Why isn't a transform_view a borrowed_range?

Question

For std::ranges, why isn't a transform_view a borrowed_range? Here is a simple example: https://godbolt.org/z/14K8Y1xMe and below:

#include <ranges>

void foo() {
    auto i = std::ranges::views::iota(10);
    auto t = i | std::ranges::views::transform([](auto v) {return v;});
    static_assert(std::ranges::borrowed_range<decltype(i)>);
    static_assert(std::ranges::borrowed_range<decltype(t)>);
}

I checked that the base is a borrowed range. Is there an issue with the function? The documentation doesn't say: https://en.cppreference.com/w/cpp/ranges/transform_view.

Answer 1

transform_view internally saves the callable as a member, and its iterator saves a pointer to transform_view for accessing the callable. So when transform_view is destroyed, the callable is also destroyed, which causes its iterator to access a dangling pointer.

That's why it's not borrowed_range.