Pointer declared as constant as well as volatile

Question

While reading I came across this type of declaration and the following line -

const volatile char *p=(const volatile char *) 0x30;

Answer 1

We can use the article Introduction to the volatile keyword which says:

A variable should be declared volatile whenever its value could change unexpectedly. In practice, only three types of variables could change:

• Memory-mapped peripheral registers
• Global variables modified by an interrupt service routine
• Global variables within a multi-threaded application

and:

Embedded systems contain real hardware, usually with sophisticated peripherals. These peripherals contain registers whose values may change asynchronously to the program flow. As a very simple example, consider an 8-bit status register at address 0x1234. It is required that you poll the status register until it becomes non-zero. The nave and incorrect implementation is as follows:

UINT1 * ptr = (UINT1 *) 0x1234;

// Wait for register to become non-zero.
while (*ptr == 0);
// Do something else.

This will almost certainly fail as soon as you turn the optimizer on, since the compiler will generate assembly language that looks something like this:

mov    ptr, #0x1234     mov    a, @ptr loop     bz    loop

The const says your program won't change the variable but as noted in the article outside sources could and volatile prevents it being optimized away.