On Jun 12, 1:22 pm, amit <amitjaspal...@gmail.com> wrote: > OS doubt: > > I have read many times that say a 24 KB process enters the Main Memory > selected by the Long Term Scheduler. > But I don't understand what it exactly means. > As far as I know Process consists of ( Code + Data(Static) + > Stack(Local Data) + Heap) > > So doubt1: Is this 24 KB the size of this whole process or just the > size of the code segment. >
I don't know buddy...... > doubt2: Now lets say this process starts getting executed by the > CPU ,Suppose the main() contains > main(){ > int x; > int y; > x=10; > ....... > } > So x,y will be allocated the memory in the Stack. > But when x=10 is encountered , how will the CPU know the > address of > x. In short how is x accessed?? buddy....cpu will never encounter what u have written(means....x). Compiler converts this to machine languages, cpu can easily access using relative address. > > doubt 3: If x and y are just address of a memory location in the > stack , can their logical address be determined by the compiler or it > will be generated by the CPU?? compiler can generate logical address .... -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.