 1 from ‘main()::<lambda()>’ to ‘const std::function<int>&’
The type of the template arg to std::function is wrong. Try using
Nisse(std::function<R()> const& func) {}
Specifically, the template argument needs to be a function type, but all you were passing was the desired return type.