在 C++ 中访问类型成员

给定一个容器，例如vector<int>

#include <vector>
using namespace std;
vector<int> v{1, 2, 3};

为什么访问公共类型成员（例如iterator and const_iterator？据我了解，这些名称是类（而不是对象）的一部分，必须通过::指定范围，但有理由禁止吗v.const_iterator when v是已知的吗？ 例子：

int f(v.iterator it) {
    return *it;
}
// or
int g(v::iterator it) {
    return *it;
}

解决方法是使用decltype as in:

int h(decltype(v)::iterator it) {
    return *it;
}

但这种方法甚至在课堂上不起作用，因为以下失败：

class A
{
public:
    int h(decltype(x)::iterator it) {
        return *it;
    }
private:
    vector<int> x;
};

Edit

只是一点旁注。 正如所指出的，其含义是v.iterator取决于类型v在使用时（编译时）忽略运行时多态性。但静态类成员也是如此。 例子：

struct A
{
    static const int x = 1;
};
struct B : public A
{
    static const int x = 2;
};
void eval()
{
    B b;
    A& ar = b;
    b.x; // 2
    ar.x; // 1, even though ar refers to the same underlying object (by the base type)
}

正如@Slava 在评论中指出的那样，decltype(x)是这样做的方法：

#include <vector>
using namespace std;
vector<int> v{1, 2, 3};
int f(decltype(v)::iterator it) {
    return *it;
}

int g(decltype(v)::iterator it) {
    return *it;
}

class A
{
private:
    vector<int> x;
public:
    int h(decltype(x)::iterator it) {
        return *it;
    }
};

会员访问.运算符和范围解析运算符::不得超载。正如你可以从名字中推断出的那样，.习惯于访问成员, while ::用于访问scope.

#include <iostream>

struct B {
    class iterator { };

    // no need for typename, compiler knows that we mean typedef B::iterator, as he can only find it
    iterator iterator1;

    // member named the same as class, ops!
    int iterator;

    // we need to use typename here, B::iterator is resolved as member
    // iterator iteartor3;
    typename B::iterator iterator2;
};

int main() {
    B bobj;

    // we access the member iterator inside b
    bobj.iterator = 1;

    // we declare object of B::iterator type
    // we need to tell compiler that we want only types
    typename B::iterator iterator;

    // this will work too
    typename decltype(bobj)::iterator iterator2;

    // we declare a member pointer to the iterator member inside some B class
    // no typename, as I want pointer to member, not pointer to... type
    int B::* pointer = &B::iterator;

    // this is just a pointer to the iterator specifically in bobj class
    int * pointer2 = &bobj.iterator;

    // foo(bar) 
    bobj.*pointer = 1;

    // this will work as expected
    int decltype(bobj)::* pointer3 = &B::iterator;
}

另外，C++ 中没有“类型成员”（至少我在 C++ 标准中找不到它们）。在类中声明为成员的类和枚举以及 typedef 声明称为“嵌套类型”或“嵌套类”。