可以使用 std::bit_cast 从 std::span 转换<A>到 std::span<B>并像访问对象 B 一样访问吗？</a>

Question

#include <array>
#include <bit>
#include <span>

struct A {
    unsigned int size;
    char* buf;
};

struct B {
    unsigned long len;
    void* data;
};

int main() {
    static_assert(sizeof(A) == sizeof(B));
    static_assert(alignof(A) == alignof(B));
    std::array<A, 10> arrayOfA;
    std::span<A> spanOfA{arrayOfA};
    std::span<B> spanOfB = std::bit_cast<std::span<B>>(spanOfA);
    // At this point, is using spanOfB standard compliant?
}

I've tried accessing bit_cast ed span on 3 major compilers and they seem to be working as expected, but is this standard compliant?

Answer 1

No. While std::span in C++23 will be defined such that it must be trivially copyable, there is no requirement that any particular span<T> has the same layout of span<U> . And even if it did, you'd still be accessing the objects of type A through a glvalue of type B , which violates strict aliasing if A and B aren't allowed to be accessed that way. And in your example, they are not.