Memory, big endian and sizes

Registers give the processor a small amount of working space. Memory provides much more space for the program's instructions and data. The instructions and data are stored in memory as bytes.

One address, one byte

Memory is divided into small locations. Each location holds one byte, and each has its own numeric address.

Neighbouring locations have neighbouring addresses. If one byte is at address 2000, the next byte is at address 2001, then 2002, and so on. The address identifies the location; the byte stored there is its contents.

This first diagram uses ordinary decimal numbers for both the addresses and their contents:

address	byte stored there
2000	25
2001	7
2002	104
2003	3

For example, 2001 is an address in this diagram, while 7 is the data at that address. Because each address identifies one byte, M68K memory is called byte-addressed.

When the processor reads from or writes to memory, that operation is called a memory access.

Bits, bytes, words and longs

A bit is a single binary digit: either 0 or 1. Bits are grouped into the sizes used here:

name	size in bytes	size in bits
byte	1	8
word	2	16
long	4	32

A word therefore occupies two consecutive byte locations. A long occupies four consecutive byte locations. The locations still hold individual bytes; a word or long is formed by treating several of those neighbouring bytes as one value.

Reading hexadecimal

Memory addresses and bytes are commonly written in hexadecimal, or hex. A dollar sign marks a hexadecimal number in M68K material, so $2000 is an address written in hex.

Hex uses sixteen digits instead of ten:

0 1 2 3 4 5 6 7 8 9 A B C D E F

After 9 comes A, and after F the next position increases. For example, the address after $2009 is $200A, and the address after $200F is $2010.

Each hex digit represents four bits, so a pair of hex digits can show one byte. The examples below are deliberately written with one, two and four pairs:

example	pairs written	bytes shown
$12	1	1
$1234	2	2
$12345678	4	4

Read a memory diagram by grouping hex digits into pairs, one pair per byte. Leading zero pairs may also be written: $0012 is the same number as $12, displayed with two pairs instead of one. The number of written pairs describes this display; a memory access still chooses whether it uses one, two or four bytes.

A long in four locations

Suppose the long value $12345678 begins at address $2000. It is split into four bytes: 12, 34, 56 and 78. Those bytes occupy four consecutive addresses:

address	byte
$2000	12
$2001	34
$2002	56
$2003	78

The lowest address is $2000; the addresses increase by one as you move down the table.

The M68K is big endian. For a value that occupies several bytes, it stores the leftmost pair of hex digits at the lowest address, followed by the remaining pairs in the same order. The leftmost pair is also called the most significant byte because it contributes the largest part of the value.

From the memory above:

• a one-byte read beginning at $2000 gets $12;
• a two-byte word read beginning at $2000 gets $1234;
• a two-byte word read beginning at $2002 gets $5678;
• a four-byte long read beginning at $2000 gets $12345678.

Memory itself only stores the four bytes. It does not attach a note saying that they are one long. The size of a particular access decides whether the processor uses one byte, two adjacent bytes or four adjacent bytes.

For example, the same bytes at $2000 through $2003 can be viewed as four separate bytes, as the two words $1234 and $5678, or as the long $12345678. Big-endian order tells the processor how to combine the bytes for either word or long view.

Even addresses for words and longs

On the 68000 used in this course, a byte access can begin at any address. A word or long access must begin at an even address.

In hex, an address is even when its last digit is 0, 2, 4, 6, 8, A, C or E. It is odd when its last digit is 1, 3, 5, 7, 9, B, D or F.

Here is a word access beginning at the even address $2002:

address	byte	part of the word?
$2001	34
$2002	56	yes
$2003	78	yes

It uses the bytes at $2002 and $2003, so the big-endian word is $5678. This access is allowed because its starting address, $2002, is even.

A word beginning at $2001 would use the bytes at $2001 and $2002:

address	byte	part of the word?
$2001	34	yes
$2002	56	yes
$2003	78

Those two bytes could be combined as $3456, but the 68000 used in this course does not allow the word access because $2001 is odd. The same starting-address rule applies to a long. Only the starting address must be even; the following bytes naturally include both odd and even addresses.

Check your understanding

Use this memory for questions 1–3:

address	byte
$3000	A1
$3001	B2
$3002	C3
$3003	D4

1. What value does a one-byte read beginning at $3001 get?
2. What value does a word read beginning at $3002 get?
3. What value does a long read beginning at $3000 get?
4. How many bytes and bits are in a word? How many are in a long?
5. Which of these starting points are allowed on the 68000 used in this course: a byte at $3001, a word at $3001, a word at $3002, and a long at $3002?

Show answers

1. $B2.
2. $C3D4.
3. $A1B2C3D4.
4. A word is 2 bytes or 16 bits. A long is 4 bytes or 32 bits.
5. The byte at $3001, the word at $3002 and the long at $3002 are allowed. The word at $3001 is not allowed because it begins at an odd address.