Question

We transit from ring3 to ring0 using ‘int’ or the new ‘syscall/sysenter’ instruction. Does that mean that the page tables and other stuffs that needs to be modified for the kernel is automatically done by the ‘int’ instruction or the interrupt handler for the ‘int 0x80’ will do the required stuff and jump to the respective system call.

Also when returning from a system call, we again need to go to user space. For this we need to know the instruction address in the user space to continue the user application. Where is that address stored. Does the ‘ret’ instruction automatically changes the ring from ring3 to ring0 or where/how this ring changing mechanism takes place?

Then, i read that changing from ring3 to ring0 is not as costly as changing from ring0 to ring3. Why is this so??

Thanks,
Bala

Answer 1

The pagetables aren’t changed when switching to kernel mode – the kernel part of the virtual address space is simply marked as only being accessible in ring0, so it just becomes accessible. The kernel changes pagetables when it changes the current process.

The int 0x80 instruction is served by a trap gate, which supplies the address for the processor to jump to as a CS:EIP pair. The new CS (code segment selector) includes a CPL (current privilege level) of 0, which effects the transition to ring0.

Due to the transition from ring3 to ring0, the processor also picks up new values for SS:ESP from the TSS (task segement selector), and saves the old ones in the TSS. This switches from the user mode stack to the kernel’s stack.

The previous CS:EIP are then pushed onto the kernel stack (this is the return address in user space). All of this is done by the processor due to the int 0x80 instruction itself.

The IRET instruction can be used to return to userspace – it pops the CS:EIP from the kernel stack. Since the CS includes a CPL of 3, the processor is switching back to ring3, which causes it to switch back to the ring3 stack as well.