How does the compiler know that msleep() has got barrier()s?
Because msleep_interruptible() is in a separate compilation unit,
the compiler has to assume that it might modify any arbitrary global.
No; compilation units have nothing to do with it, GCC can optimise
across compilation unit boundaries just fine, if you tell it to
compile more than one compilation unit at once.
What you probably mean is that the compiler has to assume any code
it cannot currently see can do anything (insofar as allowed by the
relevant standards etc.)
In many cases, the compiler also has to assume that
msleep_interruptible()
might call back into a function in the current compilation unit, thus
possibly modifying global static variables.
It most often is smart enough to see what compilation-unit-local
variables might be modified that way, though :-)
Segher
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