[Qt-interest] dereferencing pointer '<anonymous>' does break strict-aliasing rules error

[Anton Chernov]

Seems an interesting question, i didn't get it quite well, did you meaned

int i = 42;

ushort* u = (ushort *)&i;

*u = 43;


? And what can went wrong here? Or was it a joke, right?

2011/5/11 Thiago Macieira <thiago at kde.org>

> On Wednesday, 11 de May de 2011 12:29:45 Konrad Rosenbaum wrote:
> > On Tuesday 10 May 2011 14:38:23 mmiacca wrote:
> > > While i try with Qt 2010.05 (Qt 4.7.0 32 bits Windows) to compile
> > > C:\Qt\2010.05\qt\examples\qtestlib\tutorial4 i get this error
> > > ..\..\..\include/QtTest/../../src/testlib/qtestspontaneevent.h:77:
> > > warning: dereferencing pointer '<anonymous>' does break strict-aliasing
> > > rules
> >
> > It's just a warning. Unless there is more (i.e. a real error) the
> > application should still build and be executable.
>
> That doesn't mean it will work right. Then again, there's warning-free code
> that doesn't work right either :-)
>
> The C and C++ languages forbid aliasing of types (except for aliasing with
> char). So the following code is malformed:
>
>        int i = 42;
>        ushort u = (ushort *)&i;
>        u = 43;
>
>        printf("%d", i);
>
> The types "i" and "u" alias each other and this is not allowed.
>
> Undefined behaviour means anything could happen. The possible outcomes are
> (assuming little-endian machine):
>
> 1) it prints "43" because it did modify the memory located by i.
>
> 2) it prints "42" because the compiler concludes that the variable i didn't
> get modified between its assignment and its use. The two intermediary lines
> are
> considered dead code and eliminated.
>
> 3) it calls your ex-mother-in-law and gives her your new girlfriend's phone
> number.
>
> --
> Thiago Macieira - thiago (AT) macieira.info - thiago (AT) kde.org
>  Senior Product Manager - Nokia, Qt Development Frameworks
>      PGP/GPG: 0x6EF45358; fingerprint:
>      E067 918B B660 DBD1 105C  966C 33F5 F005 6EF4 5358
>
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>
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