Guaranteed read from volatile variable

If I have something like

Code:

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volatile int *ptr = 0xsomeaddress;
...

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Would

Code:

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*ptr;

---

guarantee a read on someaddress?

I saw something like this in an embedded application, and was wondering if this actually works.

(someaddress is a hardware register, on which a read operation has side effects)

Thanks!

Yes. (If you've got the optimizer set to "kill", I suppose it might get blasted, but it really shouldn't.)

Here are some short but informative reads on the subject: How to Use Volatile Keyword and Volatile Myths

I have always understood volatile as meaning the variable can be changed outside of program flow (the links suggest that also).

In that case, the optimizer could argue that it can omit the read because the program doesn't care what value it has? since the result of the read is not assigned to anything, or used to make any decision.

Somewhere, the standard says something like "the purpose of evaluating statements is for side effects to occur", so that's not really a reason to not evaluate.

(EDIT: Of course, I can't speak for your optimizer. But if it will optimize what you have away, it will optimize just about anything else away too.)

Quote:

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Originally Posted by cyberfish

In that case, the optimizer could argue that it can omit the read because the program doesn't care what value it has? since the result of the read is not assigned to anything, or used to make any decision.

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Yes, I do believe that per the standard 5.1.2.3 when it says :"An actual implementation need not evaluate part of an expression if it can deduce that its value is not used and that no needed side effects are produced (including any caused by calling a function or accessing a volatile object)." that means if the compiler doesn't see the code get used at all it would still be optimized out regardless of whether or not it was volatile.

The way I have seen around this is to use the preprocessor directive:

Code:

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#pragma optimize("", off)
//code here
#pragma optimize("", on)

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EDIT: I wasn't suggesting you didn't know about volatile. Just giving you some things I have read in the past. :biggrin:

The definition of volatile is that needed side effects are produced.

(EDIT: I think that should say "that needed side effects are always produced" by a volatile variable. The actual language says that accessing a volatile variable is an observable (like input and output), so can't be elided.)

Quote:

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Somewhere, the standard says something like "the purpose of evaluating statements is for side effects to occur", so that's not really a reason to not evaluate.

(EDIT: Of course, I can't speak for your optimizer. But if it will optimize what you have away, it will optimize just about anything else away too.)

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Thanks I didn't know that!

I am using GCC (cross-compiling for ARM), so I'm guessing that's more or less sane.

Quote:

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(including any caused by calling a function or accessing a volatile object)

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I think that means the compiler needs to be able to prove that accessing the volatile object has no side effect. For example

Code:

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volatile int x;

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The compiler knows that reading x has no side effect, because it allocated memory for that variable.

I don't think it can make that assumption if it's a pointer to some random address it doesn't know.

I see, I am still learning how to interpret this darn standard. :confused:

Quote:

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Originally Posted by cyberfish

If I have something like

Code:

---

volatile int *ptr = 0xsomeaddress;
...

---

Would[code]
*ptr;
[/code
guarantee a read on someaddress?

---

That should guarantee that reads ot write to ptr are not simply done in a register for example. I.e

Code:

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ptr = &foo;
ptr = &bar;

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but I don't know what guarantees if any, apply to using *ptr

Normally your volatile variable would not be a pointer.

I was under the impression that "volatile int *ptr" is a pointer to a volatile int, whereas "int * volatile ptr" is a volatile pointer to an int.

Quote:

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Originally Posted by cyberfish

I have always understood volatile as meaning the variable can be changed outside of program flow (the links suggest that also).

In that case, the optimizer could argue that it can omit the read because the program doesn't care what value it has? since the result of the read is not assigned to anything, or used to make any decision.

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Quite the opposite, actually. The volatile keyword tells the compiler that the variable may change as a result of something invisible to the compiler, and that the compiler needs to care about the changes that occur. If code has multiple reads, and the compiler can see nothing that changes the value, then "volatile" tells the compiler it cannot omit reads.

To answer the original question though, volatile (like const) is always an attribute of the thing immediately to its left in an expression, unless it is on the left, in which case it refers to the right.

So

Code:

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volatile int *ptr = 0xsomeaddress;

---

guarantees that *ptr (if it is accessed) will always dereference 0xsomeaddress. It is equivalent to

Code:

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int volatile *ptr = 0xsomeaddress;

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If you care about ptr itself being changed by something invisible to the compiler, then you need to do

Code:

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volatile int * volatile ptr = 0xsomeaddress;

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(noting that the first usage of volatile can be either before or after the int keyword).

