Pointer to volatile data (*((volatile uint32_t *)0x40000000)) [duplicate]

Question

I am analyzing a peripheral driver's files and found some register mapping code. I have basic knowledge about pointers, but am unable to understand the below code.

#define WATCHDOG0_LOAD_R        (*((volatile uint32_t *)0x40000000))

I have understood that it defines the identifier WATCHDOG0_LOAD_R to the register's memory address. But I am not able to understand the syntax on the right side. Could anyone explain me in detail why this pointer is written in such a way?

Answer 1

Let's take it one step at the time:

0x40000000

is your memory address.

(uint32_t *)0x40000000

casts this to a pointer to that memory address, of type uint32_t, meaning 32 bit without sign.

(volatile uint32_t *)0x40000000

volatile means, basically, "hey compiler, don't do any optimization; I really want to go every time to that memory address and fetch it, without any prefetch or anything particular".

*((volatile uint32_t *)0x40000000)

means: take the value contained at the address identified by that pointer, so the four bytes starting at 0x40000000.

Answer 2

Analyzing (*((volatile uint32_t *)0x40000000))

• 0x40000000 is the address of register in your micro memory map
• the register is 32 bits wide, that means must be uint32_t *
• volatile is added to tell compile to avoid to optimize that variable because of could change, for example, in an interrupt routine.
• last: the * dereference the pointer: make you able to access the content of that specific register.