Word alignment

32-bit word: 4 bytes

Suppose we want to store the word 0123ABCD_hex

Start at address 1000

big-endian

data

address

1000

1001

1002

1003

little-endian

data

address

1000

1001

1002

1003

We say the word is stored at address 1000, meaning it's stored beginning at address 1000

Could we store these same 4 bytes starting at address 1001, for example?

Yes, but the hardware for accessing the data in memory is simpler if the data is aligned

A word begins on a word boundary (address divisible by 4)

What's a good way to tell if an address is a word boundary?

If its address in binary ends in 00

A halfword is aligned on an address divisible by 2

What is the effect on high-level language?

Consider the structure

struct Foo {

char x ; // 1 byte

int y ; // 4 bytes

char z ; // 1 byte

int w ; // 4 bytes

} ;

What is the size of a variable of type struct Foo?

1 + 4 + 1 + 4 = 10 bytes

Not necessarily! If the ints are aligned on word boundaries, there must be 3 bytes between

the chars and the ints.

This means that the size of the struct is 16 bytes, if alignment is required.

The extra bytes are called padding or holes.

This is the main reason struct variables can't be directly compared in C,

but they can be assigned directly.

The efficient way to compare would be to compare all bits in each struct, but

the pad bytes, if any, are undefined, and may be any value.

Why is assignment OK?

Try sizeof operator on this struct

Summary

address

binary address

divisible by

ends in

byte

anything

halfword

word

doubleword

000