C++ -隐式转换无符号长长到有符号长长？

Question

我有一个奇怪的警告是由clang 12.0.1报告的。在以下代码中：

#include <vector>

int main()
{
    std::vector<int> v1;

    const auto a = v1.begin() + v1.size();

    return 0;
}

我看到这一警告被触发：

error: narrowing conversion from 'std::vector<int>::size_type' (aka 'unsigned long long') to signed type 'std::_Vector_iterator<std::_Vector_val<std::_Simple_types<int>>>::difference_type' (aka 'long long') is implementation-defined [bugprone-narrowing-conversions,cppcoreguidelines-narrowing-conversions,-warnings-as-errors]
    const auto a = v1.begin() + v1.size();
                                ^

我的理解是，当操作两个大小相同但符号不同的整数时，有符号整数被转换为无符号，而不是相反。我是不是漏掉了什么？

Answer 1

要显示所涉及的实际类型：

// Operator+ accepts difference type
// https://en.cppreference.com/w/cpp/iterator/move_iterator/operator_arith
// constexpr move_iterator operator+( difference_type n ) const;

#include <type_traits>
#include <vector>
#include <iterator>

int main()
{
    std::vector<int> v1;

    auto a = v1.begin();
    auto d = v1.size();

    // the difference type for the iterator is a "long long"
    static_assert(std::is_same_v<long long, decltype(a)::difference_type>);

    // the type for size is not the same as the difference type
    static_assert(!std::is_same_v<decltype(d), decltype(a)::difference_type>);
    
    // it is std::size_t
    static_assert(std::is_same_v<decltype(d), std::size_t>);
    
    return 0;
}

Answer 2

由于C++20，一个简单的修复就是使用std::sszie

const auto a = v1.begin() + std::ssize(v1);