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If we declare a variable as volatile every time the fresh value is updated
If we declare a variable as const then the value of that variable will not be changed

Then const volatile int temp;
What is the use of declaring the variable temp as above?
What happens if we declare as const int temp?

asked Jan 4, 2011 at 10:48
1
  • You wouldn't use const volatile int temp; at block scope (i.e. inside { }), it has no use there. Commented Feb 15, 2016 at 21:57

10 Answers 10

180

An object marked as const volatile will not be permitted to be changed by the code (an error will be raised due to the const qualifier) - at least through that particular name/pointer.

The volatile part of the qualifier means that the compiler cannot optimize or reorder access to the object.

In an embedded system, this is typically used to access hardware registers that can be read and are updated by the hardware, but make no sense to write to (or might be an error to write to).

An example might be the status register for a serial port. Various bits will indicate if a character is waiting to be read or if the transmit register is ready to accept a new character (ie., - it's empty). Each read of this status register could result in a different value depending on what else has occurred in the serial port hardware.

It makes no sense to write to the status register (depending on the particular hardware spec), but you need to make sure that each read of the register results in an actual read of the hardware - using a cached value from a previous read won't tell you about changes in the hardware state.

A quick example:

unsigned int const volatile *status_reg; // assume these are assigned to point to the 
unsigned char const volatile *recv_reg; // correct hardware addresses
#define UART_CHAR_READY 0x00000001
int get_next_char()
{
 while ((*status_reg & UART_CHAR_READY) == 0) {
 // do nothing but spin
 }
 return *recv_reg;
}

If these pointers were not marked as being volatile, a couple problems might occur:

  • the while loop test might read the status register only once, since the compiler could assume that whatever it pointed to would never change (there's nothing in the while loop test or loop itself that could change it). If you entered the function when there was no character waiting in UART hardware, you might end up in an infinite loop that never stopped even when a character was received.
  • the read of the receive register could be moved by the compiler to before the while loop - again because there's nothing in the function that indicates that *recv_reg is changed by the loop, there's no reason it can't be read before entering the loop.

The volatile qualifiers ensures that these optimizations are not performed by the compiler.

answered Jan 4, 2011 at 18:44
8
  • 5
    +1 for explanation. And I have a question: what about const volatile methods? If I have a class, which is accessed by many threads (although access is synchronized with mutex) does my const methods also have to be volatile (since some variable could be changed by other thread) Commented Aug 8, 2012 at 23:31
  • A simple question: why const volatile objects can never be placed in read-only memory? See C11, footnote 132. Commented Sep 20, 2022 at 20:04
  • @pmor: The semantics of volatile-qualified objects are implementation-defined. The Standard gives implementations blanket permission to interpret a const qualifier on an object that isn't volatile as an invitation to place the object in read-only storage, with no obligation to document such treatment. An implementation could specify that the semantics of volatile objects vary depending upon whether they are const, but if an implementation doesn't document that const affects the behavior of volatile objects, then const volatile objects must behave like volatile ones. Commented Jun 6, 2023 at 14:55
  • 1
    @supercat Since "An object that has volatile-qualified type may be modified in ways unknown to the implementation or have other unknown side effects" AND if const volatile object is placed in read-only memory, then one may ask: "how this object can be modified in ways unknown to the implementation if it is placed in read-only memory?". However, such a modification is outside of C. Perhaps OS (if any) temporarily removes "read only" attribute, etc. Have a look: reviews.llvm.org/D131012. Commented Jun 7, 2023 at 11:07
  • @pmor: I see nothing in the text that would forbid an implementation from treating volatile-qualified objects in any manner they see fit, provided they document it. For clang to refuse to place objects there because the Standard supposedly forbids it would be a gross abuse of the Standard, though in line with what I've come to expect from the maintainers of clang. Commented Jun 7, 2023 at 14:58
48
  • volatile will tell the compiler not to optimise code related the variable, usually when we know it can be changed from "outside", e.g. by another thread.
  • const will tell the compiler that it is forbidden for the program to modify the variable's value.
  • const volatile is a very special thing you'll probably see used exactly 0 times in your life (tm). As is to be expected, it means that the program cannot modify the variable's value, but the value can be modified from the outside, thus no optimisations will be performed on the variable.
answered Jan 4, 2011 at 11:02
4
  • 14
    I'd have thought that volatile variables are usually what happens when you start messing with hardware, not with other threads. Where I've seen const volatile used is in things like memory-mapped status registers or the like. Commented Jan 4, 2011 at 12:33
  • 2
    Of course, you're absolutely right, multithreading is just one example, but not the only one :). Commented Jan 4, 2011 at 15:16
  • 34
    If you work with embedded systems you will see this very often. Commented Oct 11, 2013 at 12:04
  • 1
    Its not actually that special, I see it all the time in embedded work Commented Jul 8, 2022 at 22:31
34

It is not because the variable is const that it may not have changed between two sequence points.