Quote:

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To answer the original question though, volatile (like const) is always an attribute of the thing immediately to its left in an expression, unless it is on the left, in which case it refers to the right.

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I don't think that was the original question :D. I understand the left/right syntax.

Quote:

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Quite the opposite, actually. The volatile keyword tells the compiler that the variable may change as a result of something invisible to the compiler, and that the compiler needs to care about the changes that occur. If code has multiple reads, and the compiler can see nothing that changes the value, then "volatile" tells the compiler it cannot omit reads.

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The thing is, in this case, the code doesn't actually care what that value is. The variable can change all it wants, and the compiler doesn't really care. By your definition, the read can be omitted (in the compiler's point of view, the code will still work correctly no matter how the variable changes).

A similar question would be, do writes to volatile variables have to happen?

If they don't, using them to share variables in multithreading would be very bad...

The point is that the compiler is generally not allowed to eliminate accesses (read or writes) of a variable. It cannot exclude the possibility that accessing has some effect not visible to it. The clauses in the standard relate to something in the implementation (for example, a linker) that has visibility of a complete program, and can determine that a variable declared as volatile is never accessed.

While it is true that an optimiser can determine that the value read from a volatile variable is never used (within the function that accessed the value), it is practically a pretty aggressive optimiser that will remove the read. Why? Because a fair few programmers have used the trick you asked about, and screamed rather loudly about "over-aggressive optimisers".

Generally writes to volatile variables have to happen. Very few compiler vendors would be brave enough to eliminate them.

However, if a variable is to be accessed by multiple threads, declaring it volatile is neither necessary nor sufficient to ensure reads/writes work correctly. Declaring a variable as volatile certainly does not stop a thread that is reading or writing a volatile variable from being preempted part way through the operation. (Basic operations, such as reading or writing an int or pointer are not guaranteed to be atomic).

Oh so it is the same as const. Well I guess I learnt something today.

Quote:

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Originally Posted by grumpy

Quite the opposite, actually. The volatile keyword tells the compiler that the variable may change as a result of something invisible to the compiler, and that the compiler needs to care about the changes that occur. If code has multiple reads, and the compiler can see nothing that changes the value, then "volatile" tells the compiler it cannot omit reads.

To answer the original question though, volatile (like const) is always an attribute of the thing immediately to its left in an expression, unless it is on the left, in which case it refers to the right.

So

Code:

---

volatile int *ptr = 0xsomeaddress;

---

guarantees that *ptr (if it is accessed) will always dereference 0xsomeaddress. It is equivalent to

Code:

---

int volatile *ptr = 0xsomeaddress;

---

If you care about ptr itself being changed by something invisible to the compiler, then you need to do

Code:

---

volatile int * volatile ptr = 0xsomeaddress;

---

(noting that the first usage of volatile can be either before or after the int keyword).

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But in this case, that's exactly what we want, isn't it? We have a pointer-to-volatile, and we are dereferencing the pointer; therefore we are accessing the volatile object that is being pointed to. And accessing a volatile object is exactly the thing that can't be omitted.

The second article in post #3 has a section:
>> 3. Global variables accessed by multiple tasks within a multi-threaded application
which is completely false and incorrect - for reasons grumpy has already mentioned (and the first article).
volatile does not provide memory visibility and ordering guarantees (at the hardware level) which only synchronization primitives from your threading library can provide. Thankfully, this is now in the C++0x standard, including a threading library (yay!).

>> I have always understood volatile as meaning the variable can be changed outside of program flow
I've never liked that explanation. In my mind, it only implies what really should happen instead of just stating is explicitly. That is, 'all volatile accesses must be emitted by the compiler as accesses that hit the address bus'.

It doesn't have to be the case that the "variable can change outside of the program". Writing to a memory location may turn on an LED (or some other external side-effect). By making that write a volatile access, the compiler must emit a write access on the address bus, even though it may not see any in-program side-effects for doing so (compiler doesn't know about the LED).

Another (contrived) example: Three reads from memory will turn on an LED (or set a centrifuge to its maximum RPM):

Code:

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dummy = *ptr;
dummy = *ptr;
dummy = *ptr;

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Most self-respecting compilers would optimize these down to 1 or 0 memory accesses. But if the accesses are volatile, then all 3 accesses must be emitted.

Another (contrived) example: Two reads and a write will turn on an LED:

Code:

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a = *ptr;
b = *ptr;
*ptr = 1;

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As volatile accesses, this must be emitted as 2 reads followed by a write. The compiler is not allowed to change the ordering of volatile accesses. However, the compiler is allowed to re-order all non-volatile accesses.

gg