# Virtual Memory

## A bit of information theory

A bit is a binary digit; it is also a unit of information. If you have one bit, you can specify one of two possibilities, usually written 0 and 1. If you have two bits, there are 4 possible combinations, 00, 01, 10, and 11. In general, if you have $$b$$ bits, you can indicate one of $$2^b$$ values. A byte is 8 bits, so it can hold one of 256 values.

Going in the other direction, suppose you want to store a letter of the alphabet. There are 26 letters, so how many bits do you need? With 4 bits, you can specify one of 16 values, so that’s not enough. With 5 bits, you can specify up to 32 values, so that’s enough for all the letters, with a few values left over.

In general, if you want to specify one of $$N$$ values, you should choose the smallest value of $$b$$ so that $$2^b \ge N$$. Taking the log base 2 of both sides yields $$b \ge log_2 N$$.

Suppose I flip a coin and tell you the outcome. I have given you one bit of information. If I roll a six-sided die and tell you the outcome, I have given you $$log_2 6$$ bits of information. And in general, if the probability of the outcome is 1 in $$N$$, then the outcome contains $$log_2 N$$ bits of information.

Equivalently, if the probability of the outcome is $$p$$, then the information content is $$-log_2 p$$. This quantity is called the self-information of the outcome. It measures how surprising the outcome is, which is why it is also called surprisal. If your horse has only one chance in 16 of winning, and he wins, you get 4 bits of information (along with the payout). But if the favorite wins 75% of the time, the news of the win contains only 0.42 bits.

Intuitively, unexpected news carries a lot of information; conversely, if there is something you were already confident of, confirming it contributes only a small amount of information.

For several topics in this book, we will need to be comfortable converting back and forth between the number of bits, $$b$$, and the number of values they can encode, $$N = 2^b$$.

## Memory and storage

While a process is running, most of its data is held in main memory, which is usually some kind of random access memory (RAM). On most current computers, main memory is volatile, which means that when the computer shuts down, the contents of main memory are lost. A typical desktop computer has 2–8 GiB of memory. GiB stands for “gibibyte,” which is $$2^{30}$$ bytes.

If the process reads and writes files, those files are usually stored on a hard disk drive (HDD) or solid state drive (SSD). These storage devices are non-volatile, so they are used for long-term storage. Currently a typical desktop computer has a HDD with a capacity of 500 GB to 2 TB. GB stands for “gigabyte,” which is $$10^9$$ bytes. TB stands for “terabyte,” which is $$10^{12}$$ bytes.

You might have noticed that I used the binary unit GiB for the size of main memory and the decimal units GB and TB for the size of the HDD. For historical and technical reasons, memory is measured in binary units, and disk drives are measured in decimal units. In this book I will be careful to distinguish binary and decimal units, but you should be aware that the word “gigabyte” and the abbreviation GB are often used ambiguously.

In casual use, the term “memory” is sometimes used for HDDs and SSDs as well as RAM, but the properties of these devices are very different, so we will need to distinguish them. I will use storage to refer to HDDs and SSDs.

Each byte in main memory is specified by an integer physical address. The set of valid physical addresses is called the physical address space. It usually runs from 0 to $$N-1$$, where $$N$$ is the size of main memory. On a system with 1 GiB of physical memory, the highest valid address is $$2^{30}-1$$, which is 1,073,741,823 in decimal, or 0x3fff ffff in hexadecimal (the prefix 0x indicates a hexadecimal number).

However, most operating systems provide virtual memory, which means that programs never deal with physical addresses, and don’t have to know how much physical memory is available.

Instead, programs work with virtual addresses, which are numbered from 0 to $$M-1$$, where $$M$$ is the number of valid virtual addresses. The size of the virtual address space is determined by the operating system and the hardware it runs on.

You have probably heard people talk about 32-bit and 64-bit systems. These terms indicate the size of the registers, which is usually also the size of a virtual address. On a 32-bit system, virtual addresses are 32 bits, which means that the virtual address space runs from 0 to 0xffff ffff. The size of this address space is $$2^{32}$$ bytes, or 4 GiB.

