如何计算类成员在编译时的偏移量？ [英] How to calculate offset of a class member at compile time?

问题描述

在C ++中给定类定义

  class A 
 {
 public：
 //方法定义
 .... 
 
 private：
 int i; 
 char * str 
 .... 
} 
  

一个类成员在编译时使用C ++模板元编程？类不是POD，并且可以有虚方法，基本和对象数据成员。

解决方案

基于Matthieu M.的回答，但更短，没有宏：

  template< typename T，typename U> constexpr size_t offsetOf（U T :: * member）
 {
 return（char *）&（（T *）nullptr-> * member） - （char *）nullptr; 
} 
  

这样调用：

  struct X {int a，b，c，d; } 
 
 std :: cout<< offset of c in X ==<< offsetOf（& X :: c）; 
   
 
 
 
 编辑：
 
 
 
 Jason Rice是正确的。这不会在C ++ 11中产生实际的常量表达式。鉴于 http://en.cppreference.com/w中的限制，它看起来不太可能/ cpp / language / constant_expression   - 特别是没有指针差异， reinterpret_cast 可以在常量表达式中。
 
Given a class definition in C++
class A
{
  public:
    //methods definition
    ....

  private:
    int i;
    char *str;
    ....
}
Is it possible to calculate the offset of a class member at compile time using C++ template meta-programming? The class is not POD, and can have virtual methods, primitive and object data member. 
解决方案
Based on Matthieu M.'s answer but shorter and with no macros:
template<typename T, typename U> constexpr size_t offsetOf(U T::*member)
{
    return (char*)&((T*)nullptr->*member) - (char*)nullptr;
}
And it's called like this:
struct X { int a, b, c, d; }

std::cout << "offset of c in X == " << offsetOf(&X::c);
Edit:

Jason Rice is correct. This will not produce an actual constant expression in C++11. It doesn't look possible given the restrictions in http://en.cppreference.com/w/cpp/language/constant_expression -- in particular no pointer difference and reinterpret_castcan be in a constant expression.

                        
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