C语言如何手动分配指针地址？

Question

如何在 C 编程语言中手动分配指针地址（例如分配给内存地址0x28ff44 ）？

Answer 1

Like this:

void * p = (void *)0x28ff44;

Or if you want it as a char * :

char * p = (char *)0x28ff44;

...etc.

If you're pointing to something you really, really aren't meant to change, add a const :

const void * p = (const void *)0x28ff44;
const char * p = (const char *)0x28ff44;

...since I figure this must be some kind of "well-known address" and those are typically (though by no means always) read-only.

Answer 2

Your code would be like this:

int *p = (int *)0x28ff44;

int needs to be the type of the object that you are referencing or it can be void .

But be careful so that you don't try to access something that doesn't belong to your program.

Answer 3

int *p=(int *)0x1234 = 10; //0x1234 is the memory address and value 10 is assigned in that address


unsigned int *ptr=(unsigned int *)0x903jf = 20;//0x903j is memory address and value 20 is assigned 

Basically in Embedded platform we are using directly addresses instead of names

Answer 4

Writing:

int *p=(int *)0x1234 = 10;

is not legal for some compilers. You need to split in two instructions:

int *p=(int *)0x1234; *p = 10;

Answer 5

let's say you want a pointer to point at the address 0x28ff4402, the usual way is

uint32_t *ptr;
ptr = (uint32_t*) 0x28ff4402 //type-casting the address value to uint32_t pointer
*ptr |= (1<<13) | (1<<10); //access the address how ever you want

So the short way is to use a MACRO,

#define ptr *(uint32_t *) (0x28ff4402)