Constness is a promise you make not to change the value, not that the value won't be changed.

answered Jan 4, 2011 at 10:52
2
  • 10
    Plus one for pointing out that const data is not "constant". Commented Jan 26, 2015 at 7:47
  • Is there any type of qualifier that completely prevents modification of a variable, both by program and external? Commented Feb 5, 2024 at 9:16
20

In C, const and volatile are type qualifiers and these two are independent.

Basically, const means that the value isn’t modifiable by the program.

And volatile means that the value is subject to sudden change (possibly from outside the program).

In fact, the C Standard gives an example of a valid declaration which is both const and volatile. The example is:

extern const volatile int real_time_clock;

where real_time_clock may be modifiable by hardware, but cannot be assigned to, incremented, or decremented.

So we should already treat const and volatile separately. These type qualifiers can be applied to struct, union, enum and typedef as well.

Adrian Mole
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answered Aug 9, 2016 at 11:19
7

I've needed to use this in an embedded application where some configuration variables are located in an area of flash memory that can be updated by a bootloader. These config variables are 'constant' during runtime, but without the volatile qualifier the compiler would optimise something like this...

cantx.id = 0x10<<24 | CANID<<12 | 0;

...by precomputing the constant value and using an immediate assembly instruction, or loading the constant from a nearby location, so that any updates to the original CANID value in the config flash area would be ignored. CANID has to be const volatile.

answered Jun 28, 2015 at 10:55
7

You can use const and volatile together. For example, if 0x30 is assumed to be the value of a port that is changed by external conditions only, the following declaration would prevent any possibility of accidental side effects:

const volatile char *port = (const volatile char *)0x30;
Vishnu CS
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answered Dec 23, 2016 at 17:08
6

const means that the variable cannot be modified by the c code, not that it cannot change. It means that no instruction can write to the variable, but its value might still change.

volatile means that the variable may change at any time and thus no cached values might be used; each access to the variable has to be executed to its memory address.

Since the question is tagged "embedded" and supposing temp is a user declared variable, not a hardware-related register (since these are usually handled in a separate .h file), consider:

An embedded processor which has both volatile read-write data memory (RAM) and non-volatile read-only data memory, for example FLASH memory in von-Neumann architecture, where data and program space share a common data and address bus.

If you declare const temp to have a value (at least if different from 0), the compiler will assign the variable to an address in FLASH space, because even if it were assigned to a RAM address, it still needs FLASH memory to store the initial value of the variable, making the RAM address a waste of space since all operations are read-only.

In consequence:

int temp;is a variable stored in RAM, initialized to 0 at startup (cstart), cached values may be used.

const int temp;is a variable stored in (read-ony)FLASH, initialized to 0 at compiler time, cached values may be used.

volatile int temp; is a variable stored in RAM, initialized to 0 at startup (cstart), cached values will NOT be used.

const volatile int temp; is a variable stored in (read-ony)FLASH, initialized to 0 at compiler time, cached values will NOT be used

Here comes the usefull part:

Nowadays most Embedded processors have the ability to make changes to their read-only non-volatile memory by means of a special function module, in which case const int temp can be changed at runtime, altought not directly. Said in another way, a function may modify the value at the address where temp is stored.

A practical example would be to use temp for the device serial number. The first time the embedded processor runs, temp will be equal to 0 (or the declared value) and a function can use this fact to run a test during production and if sucessfull, ask to be assigned a serial number and modify the value of temp by means of a special function. Some processors have a special address range with OTP (one-time programmable) memory just for that.

But here comes the difference:

If const int temp is a modifiable ID instead of a one-time-programmable serial number and is NOT declared volatile, a cached value might be used untill the next boot, meaning the new ID might not be valid untill the next reboot, or even worse, some functions might use the new value while other might use an older cached value untill reboot. If const int temp IS declared voltaile, the ID change will take effect immediately.

answered Sep 30, 2015 at 16:38
1
  • Another that you might add to your list is const volatile int * const p; which means that p is a const pointer so can't be changed and it points to a data item that is const volatile so you can't change that data item through the pointer p either. Commented Dec 20, 2024 at 18:02
3

In simple terms, Value in 'const volatile' variable cannot be modified programmatically but can be modified by hardware. Volatile here is to prevent any compiler optimisation.

answered Jul 30, 2018 at 15:06
2

We use 'const' keyword for a variable when we don't want to the program to change it. Whereas when we declare a variable 'const volatile' we are telling the program not to change it and the compiler that this variable can be changed unexpectedly from input coming from the outside world.

answered Nov 17, 2015 at 3:30
0
0

Example:

const volatile uint16_t variable;

This is a possible declaration of a 16-bit read-only register.

The const keyword says the program cannot change that memory location. The volatile keyword says that location might change at any moment, independently of the program course.

So, their combination is usual for read-only registers, for example a status register. Because the software cannot change the content of that register we use const, but the hardware might change its value at any time, independently on software, so we use volatile.

Moreover, the volatile keyword instructs the compiler to not optimize and not make any shadow copy of that memory location.

answered Feb 24 at 21:24

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