CODEWITHSUNDEEP
c++c++-cli
Tobias Langner
Tobias Langner

Reputation: 10808

operator wchar_t**() failing to provide implicit conversion to const wchar_t ** - why?

I have a problem understanding why a certain implicit conversion is not working as I expect it. I have the following class

ref class ManagedWStringArrayWrapper
{
    wchar_t** m_pointer;

public:

    operator wchar_t**()
    {
        return m_pointer;
    }
};

and I thought this would implicitly convert to const wchar_t ** as well - but it doesn't. Can someone tell me why?

Upvotes: 3

Views: 581

Answers (2)

Mr Lister
Mr Lister

Reputation: 46579

Because those are different things. If you look, for instance, at the C++ libraries, you'll find that there are often two functions for doing the same thing, one operating on const pointers and the other on non-const. Example.

But you can easily add an explicit operator const wchar_t**.

Upvotes: 0

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385204

Conversion from T** to T const** is not as intuitive as you might expect — in fact, it's given as an example in the standard itself as const-incorrect code.

The example given is:

#include <cassert>  

int main() {  
  char* p = 0;  

  //char const** a = &p; // not allowed, but let's pretend it is  
  char const** a = (char const**)&p; // instead force the cast to compile  

  char const* orig = "original";  
  *a = orig; // type of *a is char const*, which is the type of orig, this is allowed  

  assert(p == orig); // oops! char* points to a char const*  
}

Study the above scenario for a while and it will become clear that this is not the same as a conversion T* → T const*. Not at all!

I asked the same question in a blog post, and the FAQ has an entry on it too.

Upvotes: 3

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