从其成员子对象之一的地址计算对象的地址 [英] Computing the address of an object from the address of one of its member subobject

问题描述

我处于以下情况：

  //这是Public 
 class B {/ * usefull stuff * /}; 
 B * f（）; 
 void g（B * b）（）; 
 
 //这些类只声明为f和g的翻译单位。 
 class Whatever1 {/ *实现细节只对f和g * /}有用; 
 class Whatever2 {/ *实现细节只对f和g * /}有用; 
 class A {
 public：
 Whatever1 w1; 
 Whatever2 w2; 
 B b; 
}; 
 

在函数g中，我想将参数（指向B的指针）转换为指向A的指针。

与此：

  ptrdiff_t lag（）{
 A a; 
 return（char *）& ab-（char *）& a;} 
 
 void g（B * b）{
 A * a = （（char *）b-lag（））; 
 //使用
} 
  

这个解决方案让我非常不舒服。

是否100％正确且可移植以执行此类偏移计算？

它以任何方式触发未定义的行为吗？

编辑：std :: is_standard_layout< A> :: value is 1。

解决方案

如果 B 总是包含在 A 中，那么在A中向B的父代添加一个反向引用可能会更加干净。如果这对于任何原因是不切实际的，那么 offsetof 会清除它一点，但否则的方法是有效的，如果有点c-ish。

  void g（B * b）{
 A * a =（A *）（（char *）b）-offsetof 
 //使用
} 
  

I am in the following situation:

//This is Public
class B{/*usefull stuff*/};
B*f();
void g(B*b)();

//Those classes are only declared the translation unit of f and g.
class Whatever1{/*Implementation details only useful to f and g*/};
class Whatever2{/*Implementation details only useful to f and g*/};
class A{
public:
    Whatever1 w1;
    Whatever2 w2;
    B b;
};

In function g, I want to convert the parameter (a pointer to B) to a pointer to A.

B instances are always wrapped in A instances.

I ended up with this:

ptrdiff_t lag(){
    A a;
    return (char*)&a.b-(char*)&a;}

void g(B*b){
    A*a=(A*)((char*)b-lag());
    //Work with a
}

This solution makes me very unconfortable.

Is it 100% correct and portable to do offset computations like this ?

Does it triggers Undefined Behavior in any way ?

Edit: std::is_standard_layout< A >::value is 1.

解决方案

If a B is always contained within an A, then it might be much cleaner to add a back reference to B's parent inside A. If that is impractical for whatever reason, then offsetof will clean it up a little, but otherwise the approach is valid, if somewhat c-ish.

void g(B*b){
    A* a = (A*)(((char*)b)-offsetof(class A,b));
    //Work with a
}

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