[Libreoffice] Same expression on both sides of '!=' in calc.

Michael Meeks michael.meeks at novell.com
Sat May 7 12:15:37 PDT 2011

On Sat, 2011-05-07 at 13:03 -0430, Rafael Dominguez wrote:
> Well it does look weird, but its the same object getting compared on
> both sides, so i think theres a problem there, but better wait for
> someone with more experience in the code say something about it.

	In libreoffice-3-3 we had:

        ::formula::FormulaGrammar::AddressConvention eConvUI = pDocument->GetAddressConvention();
        if (eConv != eConvUI)
            return ((rAddress.Parse(sToken, const_cast<ScDocument*>(pDocument), eConvUI) & SCA_VALID) == SCA_VALID);

	Around here instead of:

            if (!bResult && eConv != eConv)
                bResult = ((rRange.aStart.Parse(
                    aUIString, const_cast<ScDocument*>(pDocument), eConv) & SCA_VALID) == SCA_VALID);

	It seems pretty bogus doing the same parse again - you would hope the
function would fail in the same way :-)

	Best to ask Kohei to de-louse it though :-) nice catch ...



 michael.meeks at novell.com  <><, Pseudo Engineer, itinerant idiot

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