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.

Answer 1

这很简单：

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