从类型而不是从变量获取指针级别

Question

我尝试将指针深度显示为整数个星号 - 例如 int*** 将是 3。但是我没有从类型计算它，而是编写了要求每个级别的指针都必须正确初始化的代码。这是我尝试过的：

#include <iostream>
#include <utility>

// no pointers involved
template <class T>
std::size_t get_pointer_level(T)
{
    return 0;
}

// final value reached, returning depth
template <class T>
std::size_t get_pointer_level(std::pair<std::size_t, T> arg)
{
    return arg.first;
}

// function that accummulates depth
template <class T>
auto get_pointer_level(std::pair<std::size_t, T*> arg)
{
    return get_pointer_level(std::make_pair(arg.first+1, *arg.second));
}

// initial function called for pointer argument
template <class T>
auto get_pointer_level(T* arg)
{
    return get_pointer_level(std::make_pair(std::size_t(1), *arg));
}


int main(void)
{
    int a = 10;
    auto b = &a; //int*
    auto c = &b; //int**
    auto d = &c; //int***
    auto e = &d; //int****
    std::cout << get_pointer_level(e) << std::endl; //4
}

我很确定它很可能仅与类型一起工作。我想语法会是这样的：

get_pointer_level<int****>::value

有什么想法吗？

---

编辑： 感谢您的解决方案！这是我想要完成的最终功能：

template <std::size_t A, std::size_t B>
struct is_smaller
{
    enum {value = (A < B)? 1 : 0};
};
template<std::size_t Target, typename T, std::size_t Actual = get_pointer_level<T>::value>
T value_at_level(T pointer)
{
    static_assert(Actual==Target, "Invalid target level!");
    return pointer;
}

template<std::size_t Target, typename T, std::size_t Actual = get_pointer_level<T*>::value,
         typename = std::enable_if<is_smaller<Target, Actual>::value>::type>
auto value_at_level(T* pointer)
{
    return value_at_level<Target>(*pointer);
}

int main()
{
    int a = 5;
    auto b = &a; //int*
    auto c = &b; //int**
    auto d = &c; //int***
    auto e = &d; //int****

    std::cout << "int from int****: " << value_at_level<0>(e) << std::endl; //ok
    std::cout << "int* from int***: " << value_at_level<1>(d) << std::endl; //ok
    std::cout << "int** from int**: " << value_at_level<2>(c) << std::endl; //ok
    std::cout << "int*** from int*: " << value_at_level<3>(b) << std::endl; //error
    std::cout << "int**** from int: " << value_at_level<5>(a) << std::endl; //error
}

Answer 1

部分专业化练习：

#include <cstddef>

template<typename T, std::size_t S>
struct get_pointer_level_impl
{
    static const std::size_t value = S;
};

template<typename T, std::size_t S>
struct get_pointer_level_impl<T*, S> : get_pointer_level_impl<T, S+1>
{

};

template<typename T>
struct get_pointer_level : get_pointer_level_impl<T, 0>
{

};

#include <iostream>

int main()
{
    std::cout << get_pointer_level<int>::value << "\n";
    std::cout << get_pointer_level<int*>::value << "\n";
    std::cout << get_pointer_level<int**>::value << "\n";
    std::cout << get_pointer_level<int***>::value << "\n";
}

输出：

0
1
2
3

Answer 2

如果您可以将其称为get_pointer_level<int*>() 而不是get_pointer_level<int*>::value，那么递归constexpr 函数就可以正常工作：

#include <iostream>
#include <type_traits>

template <typename T>
constexpr int get_pointer_level()
{
    return !std::is_pointer<T>::value ? 0 : 1 + get_pointer_level<typename std::remove_pointer<T>::type>();
}

int main() {
    std::cout << get_pointer_level<int>() << '\n';
    std::cout << get_pointer_level<int*>() << '\n';
    std::cout << get_pointer_level<int**>() << '\n';
    std::cout << get_pointer_level<int***>() << '\n';
    std::cout << get_pointer_level<int****>() << '\n';
}

LIVE DEMO

Answer 3

使用帮助类来确定指针级别。

#include <iostream>

template <class T> struct PointerLevel
{
   static const std::size_t value = 0;
};

template <class T> struct PointerLevel<T*>
{
   static const std::size_t value = PointerLevel<T>::value + 1;
};

// no pointers involved
template <class T>
std::size_t get_pointer_level(T)
{
   return PointerLevel<T>::value;
}

int main(void)
{
   int a = 10;
   auto b = &a; //int*
   auto c = &b; //int**
   auto d = &c; //int***
   auto e = &d; //int****
   std::cout << get_pointer_level(a) << std::endl;
   std::cout << get_pointer_level(b) << std::endl;
   std::cout << get_pointer_level(c) << std::endl;
   std::cout << get_pointer_level(d) << std::endl;
   std::cout << get_pointer_level(e) << std::endl; //4
}

输出：

0
1
2
3
4

Answer 4

你觉得使用 typeid(your type).name() 怎么样？？

std::string s( typeid(int***).name()); 
size_t n = std::count(s.begin(), s.end(), "*");