void foobar(Base* base)
{
Derived* derived = dynamic_cast<Derived*>(base); // or static_cast
derived->blabla = 0xC0FFEE;
if (base->blabla == 0xC0FFEE)
...
}
在具有严格别名 的编译器上,“derived”是“base”的别名吗?
最佳答案
只要有可能通过它们访问同一个对象,两个指针就被别名化了。该标准的第 3.10/15 段指定了对对象的访问何时有效。
If a program attempts to access the stored value of an object through an lvalue of other than one of the following types the behavior is undefined:
- the dynamic type of the object,
- a cv-qualified version of the dynamic type of the object,
- a type that is the signed or unsigned type corresponding to the dynamic type of the object,
- a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic type of the object,
- an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union),
- a type that is a (possibly cv-qualified) base class type of the dynamic type of the object,
- a char or unsigned char type.
在您的情况下,*derived
要么是对象动态类型的左值,要么是对象动态类型的基类类型。 *base
是对象动态类型的基类类型。
因此,您可以通过derived
和base
访问对象,使这两个指针成为别名。
关于c++ - 通过指向其派生类的指针访问类对象是否会违反严格的别名规则?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1546997/