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Asked: 2026-05-16T18:06:54+00:00

Given a literal memory address in hexadecimal format, how can I create a pointer

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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.

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1 Answer

  1. Editorial Team

    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.

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