Convert a _bstr_t to a char array in place

Question

I'm trying to convert a _bstr_t object to a char array. More specifically I've got a preallocated char array, which size I know and I want to copy the _bstr_t in there.

For a while my code looked like this:

  bool ReadCharPtr(_bstr_t const& input, char* output, unsigned int max_len)
  {
    //  THIS CODE IS WRONG !
    const char * temp (input );
    return ::strcpy_s(output, max_len, temp) == 0;
  }

But... I'm not sure and it's not clear in the windows documentation, what happens when I call operator char*() const. Is something allocated in the heap, that I have to deallocate later ?

Is there a way to avoid this step, and actually convert the _bstr_t in character into the preallocated array ? That would avoid allocating chars on the heap for nothing...

http://stackoverflow.com/questions/626049/bstr-t-to-utf-8-possible — LihO, Jun 19 '12 at 14:33

score 1 · Accepted Answer · answered Jun 19 '12 at 14:48

No. This is not required. The documentation says that " The operators return the pointer to the actual internal buffer, so the resulting string cannot be modified." which would imply that _bstr_t takes care of the deletion.

Additionally you could check the implementation of _bstr_t :

In MSVC 2010 :

_bstr_t uses _com_util::ConvertBSTRToString() to get the character pointer from the BSTR. _com_util::ConvertBSTRToString() documentation says that this needs to be deleted by the callers. However, _bstr_t caches this pointer and deletes in the destructor.

Convert a _bstr_t to a char array in place

1 Answers1