Question

Empirically the following works (gcc and VC++), but is it valid and portable code?

typedef struct 
{
    int w[2];
} A;

struct B
{
    int blah[2];
};

void my_func(B b)
{
    using namespace std;
    cout << b.blah[0] << b.blah[1] << endl;
}

int main(int argc, char* argv[])
{

    using namespace std;

    A a;
    a.w[0] = 1;
    a.w[1] = 2;

    cout << a.w[0] << a.w[1] << endl;

    // my_func(a);                     // compiler error, as expected
    my_func(reinterpret_cast<B&>(a));  // reinterpret, magic?
    my_func(  *(B*)(&a) );             // is this equivalent? 

    return 0;
}
// Output:
// 12
// 12
// 12

• Is the reinterpret_cast valid?
• Is the C-style cast equivalent?
• Where the intention is to have the bits located at &a interpreted as a
  type B, is this a valid / the best approach?

(Off topic: For those that want to know why I’m trying to do this, I’m dealing with two C libraries that want 128 bits of memory, and use structs with different internal names – much like the structs in my example. I don’t want memcopy, and I don’t want to hack around in the 3rd party code.)

Answer 1

In C++11, this is fully allowed if the two types are layout-compatible, which is true for structs that are identical and have standard layout. See this answer for more details.

You could also stick the two structs in the same union in previous versions of C++, which had some guarantees about being able to access identical data members (a “common initial sequence” of data members) in the same order for different structure types.

In this case, yes, the C-style cast is equivalent, but reinterpret_cast is probably more idiomatic.