以下代码基于 Modern C++ programming cookbook 中的代码,并在 VS 2017 中编译:
#include <iostream>
using namespace std;
template <typename T, size_t const Size>
class dummy_array
{
T data[Size] = {};
public:
T const & GetAt(size_t const index) const
{
if (index < Size) return data[index];
throw std::out_of_range("index out of range");
}
// I have added this
T & GetAt(size_t const index)
{
if (index < Size) return data[index];
throw std::out_of_range("index out of range");
}
void SetAt(size_t const index, T const & value)
{
if (index < Size) data[index] = value;
else throw std::out_of_range("index out of range");
}
size_t GetSize() const { return Size; }
};
template <typename T, typename C, size_t const Size>
class dummy_array_iterator_type
{
public:
dummy_array_iterator_type(C& collection,
size_t const index) :
index(index), collection(collection)
{ }
bool operator!= (dummy_array_iterator_type const & other) const
{
return index != other.index;
}
T const & operator* () const
{
return collection.GetAt(index);
}
// I have added this
T & operator* ()
{
return collection.GetAt(index);
}
dummy_array_iterator_type const & operator++ ()
{
++index;
return *this;
}
private:
size_t index;
C& collection;
};
template <typename T, size_t const Size>
using dummy_array_iterator = dummy_array_iterator_type<T, dummy_array<T, Size>, Size>;
// I have added the const in 'const dummy_array_iterator_type'
template <typename T, size_t const Size>
using dummy_array_const_iterator = const dummy_array_iterator_type<T, dummy_array<T, Size> const, Size>;
template <typename T, size_t const Size>
inline dummy_array_iterator<T, Size> begin(dummy_array<T, Size>& collection)
{
return dummy_array_iterator<T, Size>(collection, 0);
}
template <typename T, size_t const Size>
inline dummy_array_iterator<T, Size> end(dummy_array<T, Size>& collection)
{
return dummy_array_iterator<T, Size>(collection, collection.GetSize());
}
template <typename T, size_t const Size>
inline dummy_array_const_iterator<T, Size> begin(dummy_array<T, Size> const & collection)
{
return dummy_array_const_iterator<T, Size>(collection, 0);
}
template <typename T, size_t const Size>
inline dummy_array_const_iterator<T, Size> end(dummy_array<T, Size> const & collection)
{
return dummy_array_const_iterator<T, Size>(collection, collection.GetSize());
}
int main(int nArgc, char** argv)
{
dummy_array<int, 10> arr;
for (auto&& e : arr)
{
std::cout << e << std::endl;
e = 100; // PROBLEM
}
const dummy_array<int, 10> arr2;
for (auto&& e : arr2) // ERROR HERE
{
std::cout << e << std::endl;
}
}
现在,错误指向这条线
T & operator* ()
说明
“return”:无法从“const T”转换为“T &””
...这是从我在 arr2
上基于范围的 for
循环中产生的。
为什么编译器选择 operator*() 的非常量版本?
。我已经看了很长时间了;我认为这是因为它认为调用此运算符的对象不是常量:这应该是一个 dummy_array_const_iterator
。但是,这个对象已经通过
template <typename T, size_t const Size>
using dummy_array_const_iterator = const dummy_array_iterator_type<T, dummy_array<T, Size> const, Size>;
...所以我真的不明白发生了什么。有人可以澄清一下吗?
TIA
最佳答案
dummy_array_const_iterator::operator *
应该总是返回 T const &
而不管迭代器对象本身的常量性。
实现这一点的最简单方法可能只是用 T const
声明它作为基础迭代器值类型:
template <typename T, size_t const Size>
using dummy_array_const_iterator = dummy_array_iterator_type<T const, dummy_array<T, Size> const, Size>;
由于您按值返回迭代器,因此它的常量很容易是 lost通过 C++ 类型推导规则并仅将 dummy_array_const_iterator
声明为 const dummy_array_iterator_type
的别名是不够的。即以下失败:
#include <type_traits>
struct I { };
using C = I const;
C begin();
int bar()
{
auto x = begin(); // type of x is deduced as I
static_assert(std::is_same<I, decltype(x)>::value, "same"); // PASS
static_assert(std::is_same<decltype(begin()), decltype(x)>::value, "same"); // ERROR
}
关于c++ - 基于 : issue with constness 范围内的自定义迭代器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54690996/