On a 64-bit system, the size of the virtual address space is $$2^{64}$$ bytes, or $$2^4 \cdot 1024^6$$ bytes. That’s 16 exbibytes, which is about a billion times bigger than current physical memories. It might seem strange that a virtual address space can be so much bigger than physical memory, but we will see soon how that works.

When a program reads and writes values in memory, it generates virtual addresses. The hardware, with help from the operating system, translates to physical addresses before accessing main memory. This translation is done on a per-process basis, so even if two processes generate the same virtual address, they would map to different locations in physical memory.

Thus, virtual memory is one important way the operating system isolates processes from each other. In general, a process cannot access data belonging to another process, because there is no virtual address it can generate that maps to physical memory allocated to another process.

## Memory segments

The data of a running process is organized into five segments:

• The code segment contains the program text; that is, the machine language instructions that make up the program.

• The static segment contains immutable values, like string literals. For example, if your program contains the string "Hello, World", those characters will be stored in the static segment.

• The global segment contains global variables and local variables that are declared static.

• The heap segment contains chunks of memory allocated at run time, most often by calling the C library function malloc.

• The stack segment contains the call stack, which is a sequence of stack frames. Each time a function is called, a stack frame is allocated to contain the parameters and local variables of the function. When the function completes, its stack frame is removed from the stack.

The arrangement of these segments is determined partly by the compiler and partly by the operating system. The details vary from one system to another, but in the most common arrangement:

• The text segment is near the “bottom” of memory, that is, at addresses near 0.

• The static segment is often just above the text segment, that is, at higher addresses.

• The global segment is often just above the static segment.

• The heap is often above the global segment. As it expands, it grows up toward larger addresses.

• The stack is near the top of memory; that is, near the highest addresses in the virtual address space. As the stack expands, it grows down toward smaller addresses.

To determine the layout of these segments on your system, try running this program, which is in aspace.c in the repository for this book.

#include <stdio.h>
#include <stdlib.h>

int global;

int main ()
{
int local = 5;
void *p = malloc(128);
char *s = "Hello, World";

printf ("Address of main is %p\n", main);
printf ("Address of global is %p\n", &global);
printf ("Address of local is %p\n", &local);
printf ("p points to %p\n", p);
printf ("s points to %p\n", s);
}


main is the name of a function; when it is used as a variable, it refers to the address of the first machine language instruction in main, which we expect to be in the text segment.

global is a global variable, so we expect it to be in the global segment. local is a local variable, so we expect it to be on the stack.

s refers to a “string literal", which is a string that appears as part of the program (as opposed to a string that is read from a file, input by a user, etc.). We expect the location of the string to be in the static segment (as opposed to the pointer, s, which is a local variable).

p contains an address returned by malloc, which allocates space in the heap. “malloc” stands for “memory allocate.”

The format sequence %p tells printf to format each address as a “pointer”, so it displays the results in hexadecimal.

When I run this program, the output looks like this (I added spaces to make it easier to read):

Address of main is   0x      40057d
Address of global is 0x      60104c
p points to          0x     16c3010
s points to          0x      4006a4


As expected, the address of main is the lowest, followed by the location of the string literal. The location of global is next, then the address p points to. The address of local is much bigger.

The largest address has 12 hexadecimal digits. Each hex digit corresponds to 4 bits, so it is a 48-bit address. That suggests that the usable part of the virtual address space is $$2^{48}$$ bytes.

As an exercise, run this program on your computer and compare your results to mine. Add a second call to malloc and check whether the heap on your system grows up (toward larger addresses). Add a function that prints the address of a local variable, and check whether the stack grows down.

## Static local variables

Local variables on the stack are sometimes called automatic, because they are allocated automatically when a function is called, and freed automatically when the function returns.

In C there is another kind of local variable, called static, which is allocated in the global segment. It is initialized when the program starts and keeps its value from one function call to the next.

