Is unused memory in address space protected

Question

Is the unused memory in address space of a process protected by just having read permission, so that writing to a location pointed by an unitialized pointer for example always cause a page fault to be trapped by the OS? Or is it not the case, and every memory location besides the code (which ofcourse is given read only access), is given write access?

I'm asking this because my friend was showing me his code where he didn't initialize a pointer and wrote in the memory pointed by it, but still his program wasn't crashing with mingw gcc compiler for windows but always crashing with visual c++, in mac or linux.

What I think is that the OS do not protect memory for unused areas and the crashing was being caused because in the code generated by the mingw, the random pointer value was pointing to some used area such as stack, heap or code, while in other cases it was pointing to some free area. But if the OS really doesn't protect the unused areas, wouldn't these sort of bugs, such as uninitialized pointers be difficult to debug?

I guess this is why it is advised to always assign NULL to a pointer after calling delete or free, so that when something is accessed with it, it really causes a visible crash.

Answer 1

Uninitialized pointers don't necessarily to point to unused address space. They could very well be values that happen to point to writeable memory. Such as a pointer on the stack that happened to be where a previously executed function stored a valid address.

Answer 2

In a typical, current server/desktop OS (and quite a few smaller systems such as cell phones as well) you have virtual memory. This means the OS builds a table that translates from the virtual address your code uses, to a physical address that specifies the actual memory being addressed. This mapping is normally done in "pages" -- e.g., a 4KB chunk of memory at a time.

At least in the usual case, parts of the address space that aren't in use at all simply won't be mapped at all -- i.e., the OS won't build an entry in the table for that part of the address space. Note, however, that memory that is allocated will (of necessity) be rounded to a multiple of the page size, so each chunk of memory that's in use will often be followed by some small amount that's not really in use, but still allocated and "usable". Since protection is also (normally) done on a per-page basis, if the rest of that page is (say) Read-only, the remainder at the tail end will be the same.