volatile int vfoo = 0;
void func()
{
int bar;
do
{
bar = vfoo; // L.7
}while(bar!=1);
return;
}
This code busy-waits for the variable to turn to 1. If on first pass vfoo is not set to 1, will I get stuck inside.
This code compiles without warning.
What does the standard say about this?
vfoo is declared as volatile. Therefore, read to this variable should not be optimized.
However, bar is not volatile qualified. Is the compiler allowed to optimize the write to this bar? .i.e. the compiler would do a read access to vfoo, and is allowed to discard this value and not assign it to bar (at L.7).
If this is a special case where the standard has something to say, can you please include the clause and interpret the standard's lawyer talk?
What the standard has to say about this includes:
5.1.2.3 Program execution
¶2 Accessing a volatile object, modifying an object, modifying a file, or calling a function that does any of those operations are all side effects, which are changes in the state of the execution environment. Evaluation of an expression in general includes both value computations and initiation of side effects. Value computation for an lvalue expression includes determining the identity of the designated object.
¶4 In the abstract machine, all expressions are evaluated as specified by the semantics. 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).
¶6 The least requirements on a conforming implementation are:
Accesses to volatile objects are evaluated strictly according to the rules of the abstract machine.
...
The takeaway from ¶2 in particular should be that accessing a volatile object is no different from something like calling printf - it can't be elided because it has a side effect. Imagine your program with bar = vfoo; replaced by bar = printf("hello\n");
volatile variable has to be read on any access. In your code snippet that read cannot be optimized out. The compiler knows that bar might be affected by the side effect. So the condition will be checked correctly.
https://godbolt.org/z/nFd9BB
However, bar is not volatile qualified.
Variable bar is used to hold a value. Do you care about the value stored in it, or do you care about that variable being represented exactly according to the ABI?
Volatile would guarantee you the latter. Your program depends on the former.
Is the compiler allowed to optimize the write to this bar?
Of course. Why would you possibly care whether the value read was really written to a memory location allocated to the variable on the stack?
All you specified was that the value read was tested as an exit condition:
bar = ...
}while(bar!=1);
.i.e. the compiler would do a read access to vfoo, and is allowed to
discard this value and not assign it to bar (at L.7).
Of course not!
The compiler needs to hold the value obtained by the volatile read enough time to be able to compare it to 1. But no more time, as you don't ever use bar again latter.
It may be that a strange CPU as a EQ1 ("equal to 1") flag in the condition register, that is set whenever a value equal to 1 is loaded. Then the compiler would not even store temporarily the read value and just EQ1 condition test.
Under your hypothesis that compilers can discard variable values for all non volatile variables, non volatile objects would have almost no possible uses.
Related
From ISO/IEC 9899:201x section 5.1.2.3 Program execution paragraph 4:
In the abstract machine, all expressions are evaluated as specified by
the semantics. 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).
What exactly is the allowed optimization here regarding the volatile object? can someone give an example of a volatile access that CAN be optimized away?
Since volatiles access are an observable behaviour (described in paragraph 6) it seems that no optimization can take please regarding volatiles, so, I'm curious to know what optimization is allowed in section 4.
Reformatting a little:
An actual implementation need not evaluate part of an expression if:
a) it can deduce that its value is not used; and
b) it can deduce that that no needed side effects are produced (including any
caused by calling a function or accessing a volatile object).
Reversing the logic without changing the meaning:
An actual implementation must evaluate part of an expression if:
a) it can't deduce that its value is not used; or
b) it can't deduce that that no needed side effects are produced (including
any caused by calling a function or accessing a volatile object).
Simplifying to focus on the volatile part:
An actual implementation must evaluate part of an expression if needed
side effects are produced (including accessing a volatile object).
Accesses to volatile objects must be evaluated. The phrase “including any…” modifies “side effects.” It does not modify “if it can deduce…” It has the same meaning as:
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 (including any caused by calling a function or accessing a volatile object) are produced.
This means “side effects” includes side effects that are caused by accessing a volatile object. In order to decide it cannot evaluate part of an expression, an implementation must deduce that no needed side effects, including any caused by calling a function or accessing a volatile object, are produced.
