How variables address RAM?

Question

I'm pretty new to this, so if the question doesn't make sense, I apologize ahead of time.

int in c# is 4 bytes if I am correct. If I have the statement:

int x;

I would assume this is taking up 4 bytes of memory. If each memory address space is 1 byte then this would take up four address slots? If so, how does x map to the four address locations?

Answer 1

If I have the statement int x; I would assume this is taking up 4 bytes of memory. How does x map to the address of the four bytes?

First off, Mike is correct. C# has been designed specifically so that you do not need to worry about this stuff. Let the memory manager take care of it for you; it does a good job.

Assuming you do want to see how the sausage is made for your own edification: your assumption is not warranted. This statement does not need to cause any memory to be consumed. If it does cause memory to be consumed, the int consumes four bytes of memory.

There are two ways in which the local variable (*) can consume no memory. The first is that it is never used:

void M()
{
    int x;
}

The compiler can be smart enough to know that x is never written to or read from, and it can be legally elided entirely. Obviously it then takes up no memory.

The second way that it can take up no memory is if the jitter chooses to enregister the local. It may assign a machine register specifically to that local variable. The variable then has no address associated with it because obviously registers do not have an address. (**)

Assuming that the local does take up memory, the jitter is responsible for keeping track of the location of that memory.

If the local is a perfectly normal local then the jitter will bump the stack pointer by four bytes, thereby reserving four bytes on the stack. It will then associate those four bytes with the local.

If the local is a closed-over outer local of an anonymous function, a local of an iterator block, or a local of an async method then the C# compiler will generate the local as a field of a class; the jitter asks the garbage collector to allocate the class instance and the jitter associates the local with a particular offset from the beginning of the memory buffer associated with that instance by the garbage collector.

All of this is implementation detail subject to change at any time; do not rely upon it.

(*) We know it is a local variable because you said it was a statement. A field declaration is not a statement.

(**) If unsafe code takes the address of a local, obviously it cannot be enregistered.

Answer 2

If the int x; is declared within a function, then the variable will be allocated on the stack, rather than the heap or global memory. The address of x in the compiler's symbol table will refer to the first byte of the four-byte integer. However since it is on the stack, the remembered address will be that of the offset on the stack, rather than a physical address. The variable will then be referenced via a instruction using that offset from the current stack pointer.

Assuming a 32-bit run-time, the offset on the stack will be aligned so the address is a multiple of 4 bytes, i.e. the offset will end in either 0, 4, 8 or 0x0c.

Furthermore because the 80x86 family is little-endian, the first byte of the integer will be the least significant, and the fourth byte will be the most significant, e.g. the decimal value 1,000,000 would be stored as the four bytes 0x40 0x42 0x0f 0x00.

Answer 3

There's a lot (and I mean a LOT) that can be said about this. Various topics you're hitting on are things like the stack, the symbol table, memory management, the memory hierarchy, ... I could go on.

BUT, since you're new, I'll try to give an easier answer:

When you create a variable in a program (such as an int), you are telling the compiler to reserve a space in memory for that data. An int is 4 bytes, so 4 consecutive bytes are reserved. The memory location you were referring to only points to the beginning. It is known afterwards that the length is 4 bytes.

Now that memory location (in the case you provided) is not really saved in the same way that a variable would be. Every time there is a command that needs x, the command is instead replaced with a command that explicitly grabs that memory location. In other words, the address is saved in the "code" section of your program, not the "data" section.

This is just a really, REALLY high overview. Hopefully it helps.

Answer 4

You really should not need to worry about these things, since there is no way in C# that you could write code that would make use of this information.

But if you must know, at the machine-code level when we instruct the CPU to access the contents of x, it will be referred to using the address of the first one of those four bytes. The machine instruction that will do this will also contain information about how many bytes to be accessed, in this case four.