InterlockedExchange64

Hi all,

I am studying the InterlockedExchange64 function behavior and i think
I did sth wrong. IMO, the line that contains teh call
to ::DebugBreak() shouldn´t neve be executed. Am I right?


//////////////////////////////////////////////////////////////////////

volatile __int64 g;


unsigned __stdcall Check(void*)
{
__int64 t;

::InterlockedExchange64(&t, g);

while ((t == 0) ||
(t == 0xffffffffffffffff))
{
::InterlockedExchange64(&t, g);
}


::DebugBreak(); // THIS SHOULDN´T HAPPENS! SHOULD IT?

cout << std::hex << t;

return 0L;
}


int main(int argc, char **argv)
{
::_beginthreadex(NULL, 0, Check, NULL, 0, NULL);

__int64 v[] = {0, 0xffffffffffffffff};

int i = 0;

while (TRUE)
{
::InterlockedExchange64(&g, v[i %= 2, i++]);
}


return 0;
}

//////////////////////////////////////////////////

Best Regards.

.

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