Why does setting a value at an arbitrary memory location not work?

Question

I have this code:

#include 
#include 
#include 
#include 

int main (int argc, char** argv)          


        

Answer 1

The quote from C11 standard 6.5.3.2p4:

4 The unary * operator denotes indirection. [...] If an invalid value has been assigned to the pointer, the behavior of the unary * operator is undefined.

You use * operator on (volatile uint8_t*)0x12345678u pointer. Is this a valid pointer? Is it invalid pointer? What is an "invalid value" of a pointer?

There is no check that allows to find out which particilar pointer values are valid, which aren't. It is not implemented in C language. A random pointer may just happen to be a valid pointer. But most, most probably it is an invalid pointer. In which case - the behavior is undefined.

Dereferencing an invalid pointer is undefined behavior. But - outside of C scope and into operating system - on *unix systems trying to access memory that you are not allowed to, should raise a signal SIGSEGV on your program and terminate your program. Most probably this is what happens. Your program is not allowed to access memory location that is behind 0x12345678 value, the operating system specifically protects against that.

Also note, that systems use ASLR, so that pointer values within your program are indeed in some degree random. There are not linear, ie. *(char*)0x01 will not access the first byte in your ram. Operating system (or more exact, the underlying hardware as configured by the operating system) translates pointer values in your program to physical location in ram using what is called virtual memory. The same pointer values may just happen to be valid on the second run of your program. But most probably, because pointers can have so many values, most probably it isn't a valid pointer. Your operating system kills your program, as it detects an invalid memory access.