Virtual Address Space - linux

Question

I have started to learn about Virtual Address Space (VAS) and I have few questions:
How much of VAS is created for each process depending on the architecture (32-bit and 64-bit)?
Is VAS for each process created on hard disk? If so, what happens if there is not enough space?
What is the difference between VAS and Virtual Memory (VM)?

Answer 1

Virtual address versus physical address
During the execution of your program, the variables (integers, arrays, strings, etc.) are stored somewhere in the main memory of your computer (RAM). Some programming languages (like C or C++) allow you to obtain the memory address at which a given variable is stored (with the & operator), and to manipulate that address (add to it, subtract from it, print it, etc.).
Here is a C program that prints the memory address of a variable:
#include <stdio.h>
int main(void) {
int variable = 1234;
void *address = &variable;
printf("Memory address of variable: %p\n", address);
return 0;
}
Output:
Memory address of variable: 0x7ffc9e9662a4
Now, if you compile and execute this program on a typical desktop computer, with a typical operating system (like GNU/Linux or Windows), the memory address that is printed by this program is not the hardware address at which the data 1234 is actually located in the memory chip. This may be surprising, but there is a level of indirection between the addresses used by your program and the hardware addresses.
Virtual address space on 64-bit computers
On a 64-bit computer, a memory address manipulated by your program is an integer between 0 and 18446744073709551615 inclusive. Such an address is called a virtual memory address. The range of those addresses is called the virtual address space of the process. You can ask the operating system to map a range of virtual memory addresses to the physical memory of your computer, so that when you try to read or write bytes at thoses addresses, your program doesn't crash for accessing unmapped virtual memory addresses.
Typically, on x86-64 computers, only 248 virtual memory addresses can be successfully mapped to physical memory, because 256 TiB of usable virtual address space is considered sufficient. In the future, processor manufacturers may raise or remove this limit if there is a need for it.
Virtual address space on 32-bit computers
On 32-bit computers, there are 232 virtual memory addresses. On those computers, a memory address manipulated by your program is an integer between 0 and 4294967295 inclusive.
On x86 32-bit computers, there is usually no restriction on the range of virtual memory addresses that can be mapped to physical memory addresses.
Mapping a range of virtual memory addresses
On GNU/Linux, you can request a mapping by calling the function mmap(). On Windows, you can request a mapping by calling the function VirtualAlloc(). Those functions take the size of the mapping as argument, and return the first virtual address that is now backed by actual physical memory. Those functions can fail to create a new mapping if the physical memory is already completely used by other processes. And again, if you try to access (read or write) the content of a virtual memory address that is outside an area mapped by mmap() or VirtualAlloc(), the operating system will terminate your program (by sending a segmentation fault signal).
On GNU/Linux, a process can examine the mappings created in its virtual address space just by reading the file /proc/self/maps. You can learn a lot by reading the output of the command cat /proc/self/maps.
Hard disk drive
On a typical computer, the main memory is a semiconductor memory, and the hard disk drive is only a secondary storage device.
On a typical operating system, a range of virtual memory addresses can only be mapped to the main memory (which is usually a semiconductor memory device). Such a range cannot be directly mapped to a secondary storage device (usually a hard disk drive) without using the main memory as intermediary.

Answer 2

On an n-bit machine, the VAS is 2n bytes large. So, on a 32-bit machine, the VAS 232 = 4 GiB large.
Virtual memory is not created on disk. In fact, the existence of a disk is not needed for implementing virtual memory. Most implementations of virtual memory are paged. So, when a 4 GiB VAS is created, only the pages that are needed are mapped into that VAS. For example, suppose a process only uses 16 pages of memory on a 32-bit system with 4k-sized pages. Despite having a 4 GiB VAS, only 16 * 4k = 216 bytes of memory are mapped into the VAS. The rest of the memory is unmapped. If the CPU tries to access this unmapped memory, a segmentation fault will occur. If a process wants to map memory at this address, then (in a POSIX-complaint OS) it can request the mapping from the OS using mmap(2). This will make a lot more sense once you learn about page tables.
Virtual memory is a concept. A virtual address space is an entity that stems from the concept of virtual memory. These terms go hand in hand, but refer to different things.
I will list a couple of caveats.
Caveat 1.1
I am not not aware of any 64-bit processor that truly supports a 64-bit VAS. The addresses themselves are 64 bits wide, but a certain number of upper bits are ignored. AMD's first implementation of x86_64 only supported 48-bit addresses. The upper 16 bits of an address were effectively ignored. In such a system, the addresses are 64 bits wide, but the real size of the VAS is limited to 248 bytes. Subsequent architectures supported 56-bit addresses.
Caveat 1.2
If a processor supports PAE, then the VAS on an n-bit machine may be larger than 2n bytes. This is how 32-bit processors can support VASs larger than 4 GiB.
Caveat 2.1
Not really a caveat, but this is related to your question. You asked what happens when there isn't enough space on disk to create a VAS. As I mentioned in the main answer, the VAS is not created on disk. However, any computer only has a finite amount of physical memory. What happens when a process requests a page be mapped, but there is no physical memory available? There are there several ways to handle this:
Swapping is done by temporarily moving a page that is mapped in virtual memory to disk. The entire contents of the page are copied to disk. Then, the process that requested the page has the physical page mapped into their memory. Eventually, the old page may be requested. If this occurs, then the OS copies the page from disk and remaps it into the corresponding VAS. This is what Linux and most modern operating systems do.
The process is simply told there is no memory available, for example, through an error number like ENOMEM.
The process is blocked until memory is available. I haven't seen this in practice.
Swapping implies the use of a disk, but virtual memory does not imply the use of swapping, hence a disk is not necessary for virtual memory.

