Question

Will overstepping of array indices beyond an array size always result in a Segmentation Fault on a reliable true POSIX system like GNU/Linux?

I think no, if the accessed location still lies in the same page, but want to be sure.

Answer 1

No, it will not necessarily. Segmentation Violations SIGSEGV is handled by the Kernel and will only be invoked on an invalid memory access.

Arrays are compile-time constructs as is POSIX. Segmentation Violation on the other hand is a run time error indicating that you tried to do something with a piece of virtual memory that you are either not allowed (read/write/execute) or that has not been mapped (the memory controller does not know what physical resource is supposed to back it). At that point the kernel will send your process a signal SiGSEGV and the default behavior is to exit, though that can be overridden.

You will even often see C code like this, which instructs the compiler to access the memory behind the struct as a variable size array:

struct s {
    // some other elements
    int id; // whatever other elements

    char appended_array[0];
};

// ....

struct s* mystruct = malloc(sizeof(struct s) + length_of_array_i_need);

// work with mystruct->appended_array[i]
// at this point mystruct->appended_array[3] is valid C and the compiler will not even issue a warning,
// though if it lies outside of your allowed VM then the kernel will issue a SIGSEGV

Only because an array is not defined beyond a certain index, it does not necessarily mean that there is no valid memory following it.