memory address literal

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 m开发者_开发技巧y syntax is correct.

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.

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.