The following code compiles properly under x64 msvc x19.30 and gcc 11 but fails to compile under clang 13.0.1:
"error: cannot pass object of non-trivial type 'std::shared_ptr<std::pair<int, std::variant<Struct1, Struct2, UnsupportedStruct>>>' through variadic function;"
Does anyone know what the problem is?
The following code produces different outputs depending on passing object:
#include <iostream>
#include <variant>
#include <memory>
struct Struct1{};
struct Struct2{};
struct UnsupportedStruct{};
using VarTypeData = std::variant<Struct1, Struct2, UnsupportedStruct>;
using VarType = std::pair<int, VarTypeData>;
namespace
{
void print(Struct1&, std::shared_ptr<VarType> v) {std::cout << v->first << ": Struct1\n";}
void print(Struct2&, std::shared_ptr<VarType> v) {std::cout << v->first << ": Struct2\n";}
void print(...) {std::cout << "no implementation";}
}
int main()
{
VarType data1 = std::make_pair(100, UnsupportedStruct{});
auto pointerData = std::make_shared<VarType>(data1);
std::visit([&pointerData](auto& c) {print(c, pointerData);}, pointerData->second);
std::cout << std::endl;
pointerData->second = Struct1{};
std::visit([&pointerData](auto& c) {print(c, pointerData);}, pointerData->second);
}
This code works fine for clang after dereferencing:
#include <iostream>
#include <variant>
#include <memory>
struct Struct1{};
struct Struct2{};
struct UnsupportedStruct{};
using VarTypeData = std::variant<Struct1, Struct2, UnsupportedStruct>;
using VarType = std::pair<int, VarTypeData>;
namespace
{
void print(const Struct1&, const VarType& v) {std::cout << v.first << ": Struct1\n";}
void print(const Struct2&, const VarType& v) {std::cout << v.first << ": Struct2\n";}
void print(...) {std::cout << "no implementation";}
}
int main()
{
VarType data1 = std::make_pair(100, UnsupportedStruct{});
auto pointerData = std::make_shared<VarType>(data1);
std::visit([&pointerData](auto& c) {print(c, *pointerData);}, pointerData->second);
std::cout << std::endl;
pointerData->second = Struct1{};
std::visit([&pointerData](auto& c) {print(c, *pointerData);}, pointerData->second);
}
thanks to #康桓瑋 for the answer.
this code does not work for clang, because of
void print(...) {std::cout << "no implementation";}
answer: void print(...) is a C function, where variadic actually means the
's parameter. It accepts only trivial types, which
std::shared_ptr is not. So the behavior is undefined or only
conditionally supported
So, the following changes fix the problem:
template<class... Args>
void print(Args&&...) {std::cout << "no implementation";}
Related
The following code compiled:
#include <iostream>
#include <functional>
typedef void (*f_type) (int a);
void say(int a)
{
std::cout << a << "!" << std::endl;
}
int main()
{
int a=5;
say(a);
std::function<void(int)> fn{say};
f_type fn_pointer = fn.target<void(int)>();
if(fn_pointer)
fn_pointer(a);
else
std::cout << "null ptr" << std::endl;
return 0;
}
but when executed prints:
5!
nullptr
I would like to understand why target returned an empty ptr, and not a pointer to the function "say".
