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Is there a size limit on _bstr_t type ?

Posted on 2004-09-09
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Hi,
 Consider the piece of code below:

_bstr_t test1("aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa");

_bstr_t test2("bbbbbbbbbb bbbbbbbbbb cccccccccc @@@@@@@@@@ bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb");

test1 += test2;



After the third step I expected the strings to be concatenated fully, but it doesnt work that way; the protion from ccccccc is not getting appended....any idea?

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Question by:manoj_johar
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drichards earned 750 total points
ID: 12024204
THere is a limit on the size of BSTR's, but it's at least 64K, and that was in 16 bit Windows.  The length is stored in a DWORD now, so if they let you use the whole thing, it's like 4 billion characters.  Much bigger than what you've got.

Anyway, try this code and look at things in the debugger.  I just added some length calculations.  When I run your code, I get the correct result.  If I run this modified version, I get "test1: 109, test2: 109, test1+test2: 218".

    _bstr_t test1("aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa aaaaaaaaaa");
    UINT l1 = ::SysStringLen(test1);

    _bstr_t test2("bbbbbbbbbb bbbbbbbbbb cccccccccc @@@@@@@@@@ bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb bbbbbbbbbb");
    UINT l2 = ::SysStringLen(test2);

    test1 += test2;
    UINT l3 = ::SysStringLen(test1);
    std::cout << "test1: " << l1 << ", test2: " << l2 << ", test1+test2: " << l3 << std::endl;

What do you get?
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