volatile pointer without variable

Question

In C, when I want to put number 5 at address 0x28, I can do it this way:

char* x = (char *) 0x28;
*x = 5;

If I want, I can do the same without declaring a variable:

*((char *) 0x28) = 5;

If I want that compiler treats this address as volatile, I can do it this way:

volatile char* x = (char *) 0x28;
*x = 5;

Can I do it without declaring a variable?

Edit: Let me explain why I want to do "*((char *) 0x28) = 5". I write a blinking LED hello world program for ATmega32U4 and I know that address 0x28 rules the pin to which my LED is connected. And it does work: the C code which you suggested compiles to correct machine code and the LED blinks.

(char *) 0x28 is undefined behavior (UB), unless 0x28 was the address of an object. — chux
– chux, Commented Feb 9, 2018 at 18:14
@chux, given that volatile is most often use for memory-mapped registers, that would be entirely plausible. — zneak
– zneak, Commented Feb 9, 2018 at 18:16
@zneak, plausible is not at all the same thing as well-defined. The C language does not guarantee that it is possible in any way to write a value to an arbitrarily-chosen machine address. In fact, it goes to some effort to avoid introducing any concept of machine addresses into the language at all. Pointers do not necessarily correspond to those. — John Bollinger
– John Bollinger, Commented Feb 9, 2018 at 18:22
@JohnBollinger, trying to drop the language lawyer hammer on people doing something so clearly invalid on a standard computer architecture that it has to be non-portable code for an embedded environment wed to its own C compiler isn't doing anyone a service. The C standard is not a ceiling: in fact, that document has a whole section on common extensions. — zneak
– zneak, Commented Feb 9, 2018 at 18:36
@zneak If the user is asking effectively how to do *((char *) 0x28) = 5;, then it is also likely the user (or others checking this post) do not realize code "doing something so clearly invalid" applies. There is no hammer involved, simply a statement that code may be UB. — chux
– chux, Commented Feb 9, 2018 at 18:40

HolyBlackCat · Accepted Answer · 2018-02-09 18:15:04Z

3

It's simple:

*((volatile char *) 0x28) = 5;

answered Feb 9, 2018 at 18:15

HolyBlackCat

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1 Comment

chux Over a year ago

Good to leave volatile in should the address point to a hardware register.

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