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.
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??
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??
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