Managing Threads in OS: User-Space vs. Kernel-Level Benefits

  • Thread starter Thread starter khdani
  • Start date Start date
  • Tags Tags
    Os Threads
AI Thread Summary
User-space threads can lead to issues like blocking the entire process, while kernel-level threads provide better management and protection. The discussion highlights concerns about protecting critical resources such as registers and stacks from user interference. Kernel-level threads are seen as beneficial for maintaining system stability and security. The conversation revolves around the complexities of managing code segments, data segments, and stacks in user-space versus kernel-space. Ultimately, the need for protection from user actions is emphasized, particularly regarding resource management.
khdani
Messages
53
Reaction score
0
Hello,
please help with this question
What of the following is difficult to manage if there are user-space threads and what
will be easier when there are kernel-level threads?
1)code segment
2)data segment
3)stack
4)registers

I don't understand the logic of that question, I think everything would be easier with kernel-threads, cause user-space threads can block the whole process !?
 
Physics news on Phys.org
The idea here is protection of the computer from users. Of course you would want to protect registers, right? A malicious or sloppy user could kill all the other user's processes if that user could control the registers right? How about the stacks? Does the computer need protection from user code? data?

Think about it and give us an answer. We will critique that...
 
i don't follow your idea about protection, i still can kill processes which run on kernel-level threads...
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top