Can we assign a value to a given memory location?

Question

I want to assign some value (say 2345) to a memory location(say 0X12AED567). Can this be done?

In other words, how can I implement the following function?

Answer 1

C99 standard draft

This is likely not possible without implementation defined behavior.

About casts like:

*(uint32_t *)0x12AED567 = 2345;

the C99 N1256 standard draft "6.3.2.3 Pointers" says:

5 An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation. 56)

GCC implementation

GCC documents its int to pointer implementation at: https://gcc.gnu.org/onlinedocs/gcc-5.4.0/gcc/Arrays-and-pointers-implementation.html#Arrays-and-pointers-implementation

A cast from integer to pointer discards most-significant bits if the pointer representation is smaller than the integer type, extends according to the signedness of the integer type if the pointer representation is larger than the integer type, otherwise the bits are unchanged.

so the cast will work as expected for this implementation. I expect other compilers to do similar things.

Answer 2

The fact that you are asking this question kind of indicates that you're in over your head. But here you go:

*(int *)0x12AED567 = 2345;

Answer 3

Just treat the memory location as a pointer

int* pMemory =  OX12AED567;
*pMemory = 2345;

Note: This will only work if that memory location is accessible and writable by your program. Writing to an arbitrary memory location like this is inherently dangerous.

Answer 4

The answer depends on some factors. Is your program running within a modern operating system?

If yes, trying to access a memory area that is not mapped will cause a SIGSEGV. To accomplish that, you have to use a system specific function to map the region of memory that contains this exact address before trying to access it.

Answer 5

With the proviso that it's not portable or safe (at all):

*((int *)0x12AED567) = 2345;

Answer 6

As far as C is concerned, that's undefined behaviour. My following suggestion is also undefined behaviour, but avoids all the type-based and aliasing-based problems: Use chars.

int a = get_value();
char const * const p = (const char * const)&a;
char * q = (char *)0x12345;

memcpy(q, p, sizeof(int));

Alternatively, you can access bytes q[i] directly. (This is the part that is UB: the pointer q was not obtained as the address-of an actual object or as the result of an allocation function. Sometimes this is OK; for instance if you're writing a free-standing program that runs in real mode and accesses the graphics hardware, you can write to the graphics memory directly at a well-known, hard-coded address.)