Can you cast a LPTSTR to a BSTR?

问题:

Is it legal to cast a LPTSTR directly to a BSTR?

Based on my understanding of BSTR, casting a LPTSTR to a BSTR directly will leave you with a corrupted length prefix. The example code explicitly states that a string literal cannot be stored to a BSTR. Can anyone confirm for me that a LPTSTR/LPCTSTR cannot be cast directly to a BSTR without corrupting the length prefix?

EDIT:

My confusion is from seeing this used in a call to a COM object. It turns out that when compiling the COM dll, a .tli file is generated that creates an intermediate method. This method takes type _bstr_t. The _bstr_t can take LPTSTR in its constructor, so everything works smoothly.

回答1:

If your program is unicode and your LPTSTR therefore is a LPWSTR, you can use SysAllocString to convert from a pointer to a wide character string to BSTR.

A direct cast is not possible because the two have different memory representations.

If you use C++, you can use the _bstr_t class to simplify the usage of BSTR strings.

回答2:

If you are trying to pass LPCTSTRs as BSTRs you will find that you will randomly blow up any interop marshalling code you get near. The marshalling will pull the 4 byte prefix (which is random data) for an LPCTSTR and try and Marshall the string.

You will find that the 4 byte prefix turns out to be the stack contents prior to your buffer, or even better character data from the string before it. You will then try to marshall megabytes of data... :-)

use the CComBSTR wrappers and use .Detach() if you need to keep the BSTR pointer.

Be nice if the compiler spat out a warning on this one.

回答3:

A LPTSTR is a pointer to a char array (or TCHAR to be exact) BSTR is a structur (or composite data) consist of * Length prefix * Data string * Terminator

so casting will not work

回答4:

No, you cannot - if the code that believes it is a BSTR calls SysStringLen() it will run into undefined behavior since that function relies on some implementation-specific service data.

回答5:

It cannot be, because then the four bytes in memory preceding the LPTSTR would be considered as the length of the resulting BSTR. This might cause a memory protection fault on the spot (not very likely), but it would certainly result in a BSTR with a length that could be way larger than the length of the original LPTSTR. So when someone tries to read or write from that they may access invalid memory.

回答6:

You cannot cast, you must convert. You can use the built-in compiler intrinsic _bstr_t (from comutil.h) to help you do this easily. Sample:

#include <Windows.h>
#include <comutil.h>

#pragma comment( lib, "comsuppwd.lib")

int main()
{
    LPTSTR p = "Hello, String";
    _bstr_t bt = p;
    BSTR bstr = bt;
    bstr;
}

回答7:

No, you cannot cast them directly. However, you can make a string that does both. A C-String doesn't have a 4-byte header. However, the bit in the middle is the same so if you find yourself needing both representations, make a wrapper class that constructs a string with a 4byte header and a null terminator but can return accessors to the BSTR portion and the C-String.

This code is intended as an incomplete example, I haven't compiled this!

class YetAnotherStringType //just what the world needs
{
  public:
  YetAnotherStringType(const char *str)
  { 
     size_t slen = strlen(str);
     allocate(slen);
     set_size_dword(slen);  
     copy_cstr(str, slen);     
  }

  const char *get_cstr() const
  {
     return &m_data[4];
  }

  const BSTR get_bstr() const
  {
     return (BSTR*)m_data;
  }

  void copy_cstr(const char *cstr, int size = -1)
  {
      if (size == -1)
         size = strlen(cstr);
      memcpy(&m_data[4], cstr, size + 1); //also copies first null terminator
      m_data[5 + size] = 0; //add the second null terminator
  }

  void set_size_dword(size_t size)
  {
     *((unsigned int*)m_data) = size;
  }

  void allocate(size_t size)
  {
     m_data = new char[size + 6]; //enough for double terminator
  }

  char *m_data;
};

回答8:

Generally speaking no, but there are ways to make them compatible to some extent via helper classes and macros (see below).

The main reason why a 1:1 mapping will never be possible is that a BSTR (and consequently CComBSTR can contain '\0' in the string, because ultimately it is a counted string type.

---

Your best choice when using C++ would be to go for the ATL class CComBSTR in place of BSTR proper. In either case you can make use of the ATL/MFC conversion macros CW2A and friends.

Also note that the documentation (MSDN) says:

The recommended way of converting to and from BSTR strings is to use the CComBSTR class. To convert to a BSTR, pass the existing string to the constructor of CComBSTR. To convert from a BSTR, use COLE2[C]DestinationType[EX], such as COLE2T.

... which applies to your use case.

Please see John Dibling's answer for an alternative (_bstr_t).