Re: behaviour of simple short -> int conversion differs in debug and release version

From:
"Igor Tandetnik" <itandetnik@mvps.org>
Newsgroups:
microsoft.public.vc.language
Date:
Wed, 20 Sep 2006 16:42:37 -0400
Message-ID:
<O$eLlWP3GHA.4932@TK2MSFTNGP02.phx.gbl>
P@ <p.hoffmann@berlin.de> wrote:

The following code prints out -1 in debug mode (correct) and 0 in
release mode (wrong) if compiled with Visual Studio.NET 2003.

#include "stdafx.h"

int _tmain(int argc, _TCHAR* argv[])
{
 short usVal;

 // enter -1
 std::cin >> usVal;

 int iVal = usVal;
 iVal = iVal >> 17;

 std::cout << iVal << std::endl;

return 0;
}


Looks like a bug. If one looks at the disassembly, the shift is elided
entirely and a constant 0 is output to cout. Apparently, the compiler
erroneously assumes the value would always be zero-extended, and thus
believes it can calculate the shift at compile time.

I don't have VC8 handy at the moment to check how it behaves in this
case.

Be aware that the code is non-portable. C++ standard:

5.8/3 The value of E1 >> E2 is E1 right-shifted E2 bit positions... If
E1 has a signed type and a negative value, the resulting value is
implementation-defined.

MSDN specifies that VC compiler is indeed supposed to propagate most
significant bit to fill vacant positions, and thus -1 is the correct
output in your example. Other implementations may differ.
--
With best wishes,
    Igor Tandetnik

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