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 ....
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