note : it compiles for c++ up to c++14, for c++17 onward, compilation fails with error (which is cryptic to me):
In file included from /usr/include/c++/7/functional:58:0,
from main.cpp:11:
/usr/include/c++/7/bits/std_function.h: In instantiation of ‘_Functor* std::function<_Res(_ArgTypes ...)>::target() [with _Functor = void(int); _Res = void; _ArgTypes = {int}]’:
<span class="error_line" onclick="ide.gotoLine('main.cpp',28)">main.cpp:28:46</span>: required from here
/usr/include/c++/7/bits/std_function.h:733:9: error: invalid use of const_cast with type ‘void (*)(int)’, which is a pointer or reference to a function type
return const_cast<_Functor*>(__func);
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Reproduced it on VS2017, seems like the target method returns a pointer to pointer (as in returning a pointer to the actual function pointer stored in the object), and expecting its template type argument accordingly. Here is a modified example that works:
#include <iostream>
#include <functional>
typedef void(*f_type) (int a);
void say(int a)
{
std::cout << a << "!" << std::endl;
}
int main()
{
int a = 5;
say(a);
std::function<void(int)> fn{say};
f_type* fn_pointer = fn.target<void(*)(int)>();
if (fn_pointer)
(*fn_pointer)(a);
else
std::cout << "null ptr" << std::endl;
return 0;
}
Confirmed target returning a pointer to the actual function pointer by running the following:
#include <iostream>
#include <functional>
typedef void(*f_type) (int a);
void say(int a)
{
std::cout << a << "!" << std::endl;
}
void say_boo(int a)
{
std::cout << "booooo" << std::endl;
}
int main()
{
int a = 5;
std::function<void(int)> fn{say};
f_type* fn_pointer = fn.target<void(*)(int)>();
(*fn_pointer)(a);
fn = say_boo;
(*fn_pointer)(a);
return 0;
}
This produced the following output:
5!
booooo
i have this piece of code (http://coliru.stacked-crooked.com/a/ee05a00fc8ab5057):
#include <type_traits>
struct unregistered;
unregistered register_type(...);
template<class T>
constexpr bool is_registered = !std::is_same_v<unregistered, decltype(register_type(std::declval<T>()))>;
template<class T>
struct test_registration
{
static_assert(is_registered<T>, "Type is not registered!");
};
struct foo{};
struct bar{};
void register_type(int);
void register_type(char);
void register_type(void*);
void register_type(foo);
void register_type(foo*);
#include <boost/core/demangle.hpp>
#include <iostream>
int main()
{
std::cout << boost::core::demangle(typeid(test_registration<foo>).name()) << "\n";
std::cout << boost::core::demangle(typeid(test_registration<foo*>).name()) << "\n";
std::cout << boost::core::demangle(typeid(test_registration<int>).name()) << "\n";
std::cout << boost::core::demangle(typeid(test_registration<char>).name()) << "\n";
std::cout << boost::core::demangle(typeid(test_registration<void*>).name()) << "\n";
std::cout << boost::core::demangle(typeid(test_registration<long>).name()) << "\n";
std::cout << boost::core::demangle(typeid(test_registration<bar>).name()) << "\n";
return 0;
}
The compiler generates errors for the calls using int, char, void*, long and bar.
I expect the errors for long and bar.
What I do not understand is:
Why are int, char, and foo treated differently?
Why are void*, and foo* treated differently? (both are ptrs)
I assume the reason has to do with foo beeing class type and int and char beeing plain types.
I would like to know the reason. Is it a compiler bug or is there some passage in the standard explaining this behaviour?
What I am asking is not how I can fix this problem (see link to coliru for a fix).
What I want to know is why it is behaving like this.
I want to use RTTI and mangled function (pointer) type strings.
Before you categorize this as an XY problem, I'm aware that there are better options to bind functions using polymorphism etc.
This is a purely academical question how to use typeid() properly with function pointers that should differ by their legally overloaded signatures.
If I use the following code, it seems I can retrieve unique typeinfo::name() values for various namespaces/types:
#include <iostream>
#include <typeinfo>
#include <string>
void foo(int) {
}
namespace woozle {
void goozle(int) {}
}
struct bar {
void baz(int) {}
static void boo(int) {}
};
int main() {
std::cout << typeid(&foo).name() << std::endl;
std::cout << typeid(&woozle::goozle).name() << std::endl;
std::cout << typeid(&bar::baz).name() << std::endl;
std::cout << typeid(&bar::boo).name() << std::endl;
}
The output is:
PFviE
PFviE
M3barFviE
PFviE
Perfectly what I expected (I assume the i in the mangled name refers to the parameter signature).
