在 T 的静态成员中使用 sizeof(T) 的 C++ 错误

Question

为什么我的编译器不允许这样做？

class Resource
{
   private:
      static const int max_instances = 10;

      // set aside memory to be used later with placement new
      static char memory[max_instances * sizeof(Resource)]; // error: invalid application of 'sizeof' to incomplete type 'Resource'
};

Answer 1

我认为这是因为您还没有完成对 class 的定义，因此编译器不知道它的大小，因此会产生此错误。我不知道你想在这里完成什么，按照 aschepler 的回答并在实际 class 定义

之外定义 static const char

Answer 2

[expr.sizeof]/1:

The sizeof operator shall not be applied to an expression that has function or incomplete type

[class.mem]/6，强调我的：

A class is considered a completely-defined object type ([basic.types]) (or complete type) at the closing } of the class-specifier. Within the class member-specification, the class is regarded as complete within function bodies, default arguments, noexcept-specifiers, and default member initializers (including such things in nested classes). Otherwise it is regarded as incomplete within its own class member-specification.

Answer 3

我的老药方，把它包在一个函数里：

class Resource{
    //...
    auto constexpr max_instances=10;

    static auto& memory(){
        static std::aligned_storage<sizeof(Resource),alignof(Resource)> storage[max_instances];
        return storage;
    };
};

在 T 的静态成员中使用 sizeof(T) 的 C++ 错误

c++ error using sizeof(T) in static member of T

c++

sizeof