Why infinite recursion leads to seg fault

Question

Why infinite recursion leads to seg fault ? Why stack overflow leads to seg fault. I am looking for detailed explanation.

int f()
{
  f();
}

int main()
{
           


        

Answer 1

At a "low" level, the stack is "maintained" through a pointer (the stack pointer), kept in a processor register. This register points to memory, since stack is memory after all. When you push values on the stack, its "value" is decremented (stack pointer moves from higher addresses to lower addresses). Each time you enter a function some space is "taken" from the stack (local variables); moreover, on many architectures the call to a subroutine pushes the return value on the stack (and if the processor has no a special register stack pointer, likely a "normal" register is used for the purpose, since stack is useful even where subroutines can be called with other mechanisms), so that the stack is at least diminuished by the size of a pointer (say, 4 or 8 bytes).

In an infinite recursion loop, in the best case only the return value causes the stack to be decremented... until it points to a memory that can't be accessed by the program. And you see the segmentation fault problem.

You may find interesting this page.

Answer 2

AFAIK: The ends of the stack are protected by addresses that aren't accessible to the process. This prevents the stack from growing over allocated data-structures, and is more efficient than checking the stack size explicitly, since you have to check the memory protection anyway.