C - Implicit conversion of unsigned long long to signed long long?-CodePudding

I'm having a rather strange warning being reported by clang-tidy 12.0.1. In the following code:

#include <vector>

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

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

    return 0;
}

I see this warning being triggered:

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();
                                ^

It was my understanding that when operating two integers with the same size but different signedness, the signed integer is converted to unsigned, not the other way around. Am I missing something here?

CodePudding user response：

To show the actual types involved :

// 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;
}

CodePudding user response：

Since C 20 a simple fix is to use std::sszie:

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