On Dec 01, 2006, at 06:24:54, David Schwartz wrote:
Imagine we change the code to read from some auto-increment hardware
at a specific MMIO address instead of a global:
static int my_func(int a)
{
return a + *(volatile int *)(0xABCD1234);
}
But you're telling me that the change in the header file (where the
new function returns the exact same series of values with every call
as the old) causes the program to enter an infinite loop?
How do you rationalize this again?
No, I'm not saying that. I'm saying it *can*.
No, it can't. If you leave the prototype alone and the function when
called in sequence returns the same list of values, then by
definition the internals can have no effect on the code which uses
that function. As further proof, if you wrapped my "my_func()" with
this in some C file:
int my_other_func(int a)
{
return my_func(a);
}
Next you stick a my_other_func declaration in a header and use
my_other_func instead of my_func() in the main function. Now the
result is that the compiler has no damn clue what my_other_func()
contains so it can't optimize it out of the loop with either
version. You cannot treat "volatile" the way you are saying it is
treated without severely violating both the C99 spec *and* common sense.
In some cases, it's very unlikely that compilers will ever become
smart enough to demonstrate that our code is broken, but that
doesn't make the code any less broken, just less likely to fail.
No, the code is not broken because the language simply isn't defined
that way. Essentially when the compiler is looking at any volatile
data it cannot ignore or optimize-away any operations on that data.
On the other hand, when you cast volatile data into non-volatile
data, the compiler must preserve linearity of program execution. If
you call a function in a loop which dereferences a pointer to a
volatile then the compiler *MUST* always dereference the pointer,
even if it later discards the result and continues on its merry way.
Actually, no. The reason for the volatile in the pointer
dereference is to force the memory access to *always* happen.
That's why it was placed there, however it was thrown away right
after it was placed, in the same step it was supposed to force a
memory access.
Doesn't matter if you throw away the result. The C standard defines
this:
(void)( *(volatile int *)0xABCD1234 );
to imply that the code reads an integer from that memory location and
then discards the result. The whole point of volatile is you still
MUST do the read. Feel free to read the bugzilla entry mentioned in
this thread as it even quotes all the pertinent sections of the C
standard for you.
The problem is that '*(volatile unsigned int *)' results in a
'volatile unsigned int'. The *assignment* occurs in the return
operation, after the 'volatile unsigned int' is *cast* to a plain
'unsigned int'. The assignment is *not* in any sense volatile or
inviolate, so neither is the return value.
No, the assignment is irrelevant but the pointer dereference in
rvalue context *is* relevant. The dereference forces a read operation.
One solution would be this:
static inline unsigned int readl(const volatile void __iomem *addr)
{
volatile unsigned int j;
j=*(volatile unsigned int __force *) addr;
return j;
}
This is no different from the current code. You cast the pointer to
a volatile unsigned int, you dereference that pointer, and you cast
the result to an unsigned int. C does not have the same "assignment"
distinctions that C++ has.
(This may or may not fix the issue though. There is at least one
known compiler issue that might be causing the breakage. However,
correct compiler optimizations should be ruled out first.)
Nope, any GCC behavior requiring code such as the above is broken,
not just in my opinion but in the opinion of the GCC developers
themselves, as you'd notice if you took the time to read down to the
end of the bugzilla discussion.
Cheers,
Kyle Moffett
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