c++ - 在 C++ 中访问类型成员

Question

给定一个容器，例如 vector<int>

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

为什么像iterator这样的public类型的成员好像很难访问和 const_iterator ？据我了解，这些名称是类的一部分(不是对象)，必须通过 :: 访问。指定范围，但是否有理由禁止 v.const_iterator什么时候v是已知的？示例:

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

解决方法是使用 decltype如:

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

但是这种方法甚至在类中都不起作用，因为以下方法失败了:

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

编辑

只是一点旁注。正如所指出的，v.iterator 的含义将取决于 v 的类型在使用时(编译时)忽略运行时多态性。但是静态类成员也是如此。示例:

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)
}

Answer 1

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

#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;
    }
};

成员访问 . 运算符和作用域解析运算符 :: 不得重载。正如您可能从名称中推断的那样，. 用于 access members ，而 :: 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;
}

此外，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”。