无法将Persistent<Function>添加到vector<Persistent<Function>>
无法将Persistent<Function>添加到vector<Persistent<Function>>
提问于 2016-01-11 19:11:26
我正在从v8 3.x迁移到4.y，并且在我的一个头文件中定义的函数中我遇到了问题。
//JavascriptBase.h
namespace Company {
class Base {
protected:
void registerHandler(v8::Persistent<v8::Function>& func) {
user_functions_.push_back(func);
}
private:
std::vector<v8::Persistent<v8::Function>> user_functions_;
}
}当我试图编译时，我会得到以下错误：
third-party2/v8/4.7.39/gcc-4.8.1-glibc-2.17-fb/281a9e6/include/v8.h:667:53: error: assigning to 'v8::Object *volatile' from incompatible type 'v8::Primitive *'
while (false) { *(static_cast<O* volatile*>(0)) = static_cast<Primitive*>(0); };
^ ~~~~~~~~~~~~~~~~~~~~~~~~~~
third-party2/v8/4.7.39/gcc-4.8.1-glibc-2.17-fb/281a9e6/include/v8.h:663:5: note: in instantiation of function template specialization 'v8::NonCopyablePersistentTraits<v8::Function>::Uncompilable<v8::Object>' requested here
Uncompilable<Object>();
^
third-party2/v8/4.7.39/gcc-4.8.1-glibc-2.17-fb/281a9e6/include/v8.h:7175:6: note: in instantiation of function template specialization 'v8::NonCopyablePersistentTraits<v8::Function>::Copy<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >' requested here
M::Copy(that, this);
^
third-party2/v8/4.7.39/gcc-4.8.1-glibc-2.17-fb/281a9e6/include/v8.h:729:5: note: in instantiation of function template specialization 'v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >::Copy<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >' requested here
Copy(that);
^
third-party2/libgcc/4.8.1/gcc-4.8.1-glibc-2.17-fb/8aac7fc/include/c++/4.8.1/ext/new_allocator.h:120:23: note: in instantiation of member function 'v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >::Persistent' requested here
{ ::new((void *)__p) _Up(std::forward<_Args>(__args)...); }
^
third-party2/libgcc/4.8.1/gcc-4.8.1-glibc-2.17-fb/8aac7fc/include/c++/4.8.1/bits/alloc_traits.h:254:8: note: in instantiation of function template specialization '__gnu_cxx::new_allocator<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > >::construct<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >, const v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > &>' requested here
{ __a.construct(__p, std::forward<_Args>(__args)...); }
^
third-party2/libgcc/4.8.1/gcc-4.8.1-glibc-2.17-fb/8aac7fc/include/c++/4.8.1/bits/alloc_traits.h:393:4: note: in instantiation of function template specialization 'std::allocator_traits<std::allocator<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > > >::_S_construct<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >, const v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > &>' requested here
{ _S_construct(__a, __p, std::forward<_Args>(__args)...); }
^
third-party2/libgcc/4.8.1/gcc-4.8.1-glibc-2.17-fb/8aac7fc/include/c++/4.8.1/bits/stl_vector.h:905:21: note: in instantiation of function template specialization 'std::allocator_traits<std::allocator<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > > >::construct<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >, const v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > &>' requested here
_Alloc_traits::construct(this->_M_impl, this->_M_impl._M_finish,
^
./Company/JavaScriptBase.h:84:21: note: in instantiation of member function 'std::vector<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> >, std::allocator<v8::Persistent<v8::Function, v8::NonCopyablePersistentTraits<v8::Function> > > >::push_back' requested here
user_functions_.push_back(func);
^
1 error generated.我已经做了一些搜索，发现Persistent对象是不可分配的，但是我不知道如何处理这些信息。任何帮助都将不胜感激。
回答 2
Stack Overflow用户
回答已采纳
发布于 2016-01-11 19:31:13
如果它们是不可分配的，那么你(显然)就不能分配它们。而且，由于std::vector<>依赖于使用赋值运算符复制值，所以您的方法将无法工作。
但是，您可以存储指向它们的指针，因为复制任何东西的指针都是很好的定义。
Stack Overflow用户
发布于 2016-01-12 02:23:17
您可以使用这个“ty胡枝子”使一个可复制的持久
using CopyablePersistent = v8::Persistent<T, v8::CopyablePersistentTraits<T>>;这假设它在一个模板化的类中，但是你可以做你想要的T。
编辑:显然这是一件“非常糟糕”的事情，但我不知道为什么。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/34729216
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