问题
Given a literal memory address in hexadecimal format, how can I create a pointer in C that addresses this memory location?
Memory addresses on my platform (IBM iSeries) are 128bits. C type long long is also 128bits.
Imagine I have a memory address to a string (char array) that is: C622D0129B0129F0
I assume the correct C syntax to directly address this memory location:
const char* const p = (const char* const)0xC622D0129B0129F0ULL
I use ULL suffix indicate unsigned long long literal.
Whether my kernel/platform/operating system will allow me to do this is a different question. I first want to know if my syntax is correct.
回答1:
Your syntax is almost correct. You don't need one of those const:
const char* const p = (const char*)0xC622D0129B0129F0ULL
The const immediately before p indicates that the variable p cannot be changed after initialisation. It doesn't refer to anything about what p points to, so you don't need it on the right hand side.
回答2:
There is no such thing like an address literal in C.
The only thing that is guaranteed to work between integers and pointers is cast from void* to uintptr_t (if it exists) and then back. If you got your address from somewhere as an integer this should be of type uintptr_t. Use something like
(void*)(uintptr_t)UINTMAX_C(0xC622D0129B0129F0)
to convert it back.
You'd have to include stdint.h to get the type uintptr_t and the macro UINTMAX_C.
来源：https://stackoverflow.com/questions/3593985/memory-address-literal