It does not mean that an implementation can discard evaluation of part of an expression even if that expression includes accesses to a volatile object.
can someone give an example of a volatile access that CAN be optimized
away?
I think that you misinterpreted the text, IMO this paragraph means that
volatile unsigned int bla = whatever();
if (bla < 0) // the code is not evaluated even if a volatile is involved
Adding another example that fits into this in my understanding:
volatile int vol_a;
....
int b = vol_a * 0; // vol_a is not evaluated
In cases where an access to a volatile object would affect system behavior in a way that would be necessary to make a program achieve its purpose, such an access must not be omitted. If the access would have no effect whatsoever on system behavior, then the operation could be "performed" on the abstract machine without having to execute any instructions. It would be rare, however, for a compiler writer to know with certainty that the effect of executing instructions to perform the accesses would be the same as the effect of pretending to do those instructions on the abstract machine while skipping them on the real one.
In the much more common scenario where a compiler writer would have no particular knowledge of any effect that a volatile access might have, but also have no particular reason to believe that such accesses couldn't have effects the compiler writer doesn't know about (e.g. because of hardware which is triggered by operations involving certain addresses), a compiler writer would have to allow for the possibility that such accesses might have "interesting" effects by performing them in the specified sequence, without regard for whether the compiler writer knows of any particular reason that the sequence of operations should matter.
The C99 standard 5.1.2.3$2 says
Accessing a volatile object, modifying an object, modifying a file, or calling a function
that does any of those operations are all side effects, 12) which are changes in the state of
the execution environment. Evaluation of an expression in general includes both value
computations and initiation of side effects. Value computation for an lvalue expression
includes determining the identity of the designated object.
I guess that in a lot of cases the compiler can't inline and possibly eliminate the functions doing I/O since they live in a different translation unit. And the parameters to functions doing I/O are often pointers, further hindering the optimizer.
But link-time-optimization gives the compiler "more to chew on".
And even though the paragraph I quoted says that "modifying an object" (that's standard-speak for memory) is a side-effect, stores to memory is not automatically treated as a side effect when the optimizer kicks in. Here's an example from John Regehrs Nine Ways to Break your Systems Software using Volatile where the message store is reordered relative to the volatile ready variable.
.
volatile int ready;
int message[100];
void foo (int i) {
message[i/10] = 42;
ready = 1;
}
How do a C compiler determine if a statement operates on a file? In a free-standing embedded environment I declare registers as volatile, thus hindering the compiler from optimizing calls away and swapping order of I/O calls.
Is that the only way to tell the compiler that we're doing I/O? Or do the C standard dictate that these N calls in the standard library do I/O and thus must receive special treatment? But then, what if someone created their own system call wrapper for say read?
As C has no statement dedicated to IO, only function calls can modify files. So if the compiler sees no function call in a sequence of statements, it knows that this sequence has not modified any file.
If only functions from the standard library are called, and if the environment is hosted, the compiler could know what they do and use that to guess what will happen.
But what is really important, is that the compiler only needs to respect side effects. It is perfectly allowed when it does not know, to assume that a function call could involve side effects and act accordingly. It will not be a violation of the standard if no side effects are actually involved, it will just possibly lose a higher optimization.
My code is written in C. I have an ISR (Interrupt Service Routine) that communicates with the main code using global variables. The ISR is in a different compilation unit from the main code.
Is there any reason I cannot use "volatile" for the main code but leave it off in the ISR?
My reasoning is as follows:
The volatile qualifier is preventing the compiler from fully optimizing the ISR. From the ISR's point of view the variable is not volatile - i.e. it cannot be externally changed for the duration of the ISR and the value does not need to be output for the duration of the ISR. Additionally, if the ISR is in its own compilation unit, the compiler MUST have the ISR read the global from memory before its first use and it MUST store changes back before returning. My reasoning for this is: Different compilation units need not be compiled at the same time so the compiler has no idea what is happening beyond the confines of the ISR (or it should pretend to) and so it must ensure that the global is read/written at the boundaries of the ISR.
