I have the following code snippet that fails to compile...seems like it should, but it evading me right now. Greatly appreciate any help/suggestions! #include <atomic> #include <memory> struct MyClass { static void free_lock(std::atomic<int>** lck) { (*lck)->store(0); } typedef std::unique_ptr<std::atomic<int>*, decltype(&MyClass::free_lock)> lock_scope;
我有下面的代码片段，无法编译......看起来应该是这样，但它现在回避我。 非常感谢任何帮助/建议！ #include <atomic> #include <memory> struct MyClass { static void free_lock(std::atomic<int>** lck) { (*lck)->store(0); } typedef std::unique_ptr<std::atomic<int>*, decltype(&MyClass::free_lock)> lock_scope; static lock_scope get_lock() { static std::atomic<
/usr/include/c++/4.4/bits/ios_base.h: In member function 'std::basic_ios >& std::basic_ios >::operator=(const std::basic_ios >&)': /usr/include/c++/4.4/bits/ios_base.h:793: error: 'std::ios_base& std::ios_base::operator=(const std::ios_base&)' is private /usr/include/c++/4.4/iosfwd:47: error: within this context /usr/include/c++/4.4/iosfwd: In member
/usr/include/c++/4.4/bits/ios_base.h：在成员函数'std :: basic_ios>＆std :: basic_ios> :: operator =（const std :: basic_ios>＆）'中： /usr/include/c++/4.4/bits/ios_base.h:793：error：'std :: ios_base＆std :: ios_base :: operator =（const std :: ios_base＆）'是私人的 /usr/include/c++/4.4/iosfwd:47：错误：在此上下文中 /usr/include/c++/4.4/iosfwd：在成员函数'std :
Studing STL I have written aa simple program to test functors and modifiers. My question is about the difference aon using CLASS or STRUCT to write a functor and try to operate on it with function adaptors. As far as I understand in C++ the difference beetween CLASS and STRUCT is that in the last case the members are public by default. This is also what I read many times in the answers in this
学习STL我写了一个简单的程序来测试函数和修饰符。 我的问题是关于使用CLASS或STRUCT编写仿函数并试图用函数适配器对其进行操作的区别。 据我所知，在C ++中CLASS和STRUCT之间的区别在于，在最后一种情况下，成员默认是公开的。 这也是我在本网站的答案中多次阅读的内容。 因此，请解释为什么即使我尝试使用not2修饰符时声明所有成员（只是函数overloading（））public，但这段代码仍然无法编译。 （我还没有尝试其他修饰符
In the case of a substitution failure involving a template alias (eg a template alias on a missing member typename, as in the code snippet below), should an error be triggered ? Clang and gcc seem to disagree on this: // some types struct bar { }; struct foo { typedef void member_type; }; // template alias template<class T> using member = typename T::member_type; template<cla
在涉及模板别名的替换失败的情况下（例如，缺少成员类型名称的模板别名，如下面的代码片段所示），是否应该触发错误？ 铿锵和海湾合作委员会似乎不同意这一点： // some types struct bar { }; struct foo { typedef void member_type; }; // template alias template<class T> using member = typename T::member_type; template<class T> void baz(... ) { } // only works for gcc, clang fails with
While compiling my program today, I noticed something very strange (which I'm sure can be explained somehow) in the memory consumption pattern of GCC (compilation step). The process called "cc1plus" was using approximately 10 GB of RAM for a program with less than 10 000 lines of code. After commenting and uncommenting lines of code I finally found the "culprit": std::b
在今天编写我的程序时，我注意到GCC的内存消耗模式（编译步骤）中有一些非常奇怪的东西（我敢肯定可以用某种方式解释）。 称为“cc1plus”的进程使用大约10 GB的RAM用于少于10000行代码的程序。 在评论和取消注释代码行后，我终于找到了“罪魁祸首”： std::bitset<UINT_MAX> 如果你想测试自己，请尝试这个很短的程序： #include <iostream> #include <bitset> #include <climits> std::bitset<UINT
Related: Function returning constexpr does not compile I feel like constexpr is limited in usefulness in C++11 because of the inability to define two functions that would otherwise have the same signature, but have one be constexpr and the other not constexpr. In other words, it would be very helpful if I could have, for example, a constexpr std::string constructor that takes constexpr argumen
相关：函数返回constexpr不编译 我觉得constexpr在C ++ 11的有用性方面有限，因为无法定义两个函数，否则它们会具有相同的签名，但一个是constexpr，另一个不是constexpr。 换句话说，如果我可以有一个constexpr std :: string构造函数，它只接受constexpr参数，而一个非constexpr std :: string构造函数用于非constexpr参数，这将非常有帮助。 另一个例子是理论上复杂的函数，可以通过使用状态来提高效率。 你不能用conste
I have wrote the following code to get the offset of a tuple element template<size_t Idx,class T> constexpr size_t tuple_element_offset() { return static_cast<size_t>( reinterpret_cast<char*>(&std::get<Idx>(*reinterpret_cast<T*>(0))) - reinterpret_cast<char*>(0)); } This is actually similar to the implementation of the off
我写了下面的代码来获取元组元素的偏移量 template<size_t Idx,class T> constexpr size_t tuple_element_offset() { return static_cast<size_t>( reinterpret_cast<char*>(&std::get<Idx>(*reinterpret_cast<T*>(0))) - reinterpret_cast<char*>(0)); } 这实际上类似于offsetof宏的实现。 它看起来很丑，但编译和工作正常gcc-4.6 typedef std::t
According to Generalized Constant Expressions—Revision 5 the following is illegal. constexpr int g(int n) // error: body not just ‘‘return expr’’ { int r = n; while (--n > 1) r *= n; return r; } This is because all 'constexpr' functions are required to be of the form { return expression; } { return expression; } . I can't see any reason that this needs to be so. I
根据广义常量表达式 - 第5版，以下是非法的。 constexpr int g(int n) // error: body not just ‘‘return expr’’ { int r = n; while (--n > 1) r *= n; return r; } 这是因为所有'constexpr'函数都必须是{ return expression; } { return expression; } 。 我看不出有什么理由需要这样做。 在我看来，唯一真正需要的是没有外部状态信息被读取/写入，并且传入的参数也是“constexpr”语句。 这意味
Recurvise call C++ template function with Variable template void int void foo() { } template <typename T, typename ...U> void foo() { foo<U...>(); } int main() { foo<int, char, int>(); return 0; } The compile meesage such as: error C2672: 'foo': no matching overloaded function found note: see reference to function template instantiation 'void foo(v
重新调整使用变量模板调用C ++模板函数 void int void foo() { } template <typename T, typename ...U> void foo() { foo<U...>(); } int main() { foo<int, char, int>(); return 0; } 编译代码如： 错误C2672：'foo'：没有匹配的重载函数找到注意：请参阅参考函数模板实例化'void foo（void）'正在编译注意：请参阅参考函数模板实例化'void foo（void）'正在编
I'm trying to define a template class that has some operations on a type. The method ToString should be implemented to call the ToString in the namespace of whatever type the Tools template class has been instantiated with. namespace X { class SomeType { }; std::wstring ToString(SomeType) { // ... } } template<class T> class Tools { static auto ToS
我试图定义一个在类型上有一些操作的模板类。 应该实现ToString方法，以便在Tools类模板实例化的任何类型的名称空间中调用ToString。 namespace X { class SomeType { }; std::wstring ToString(SomeType) { // ... } } template<class T> class Tools { static auto ToString(T& t) { return ToString(t); } } 我在ToString的Tools实现中遇到错误。 编译器