Now I want to have something like this (which is perfectly legal function overloading):
#include <iostream>
#include <typeinfo>
#include <string>
void foo(int) {
}
void foo(std::string) {
}
namespace woozle {
void goozle(int) {}
void goozle(std::string) {}
}
struct bar {
void baz(int) {}
static void boo(int) {}
void baz(std::string) {}
static void boo(std::string) {}
};
int main() {
std::cout << typeid(&foo).name() << std::endl;
std::cout << typeid(&woozle::goozle).name() << std::endl;
std::cout << typeid(&bar::baz).name() << std::endl;
std::cout << typeid(&bar::boo).name() << std::endl;
}
and of course the compiler complains about ambiguity:
main.cpp: In function 'int main()':
main.cpp:24:25: error: address of overloaded function with no contextual type information
std::cout << typeid(&foo).name() << std::endl;
^~~~
main.cpp:25:25: error: address of overloaded function with no contextual type information
std::cout << typeid(&woozle::goozle).name() << std::endl;
^~~~~~~
main.cpp:26:25: error: address of overloaded function with no contextual type information
std::cout << typeid(&bar::baz).name() << std::endl;
^~~~
main.cpp:27:25: error: address of overloaded function with no contextual type information
std::cout << typeid(&bar::boo).name() << std::endl;
^~~~
TL;DR
What is the proper syntax to specify a specific function overload with typeid() (if there is any)?
How can I provide the "contextual type information" that is demanded from the error message?
I'm coming from here in deep thinking mode.
To select a specific function from a set of overloaded functions you can use use the cast notation:
std::cout << typeid(static_cast<void (*)(int)>(foo)).name() << std::endl;
std::cout << typeid(static_cast<void (*)(std::string)>(foo)).name() << std::endl;
std::cout << typeid(static_cast<void (bar::*)(int)>(&bar::baz)).name() << std::endl;
std::cout << typeid(static_cast<void (bar::*)(std::string)>(&bar::baz)).name() << std::endl;
Specifically with typeid though, if you have the type already written down, you can skip the actual function name.
std::cout << typeid(void (*)(int)).name() << std::endl;
is shorter and does the job just as well.
I have succeeded writing a class like this one, capturing this in a lambda defined as non-static attribute of said class:
#include <memory>
#include <iostream>
#include <functional>
struct S
{
S()
{
std::cout << "S::S()[" << this << "]" << std::endl;
}
std::string y_{"hi mate"};
int x_;
std::function<void(int*)> del_{[this](int *ptr)
{
std::cout << "Deleting ptr[" << ptr << "] this[" << this << "] this->y_[" << this->y_ << "]" << std::endl;
}};
std::unique_ptr<decltype(x_), decltype(del_)> unique_{&x_, del_};
};
int main()
{
S s;
}
This compiles and seems to run just fine.
However, with a templated class, it doesn't work anymore:
#include <memory>
#include <iostream>
#include <functional>
template <typename>
struct S
{
S()
{
std::cout << "S::S()[" << this << "]" << std::endl;
}
std::string y_{"hi mate"};
int x_;
std::function<void(int*)> del_{[this](int *ptr)
{
std::cout << "Deleting ptr[" << ptr << "] this[" << this << "] this->y_[" << this->y_ << "]" << std::endl;
}};
std::unique_ptr<decltype(x_), decltype(del_)> unique_{&x_, del_};
};
int main()
{
S<int> s;
}
$> g++ -std=c++1y custom_deleter_template.cpp
~/test custom_deleter_template.cpp: In instantiation of ‘struct
S::’: custom_deleter_template.cpp:9:3: required
from ‘S< >::S() [with
= int]’ custom_deleter_template.cpp:24:10:
required from here custom_deleter_template.cpp:15:35: internal
compiler error: in tsubst_copy, at cp/pt.c:12569
std::function del_{[this](int *ptr)
^ Please submit a full bug report, with preprocessed source if appropriate. See
for instructions.