Perhaps, I am misunderstanding the significance of compilation units? One reference that I found said that GCC has made this volatile mismatch a compile time error; I am not sure how it could, if they are in different compilation units, shouldn't they be independent? Can I not compile a library function separately and link it in later?
Nine ways to break your systems code using volatile
Perhaps an argument could be made from the concept of sequence points. I do not fully understand the concepts of sequence points or side effects; but, the C99 spec states in 5.1.2.3 paragraph 2:
"... At certain specified points in the execution sequence called sequence points, all side effects of previous evaluations shall be complete and no side effects of subsequent evaluations shall have taken place."
Annex C, lists sequence points that include:
The call to a function, after the arguments have been evaluated.
Immediately before a library function returns.
Ref:WG14 Document: N1013, Date: 07-May-2003
Note: A previous question, Global Variable Access Relative to Function Calls and Returns asked whether globals are stored/written before/after function calls and and returns. But this is a different question which asks whether a global variable may be differently qualified as "volatile" in different compilation units. I used much of the same reasoning to justify my preliminary conclusions, which prompted some readers to think it is the same question.
ISO/IEC 9899:2011 (the C11 standard) says:
6.7.3 Type qualifiers
¶6 If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined. If an attempt is made to refer to an object defined with a volatile-qualified type through use of an lvalue with non-volatile-qualified type, the behavior is undefined.133)
133) This applies to those objects that behave as if they were defined with qualified types, even if they are never actually defined as objects in the program (such as an object at a memory-mapped input/output address).
The second sentence of ¶6 says that you invoke undefined behaviour if you have either of the organizations shown here:
File main.c File isr.c:
volatile int thingamyjig = 37; extern int thingamyjig; // V1
extern int thingamyjig; volatile int thingamyjig = 37; // V2
In each case of V1 or V2, you run foul of the undefined behaviour specified in that section of the standard — though V1 is what I think you're describing in the question.
The volatile qualifier must be applied consistently:
File main.c File isr.c:
volatile int thingamyjig = 37; extern volatile int thingamyjig; // V3
extern volatile int thingamyjig; volatile int thingamyjig = 37; // V4
Both V3 and V4 preserve the volatile-qualifiers consistently.
Note that one valid manifestation of 'undefined behaviour' is 'it behaves sanely and as you would like it to'. Unfortunately, that is not the only, or necessarily the most plausible, possible manifestation of undefined behaviour. Don't risk it. Be self-consistent.
I know that when a variable declaration precedes with register keyword compiler may put the variable in CPU's register for faster access. Same way I know that compiler can put const variable in ROM because const variable's value won't change throughout the execution of program.I know also that register specifier is a request to the implementation to put variable in CPU register. But where the variable will be stored if it is marked with both const & register?
Consider following program:
#include <stdio.h>
int main()
{
register const int a=3;
printf("%d",a);
return 0;
}
In this case where the variable gets stored? In the CPU register or on the stack(if compiler ignores the register request) or in the ROM if compiler optimizes it.
I don't actually know what any individual compiler does in this situation, but from logic, and given that register is more of a suggestion, and the compiler is permitted to ignore it as far as the standard goes, I'd say there are three likely outcomes:
As constants are usually pretty efficient, and can often be encoded as part of the instruction or optimized out, and therefore might not even need to be loaded from the data section into a register explicitly, it might just ignore the register keyword completely and treat it as just const.
If there is any code generation step that needs to decide which values to put in a register and which elsewhere, and a is one of these values, the compiler might take into account that you declared it as register and prefer that.
(And this may just be a variation on #2) The compiler may decide that you had a reason to specify register here and always force a into a register when it can. Worst case, this could run counter to how the optimizer would usually assign your code to registers though, and could result in sub-optimal code. Or if you really know what you're doing, it might work around an optimizer bug.
First when you declare a variable as const, the compiler will never put it in ROM, because ... neither the compiler, nor the linker, nor later loader can write anything in ROM ! It can only put it in read-only segments that will still reside in RAM even if any write access at run time will cause an error.
Then to answer more precisely your question, the compiler may always do what it wants.
when you declare a variable const, the compiler should throw an error if it detects a change and can put it in a read-only segment
when you declare a variable register, the compiler should throw en error if it detects an attempt to take its address, and can try to keep it in register.
