Thread: Not doing free()

  1. 12-13-2018 #1
    RyanC
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    Not doing free()

    Hi ;
    In abbreviation, the concept is deleting third entry of linked lists;
    if I have a linked lists (it's one direction) and lets say I have jumped from the second entry of linked lists to the fourth entry ; meaning with this I have updated the pointer of Node * next of second entry to point on the fourth entry, what happen to third entry, I'm not having access to it at all, this means implicitly I have deleted it, my question will PC free it by itself? if not doing free to it as, will the data of entry three be saved forever in the memory?! it's not logically.


    thanks !
  2. 12-13-2018 #2
    RyanC
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    I know it's not good idea for deleting entry in that approach, just wanted to figure out what's going on in that case (for those whom think it's stupid question, Yup may I'm more than stupid of what you guess, but I'm here to learn)
    Last edited by RyanC; 12-13-2018 at 07:38 AM.
  3. 12-13-2018 #3
    Salem
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    > my question will PC free it by itself?
    No, you have to do this yourself.

    > if not doing free to it as, will the data of entry three be saved forever in the memory?
    Well it will remain in memory until the program itself exits (or is killed by the OS).
    Or if you have no OS, it will stay in memory until your machine does a hard reboot - perhaps because you ran out of memory.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
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