Answer 3

Virtual Address Space - wikipedia
When a new application on a 32-bit OS is executed, the process has a 4 GiB VAS: each one of the memory addresses (from 0 to 232 − 1) in that space can have a single byte as a value. Initially, none of them have values.
For n-bit OS, these n-address lines allow address space upto 2n addresses, i.e., 0 to 2n - 1. This would mean 16 EiB for 64-bit OS. (Though in actual implementations, less space is used as this much space is unnecessary.)
CPU Cache - wikipedia
Most general purpose CPUs implement some form of virtual memory. To summarize, either each program running on the machine sees its own simplified address space, which contains code and data for that program only, or all programs run in a common virtual address space. A program executes by calculating, comparing, reading and writing to addresses of its virtual address space, rather than addresses of physical address space, making programs simpler and thus easier to write.
For example, in C++, program memory is divided in stack, heap, data, code. I'm not sure if analogy is correct (may be), but it somewhat presents an insight if you're aware.
Virtual memory - wikipedia
In computing, virtual memory is a memory management technique that provides an "idealized abstraction of the storage resources that are actually available on a given machine"3 which "creates the illusion to users of a very large (main) memory".[4]
The computer's operating system, using a combination of hardware and software, maps memory addresses used by a program, called virtual addresses, into physical addresses in computer memory. Main storage, as seen by a process or task, appears as a contiguous address space or collection of contiguous segments. The operating system manages virtual address spaces and the assignment of real memory to virtual memory.
Address translation hardware in the CPU, often referred to as a memory management unit (MMU), automatically translates virtual addresses to physical addresses. Software within the operating system may extend these capabilities to provide a virtual address space that can exceed the capacity of real memory and thus reference more memory than is physically present in the computer.
If you know about computer architecture (which I'm sure you do from the question), it'd be clarified by now.
Still, for anyone in general, I'm giving a bit of explanation.
Assume addresses as pointers in C++. If you don't know C++, closest analogy would be array/list indices in any language. Now the addresses point to the memory locations, just like pointers point to the variable. The actual data is stored in the variable. To get the variable data using pointer/index, you provide address location from where the data is to be extracted. Now in physical memory, there won't be a thing like a variable. There is memory and it's location address through which it is accessed.
The real memory is physical memory, which is the hard disks. It is accessed with physical addresses, which would be unique for each byte.
Accessing physical memory directly with physical addresses would be cumbersome. Thus the addresses are simplified by the OS to virtual addresses. These addresses may or may not be unique (these aren't physical addresses, remember). Thus, multiple virtual addresses may point to same location.
The virtual memory is not actually existent, rather it's just a concept of physical memory simplified using virtual addresses to give the user an illusionous space say where next memory location is stored at next address (virtual address to be precise).
Since multiple virtual addresses can be mapped, by using MMU, to same physical address, and thus to point to same phyical memory location, the virtual memory size can be made to exceed the physical memory size (virtually). But effectively, the memory size would still be same as physical.
Thus, to access a memory data, Virtual Addresses are specified by user/program to OS, which are converted to Physical Addresses by memory management unit (mmu) and then applied to the address lines of the computer architecture (electronics spotted!!), which yields the data at the corresponding physical location. And this concept is called Virtual Memory.
-Himanshu