I have a come across following definition in an embedded code:
#define REG_ADD 0x20081004
#define pREG ((void * volatile *)REG_ADD)
Why there are 2 * in pREG definition? What does it mean?
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doubleE
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1Pointer to pointer. – Some programmer dude Apr 19 '17 at 03:29
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This is corrrect syntax, but it still looks like a bad idea. Can you provide context? – too honest for this site Apr 19 '17 at 03:34
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@dandi-kain: do a printf() on REG_ADD and it's address, then printf() pREG, DE-refferenced pREG and the address of pREG, this should help you see what is going on... – T.T.T. Apr 19 '17 at 04:17
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You can't "printf" the address of a constant (REG_ADD) since it is not defined. Same goes for pREG. The address of pREG is not defined. – Ajay Brahmakshatriya Apr 19 '17 at 04:43
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1Cast to [pointer to volatile pointer to void](https://cdecl.org/?q=void+*+volatile+*REG_ADD) – chux - Reinstate Monica Apr 19 '17 at 04:43
1 Answers
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void** is a pointer to pointer-to-void.
void * volatile * is a pointer to volatile-qualified-pointer-to-void. (Read the declaration right-to-left, pretty much. See this.)
What this means in plain English is that pREG is likely a pointer to some sort of hardware index register, which in turn contains an address. To tell the compiler that this index register can get updated at any moment, the register itself should be treated as volatile.
A somewhat more readable way to write the same would be:
typedef void* volatile reg_add_t;
reg_add_t* pREG = (reg_add_t)0x20081004u;
Please note that the use of void* for this purpose is questionable. This register will have a defined use, possibly it should have been a uint32_t * volatile or uint8_t * volatile instead.
Lundin
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