Preprocessed source stored into /tmp/pyro/ccxfNspM.out file, please
attach this to your bugreport.
Before filing a bugreport (which I can't do, they blocked account creation), is it normal that it does not compile, based on what the standard says?
Compiler is g++ (Ubuntu 4.9.2-0ubuntu1~14.04) 4.9.2, used flag -std=c++1y. Same thing happens with flag -std=c++11.
This is indeed a bug in GCC, which is already being tracked.
It seems to affect 4.8 and 4.9. As pointed out in the comments this particular example works fine for 4.7 and 5.0. You can see that for yourself here and play with the different versions of gcc.
However this reduced version of your code with no external dependency still crashes with 5.0:
template <typename>
struct S {
int f{[this](){return 42;}()};
};
int main(){
return S<int>{}.f; // should return 42
}
I would suggest that you wait for the bug I referenced to be fixed before using your code, or switch to another compiler ;).
The following noddy test code:
#include <iostream>
#include <list>
#include <boost/any.hpp>
#include <boost/foreach.hpp>
#include <typeinfo.h>
using boost::any_cast;
using std::cout;
using std::cerr;
typedef std::list<boost::any> many;
template <typename T>
inline bool is_any(const boost::any& op)
{
return (op.type() == typeid(T));
}
int main()
{
many theStrangeList;
theStrangeList.push_back("Can you really...");
theStrangeList.push_back(std::string ("do random types in 1 container?"));
theStrangeList.push_back(6.359);
theStrangeList.push_back(7);
BOOST_FOREACH(boost::any a, theStrangeList)
{
try
{
if (is_any<const char*>(a))
{
cout << any_cast<const char*>(a) << '\n';
}
else if (is_any<std::string>(a))
{
cout << any_cast<std::string>(a) << '\n';
}
else if (is_any<double>(a))
{
cout << "double = " << any_cast<double>(a) << '\n';
}
}
catch (const boost::bad_any_cast& e)
{
cerr << e.what();
cerr << "\n";
}
}
return 0;
}
Compiles and works fine using Sun's CC compiler and default settings.
However when using g++ I get the following :
$ g++ -I$BOOST_ROOT -o myany myany.cpp
myany.cpp:5:22: typeinfo.h: No such file or directory
/ilx/boost_1_41_0/boost/any.hpp: In constructor `boost::any::holder<ValueType>::holder(const ValueType&) [with ValueType = char[18]]':
/ilx/boost_1_41_0/boost/any.hpp:47: instantiated from `boost::any::any(const ValueType&) [with ValueType = char[18]]'
myany.cpp:21: instantiated from here
/ilx/boost_1_41_0/boost/any.hpp:122: error: ISO C++ forbids assignment of arrays
This is g++ version 3.4.3, so it might be different on a 4.x version, I'll try it later. Is this the reason why there isn't a 'is_any' template included with boost any, or is it a compiler bug?
I get the same result if I remove the template, as you would expect with an inlined function.
(related question)
For the first error try
#include <typeinfo>
not
#include <typeinfo.h>
Seems I only answered the second part of the question, so here I go with the first part as well:
Is this the reason why there isn't a 'is_any' template included with boost any?
There are no actual need to is_any, do the following instead:
if (const std::string* s = boost::any_cast<std::string>(&a))
{
std::cout << "string = " << *s << '\n';
}
else if (const double* d = boost::any_cast<double>(&a))
{
std::cout << "double = " << *d << '\n';
}
But this isn't extensible, prefer using boost::variant instead.
Is it a compiler bug?
It is a compiler bug in Sun CC. gcc is correct, the type of "Can you really..." is char[18], which doesn't satisfy the requirements of boost::any:
A ValueType is CopyConstructible.
A ValueType is optionally Assignable. The strong exception-safety guarantee is required for all forms of assignment.
The destructor for a ValueType upholds the no-throw exception-safety guarantee.