The only thing that is explicitely required is that a compiler correctly processes correct code. It is requested to throw an error if it cannot compile some code, but it is allowed to accept incorrect code : that may be called extensions or special features. For example a compiler is free to declare that it fully ignores register declaration and allows taking the address of a register variable but it must at least issue a warning [edit see below for details]. Simply it must not choke on correct register (or const usage).
And it can make use of register and const declaration for its optimisations but is perfectly free to ignore them. For example the following code :
const int a = 5;
const int *b = (int *) &a;
*b = 4;
leads to Undefined Behaviour. It is implementation dependant if after that :
a = 5
a = 4
the program crashed
the computer was burned to fire (but this one should be uncommon :-) )
...
EDIT :
JensGustedt noted in comment that C language specification contains in paragraph 6.5.3.2 Address and indirection operators this constraint clause :
The operand of the unary & operator shall ..., or an lvalue that designates an object that is not a bit-field and is not declared with the register storage-class specifier. (emphasize mine)
As it in a constraint clause, paragraph 5.1.1.3 Diagnostics of same specification requires that A conforming implementation shall produce at least one diagnostic message (identified in an implementation-defined manner) if a preprocessing translation unit or translation unit contains a violation of any syntax rule or constraint.
So a conforming C compiler shall at least issue a warning if programmer tries to take the address of a register variable.
I have used a static global variable and a static volatile variable in file scope,
both are updated by an ISR and a main loop and main loop checks the value of the variable. here during optimization neither the global variable nor the volatile variable are optimized. So instead of using a volatile variable a global variable solves the problem.
So is it good to use global variable instead of volatile?
Any specific reason to use static volatile??
Any example program would be appreciable.
Thanks in advance..
First let me mention that a static global variable, is the same as a global variable, except that you are limiting the variable to the scope of the file. I.e. you can't use this global variable in other files via the extern keyword.
So you can reduce your question to global variables vs volatile variables.
Now onto volatile:
Like const, volatile is a type modifier.
The volatile keyword was created to prevent compiler optimizations that may make code incorrect, specifically when there are asynchronous events.
Objects declared as volatile may not be used in certain optimizations.
The system always reads the current true value of a volatile object at the point it is used, even if a previous instruction asked for a value from the same object. Also, the value of the object is written immediately on assignment. That means there is no caching of a volatile variable into a CPU register.
Dr. Jobb's has a great article on volatile.
Here is an example from the Dr. Jobb's article:
class Gadget
{
public:
void Wait()
{
while (!flag_)
{
Sleep(1000); // sleeps for 1000 milliseconds
}
}
void Wakeup()
{
flag_ = true;
}
...
private:
bool flag_;
};
If the compiler sees that Sleep() is an external call, it will assume that Sleep() cannot possibly change the variable flag_'s value. So the compiler may store the value of flag_ in a register. And in that case, it will never change. But if another thread calls wakeup, the first thread is still reading from the CPU's register. Wait() will never wake-up.
So why not just never cache variables into registers and avoid the problem completely?
It turns out that this optimization can really save you a lot of time overall. So C/C++ allows you to explicitly disable it via the volatile keyword.
The fact above that flag_ was a member variable, and not a global variable (nor static global) does not matter. The explanation after the example gives the correct reasoning even if you're dealing with global variables (and static global variables).
A common misconception is that declaring a variable volatile is sufficient to ensure thread safety. Operations on the variable are still not atomic, even though they are not "cached" in registers
volatile with pointers:
Volatile with pointers, works like const with pointers.
A variable of type volatile int * means that the variable that the pointer points to is volatile.
A variable of type int * volatile means that the pointer itself is volatile.
They are different things. I'm not an expert in volatile semantics. But i think it makes sense what is described here.
Global
Global just means the identifier in question is declared at file-scope. There are different scopes, called function (where goto-labels are defined in), file (where globals reside), block (where normal local variables reside), and function prototype (where function parameters reside). This concept just exist to structure the visibility of identifiers. It doesn't have anything to do with optimizations.