For example, the following function keeps track of how many times it has been called.

int times_called()
{
static int counter = 0;
counter++;
return counter;
}


The keyword static indicates that counter is a static local variable. The initialization happens only once, when the program starts.

If you add this function to aspace.c you can confirm that counter is allocated in the global segment along with global variables, not in the stack.

How does a virtual address (VA) get translated to a physical address (PA)? The basic mechanism is simple, but a simple implementation would be too slow and take too much space. So actual implementations are a bit more complicated.

Most processors provide a memory management unit (MMU) that sits between the CPU and main memory. The MMU performs fast translation between VAs and PAs.

1. When a program reads or writes a variable, the CPU generates a VA.

2. The MMU splits the VA into two parts, called the page number and the offset. A “page” is a chunk of memory; the size of a page depends on the operating system and the hardware, but common sizes are 1–4 KiB.

3. The MMU looks up the page number in the translation lookaside buffer (TLB) and gets the corresponding physical page number. Then it combines the physical page number with the offset to produce a PA.

4. The PA is passed to main memory, which reads or writes the given location.

The TLB contains cached copies of data from the page table (which is stored in kernel memory). The page table contains the mapping from virtual page numbers to physical page numbers. Since each process has its own page table, the TLB has to make sure it only uses entries from the page table of the process that’s running.

Figure 4.1 shows a diagram of this process. To see how it all works, suppose that the VA is 32 bits and the physical memory is 1 GiB, divided into 1 KiB pages.

• Since 1 GiB is $$2^{30}$$ bytes and 1 KiB is $$2^{10}$$ bytes, there are $$2^{20}$$ physical pages, sometimes called “frames.”

• The size of the virtual address space is $$2^{32}$$ B and the size of a page is $$2^{10}$$ B, so there are $$2^{22}$$ virtual pages.

• The size of the offset is determined by the page size. In this example the page size is $$2^{10}$$ B, so it takes 10 bits to specify a byte on a page.

• If a VA is 32 bits and the offset is 10 bits, the remaining 22 bits make up the virtual page number.

• Since there are $$2^{20}$$ physical pages, each physical page number is 20 bits. Adding in the 10 bit offset, the resulting PAs are 30 bits.

So far this all seems feasible. But let’s think about how big a page table might have to be. The simplest implementation of a page table is an array with one entry for each virtual page. Each entry would contain a physical page number, which is 20 bits in this example, plus some additional information about each frame. So we expect 3–4 bytes per entry. But with $$2^{22}$$ virtual pages, the page table would require $$2^{24}$$ bytes, or 16 MiB.

And since we need a page table for each process, a system running 256 processes would need $$2^{32}$$ bytes, or 4 GiB, just for page tables! And that’s just with 32-bit virtual addresses. With 48- or 64-bit VAs, the numbers are ridiculous.

Fortunately, we don’t actually need that much space, because most processes don’t use even a small fraction of their virtual address space. And if a process doesn’t use a virtual page, we don’t need an entry in the page table for it.

Another way to say the same thing is that page tables are “sparse”, which implies that the simple implementation, an array of page table entries, is a bad idea. Fortunately, there are several good implementations for sparse arrays.

One option is a multilevel page table, which is what many operating systems, including Linux, use. Another option is an associative table, where each entry includes both the virtual page number and the physical page number. Searching an associative table can be slow in software, but in hardware we can search the entire table in parallel, so associative arrays are often used to represent the page table entries in the TLB.

You can read more about these implementations at http://en.wikipedia.org/wiki/Page_table; you might find the details interesting. But the fundamental idea is that page tables are sparse, so we have to choose a good implementation for sparse arrays.

I mentioned earlier that the operating system can interrupt a running process, save its state, and then run another process. This mechanism is called a context switch. Since each process has its own page table, the operating system has to work with the MMU to make sure each process gets the right page table. In older machines, the page table information in the MMU had to be replaced during every context switch, which was expensive. In newer systems, each page table entry in the MMU includes the process ID, so page tables from multiple processes can be in the MMU at the same time.