x86 code execution after call/return/jmp etc?

Question

I hope this question isn\'t to stupid cause it may seem obvious. As I\'m doing a little research on Buffer overflows I stumble over a simple question:

After going to a n

Answer 1

The call instruction saves (onto the stack) the address to the instruction after it onto the stack. After that, it simply jumps. It doesn't explicitly tell the cpu to look for a return instruction, since that will be handled by popping (from the stack) the return address that call saved in the first place. This allows for multiple calls and returns, or to put it simply, nested calls.