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Editorial Team
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Asked: 2026-05-15T15:30:16+00:00

I know this is more heavy question, but I think its interesting too. It

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I know this is more “heavy” question, but I think its interesting too. It was part of my previous questions about compiler functions, but back than I explained it very badly, and many answered just my first question, so ther it is:

So, if my knowledge is correct, modern Windows systems use paging as a way to switch tasks and secure that each task has propriate place in memory. So, every process gets its own place starting from 0.

When multitasking goes into effect, Kernel has to save all important registers to the task´s stack i believe than save the current stack pointer, change page entry to switch to another proces´s physical adress space, load new process stack pointer, pop saved registers and continue by call to poped instruction pointer adress.

Becouse of this nice feature (paging) every process thinks it has nice flat memory within reach. So, there is no far jumps, far pointers, memory segment or data segment. All is nice and linear.

But, when there is no more segmentation for the process, why does still compilers create variables on the stack, or when global directly in other memory space, than directly in program code?

Let me give an example, I have a C code:int a=10;

which gets translated into (Intel syntax):mov [position of a],#10

But than, you actually ocupy more bytes in RAM than needed. Becouse, first few bytes takes the actuall instruction, and after that instruction is done, there is new byte containing the value 10.

Why, instead of this, when there is no need to switch any segment (thus slowing the process speed) isn´t just a value of 10 coded directly into program like this:

xor eax,eax //just some instruction
10 //the value iserted to the program
call end //just some instruction

Becouse compiler know the exact position of every instruction, when operating with that variable, it would just use it´s adress.

I know, that const variables do this, but they are not really variables, when you cannot change them.

I hope I eplained my question well, but I am still learning English, so forgive my sytactical and even semantical errors.

EDIT:

I have read your answers, and it seems that based on those I can modify my question:

So, someone told here that global variable is actually that piece of values attached directly into program, I mean, when variable is global, is it atached to the end of program, or just created like the local one at the time of execution, but instead of on stack on heap directly?

If the first case – attached to the program itself, why is there even existence of local variables? I know, you will tell me becouse of recursion, but that is not the case. When you call function, you can push any memory space on stack, so there is no program there.

I hope you do understand me, there always is ineficient use of memory, when some value (even 0) is created on stack from some instruction, becouse you need space in program for that instruction and than for the actual var. Like so: push #5 //instruction that says to create local variable with integer 5
 And than this instruction just makes number 5 to be on stack. Please help me, I really want to know why its this way. Thanks.

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1 Answer

  1. Editorial Team

    Consider:

    • local variables may have more than one simultaneous existence if a routine is called recursively (even indirectly in, say, a recursive decent parser) or from more than one thread, and these cases occur in the same memory context
    • marking the program memory non-writable and the stack+heap as non-executable is a small but useful defense against certain classes of attacks (stack smashing…) and is used by some OSs (I don’t know if windows does this, however)

    Your proposal doesn’t allow for either of these cases.

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