Static
static is a storage duration (we won't look at that here) and a way to give a name declared within file scope internal linkage. This can be done for functions or objects only required within one translation unit. A typical example might be a help function printing out the accepted parameters, and which is only called from the main function defined in the same .c file.
6.2.2/2 in a C99 draft:
If the declaration of a file scope
identifier for an object or a function
contains the storage class specifier
static, the identifier has internal
linkage.
Internal linkage means that the identifier is not visible outside the current translation unit (like the help function of above).
Volatile
Volatile is a different thing: (6.7.3/6)
An object that has volatile-qualified
type may be modified in ways unknown to
the implementation or have other
unknown side effects. Therefore any
expression referring to such an object
shall be evaluated strictly according
to the rules of the abstract machine,
as described in 5.1.2.3. Furthermore,
at every sequence point the value last
stored in the object shall agree with
that prescribed by the abstract
machine, except as modified by the
unknown factors mentioned
previously.
The Standard provides an excellent example for an example where volatile would be redundant (5.1.2.3/8):
An implementation might define a
one-to-one correspondence between
abstract and actual semantics: at
every sequence point, the values of
the actual objects would agree with
those specified by the abstract
semantics. The keyword volatile
would then be redundant.
Sequence points are points where the effect of side effects concerning the abstract machine are completed (i.e external conditions like memory cell values are not included). Between the right and the left of && and ||, after ; and returning from a function call are sequence points for example.
The abstract semantics is what the compiler can deduce from seeing only the sequence of code within a particular program. Effects of optimizations are irrelevant here. actual semantics include the effect of side effects done by writing to objects (for example, changing of memory cells). Qualifying an object as volatile means one always gets the value of an object straight from memory ("as modified by the unknown factors"). The Standard doesn't mention threads anywhere, and if you must rely on the order of changes, or on atomicity of operations, you should use platform dependent ways to ensure that.
For an easy to understand overview, intel has a great article about it here.
What should i do now?
Keep declaring your file-scope (global) data as volatile. Global data in itself does not mean the variables' value will equal to the value stored in memory. And static does only make your objects local to the current translation unit (the current .c files and all other files #include'ed by it).
The "volatile" keyword suggests the compiler not to do certain optimizations on code involving that variable; if you just use a global variable, nothing prevents the compiler to wrongly optimize your code.
Example:
#define MYPORT 0xDEADB33F
volatile char *portptr = (char*)MYPORT;
*portptr = 'A';
*portptr = 'B';
Without "volatile", the first write may be optimized out.
The volatile keyword tells the compiler to make sure that variable will never be cached. All accesses to it must be made in a consistent way as to have a consistent value between all threads. If the value of the variable is to be changed by another thread while you have a loop checking for change, you want the variable to be volatile as there is no guarantee that a regular variable value won't be cached at some point and the loop will just assume it stays the same.
Volatile variable on Wikipedia
They may not be in different in your current environment, but subtle changes could affect the behavior.
Different hardware (more processors, different memory architecture)
A new version of the compiler with better optimization.
Random variation in timing between threads. A problem may only occur one time in 10 million.
Different compiler optimization settings.
It is much safer in the long run to use proper multithreading constructs from the beginning, even if things seem to work for now without them.
Of course, if your program is not multi-threaded then it doesn't matter.
I +1 friol's answer. I would like to add some precisions as there seem to be a lot of confusions in different answers: C's volatile is not Java's volatile.
So first, compilers can do a lot of optimizations on based on the data flow of your program, volatile in C prevents that, it makes sure you really load/store to the location every time (instead of using registers of wiping it out e.g.). It is useful when you have a memory mapped IO port, as friol's pointed out.
Volatile in C has NOTHING to do with hardware caches or multithreading. It does not insert memory fences, and you have absolutely no garanty on the order of operations if two threads do accesses to it. Java's volatile keyword does exactly that though: inserting memory fences where needed.
volatile variable means that the value assinged to it is not constant, i.e if a function containing a volatile variable "a=10" and the function is adding 1 in each call of that function then it will always return updated value.
{
volatile int a=10;
a++;
}
when the above function is called again and again then the variable a will not be re-initialised to 10, it will always show the updated value till the program runs.
1st output= 10
then 11
then 12
and so on.