copy variant to char*

Question

how to copy to a character array from variant in c++. So far I have been able to come up with the snippet below. but need help in making it work.

_variant_t vt_Data(parent->CharData);

if(vt_Data.vt != VT_NULL)
{
   long ubound;
   long lbound;
   long length;
   TSTHR(SafeArrayGetLBound(vt_Data.parray, 1, &lbound));
   TSTHR(SafeArrayGetUBound(vt_Data.parray, 1, &ubound));
   if(length = ubound - lbound +1)
   {
      char *char_data = new char[length];

      TSTHR(SafeArrayAccessData(vt_Data.parray, (void **) &char_data))
      memcpy(char_data, vt_Data.parray, length);
      TSTHR(SafeArrayUnaccessData(vt_Data.parray));
   }
}

Do you realize that the expression `if(length = ubound - lbound +1)` *assigns* `ubound - lbound +1` to `length` and then checks if `length` is non zero? If not, use `==` to check for equality, `=` is the assignment operator. — Ed S., Jun 27 '12 at 17:08
I think his point is the length will almost always be non-zero due to `+1` for the null. This check will also pass `lbound > ubound` and potentially allow allocation of negative length if that's a concern. — AJG85, Jun 27 '12 at 17:20

score 0 · Accepted Answer · answered Jun 27 '12 at 23:36

If the type of parent->CharData is char*, bstr, or TCHAR* (or similar) then the following should work. (although untested)

//uses the bstr_t const assignment operator to extract the string
_bstr_t bstr = vt_Data;

This will create a new copy of the string.

copy variant to char*

1 Answers1