不明确的类型引用
为什么这有效：
template <typename T>
struct foo
{
};
struct A
{
typedef foo<A> type;
};
struct B : public A
{
typedef foo<B> type;
};
int main()
{
B::type john;
return 0;
}
但不是这样：
template <typename T>
struct foo
{
};
template <typename T>
struct Shared
{
typedef foo<T> type;
};
struct A : public Shared<A>
{
};
struct B : public A, public Shared<B>
{
};
int main()
{
// g++ 4.5 says :
// error: reference to 'type' is ambiguous
B::type john;
return 0;
}
在我的代码中， foo 实际上是 boost::shared_ptr 并且，正如你所看到的，我正在尝试考虑一些 typedef 使用 Shared 类。
Why does this works :
template <typename T>
struct foo
{
};
struct A
{
typedef foo<A> type;
};
struct B : public A
{
typedef foo<B> type;
};
int main()
{
B::type john;
return 0;
}
But not this :
template <typename T>
struct foo
{
};
template <typename T>
struct Shared
{
typedef foo<T> type;
};
struct A : public Shared<A>
{
};
struct B : public A, public Shared<B>
{
};
int main()
{
// g++ 4.5 says :
// error: reference to 'type' is ambiguous
B::type john;
return 0;
}
In my code, foo is actually boost::shared_ptr and, as you can see, I'm trying so factor some typedefs using a Shared class.
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评论（2）
因为
B继承了foo并间接继承了foo，并且都包含成员type。你指的是哪一个？您简单的第一段代码具有
B的type隐藏A的type，但在更复杂的第二段代码中不会发生这种情况，它涉及更深的继承树。Because
Binheritsfoo<B>and, indirectly,foo<A>, and both contain a membertype. Which did you mean?Your simple, first piece of code has
B'stypehidingA'stype, but that doesn't happen in the more complex second piece of code, which involves a deeper inheritance tree.因为你实际上有 2 个
typetypedef。一个来自A，它从shared获取他的，另一个来自shared。在第一种情况下，您将A基类的typetypedef 隐藏在B中。Because you actually have 2
typetypedefs. One fromA, which gets his fromshared<A>and one fromshared<B>. In your first case, you hide thetypetypedef of theAbase class with the typedef inB.