Difference between const & const volatile

Question

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?

Answer 1

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.