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

It's essentially the same principle as a buffer overflow; the OS allocates a fixed amount of memory for the stack, and when you run out (stack overflow) you get undefined behavior, which in this context means a SIGSEGV.

The basic idea:

int stack[A_LOT];
int rsp=0;

void call(Func_p fn)
    {
    stack[rsp++] = rip;
    rip = fn;
    }

void retn()
    {
    rip = stack[--rsp];
    }

/*recurse*/
for(;;){call(somefunc);}

eventually rsp moves past the end of the stack and you try to put the next return address in unallocated storage and your program barfs. Obviously real systems are a lot more complicated than that, but that could (and has) take up several large books.