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http://www.cs.umass.edu/~verts/cs32/endian.htmlAn Essay on Endian Order
Copyright (C) Dr. William T. Verts, April 19, 1996
Depending on which computing system you use, you will have to consider the byte order in
which multibyte numbers are stored, particularly when you are writing those numbers to a file.
The two orders are called "Little Endian" and "Big Endian".
The Basics
"Little Endian" means that the low-order byte of the number is stored in memory at the lowest
address, and the high-order byte at the highest address. (The little end comes first.) For
example, a 4 byte LongInt
Byte3 Byte2 Byte1 Byte0
will be arranged in memory as follows:
Base Address+0 Byte0
Base Address+1 Byte1
Base Address+2 Byte2
Base Address+3 Byte3
Intel processors (those used in PC's) use "Little Endian" byte order.
"Big Endian" means that the high-order byte of the number is stored in memory at the lowest
address, and the low-order byte at the highest address. (The big end comes first.) Our
LongInt, would then be stored as:
Base Address+0 Byte3
Base Address+1 Byte2
Base Address+2 Byte1
Base Address+3 Byte0
Motorola processors (those used in Mac's) use "Big Endian" byte order.
Which is Better?
You may see a lot of discussion about the relative merits of the two formats, mostly religious
arguments based on the relative merits of the PC versus the Mac. Both formats have their
advantages and disadvantages.
In "Little Endian" form, assembly language instructions for picking up a 1, 2, 4, or longer byte
number proceed in exactly the same way for all formats: first pick up the lowest order byte at
offset 0. Also, because of the 1:1 relationship between address offset and byte number (offset
0 is byte 0), multiple precision math routines are correspondingly easy to write.
In "Big Endian" form, by having the high-order byte come first, you can always test whether
the number is positive or negative by looking at the byte at offset zero. You don't have to
know how long the number is, nor do you have to skip over any bytes to find the byte
containing the sign information. The numbers are also stored in the order in which they are
printed out, so binary to decimal routines are particularly efficient.
What does that Mean for Us?
What endian order means is that any time numbers are written to a file, you have to know how
the file is supposed to be constructed. If you write out a graphics file (such as a .BMP file) on
a machine with "Big Endian" integers, you must first reverse the byte order, or a "standard"
program to read your file won't work.
The Windows .BMP format, since it was developed on a "Little Endian" architecture, insists on
the "Little Endian" format. You must write your Save_BMP code this way, regardless of the
platform you are using.
Common file formats and their endian order are as follows:
Adobe Photoshop -- Big Endian
BMP (Windows and OS/2 Bitmaps) -- Little Endian
DXF (AutoCad) -- Variable
GIF -- Little Endian
IMG (GEM Raster) -- Big Endian
JPEG -- Big Endian
FLI (Autodesk Animator) -- Little Endian
MacPaint -- Big Endian
PCX (PC Paintbrush) -- Little Endian
PostScript -- Not Applicable (text!)
POV (Persistence of Vision ray-tracer) -- Not Applicable (text!)
QTM (Quicktime Movies) -- Little Endian (on a Mac!)
Microsoft RIFF (.WAV & .AVI) -- Both
Microsoft RTF (Rich Text Format) -- Little Endian
SGI (Silicon Graphics) -- Big Endian
Sun Raster -- Big Endian
TGA (Targa) -- Little Endian
TIFF -- Both, Endian identifier encoded into file
WPG (WordPerfect Graphics Metafile) -- Big Endian (on a PC!)
XWD (X Window Dump) -- Both, Endian identifier encoded into file
Correcting for the Non-Native Order
It is pretty easy to reverse a multibyte integer if you find you need the other format. A single
function can be used to switch from one to the other, in either direction. A simple and not very
efficient version might look as follows:
Function Reverse (N:LongInt) : LongInt ;
Var B0, B1, B2, B3 : Byte ;
Begin
B0 := N Mod 256 ;
N := N Div 256 ;
B1 := N Mod 256 ;
N := N Div 256 ;
B2 := N Mod 256 ;
N := N Div 256 ;
B3 := N Mod 256 ;
Reverse := (((B0 * 256 + B1) * 256 + B2) * 256 + B3) ;
End ;
A more efficient version that depends on the presence of hexadecimal numbers, bit masking
operators AND, OR, and NOT, and shift operators SHL and SHR might look as follows:
Function Reverse (N:LongInt) : LongInt ;
Var B0, B1, B2, B3 : Byte ;
Begin
B0 := (N AND $000000FF) SHR 0 ;
B1 := (N AND $0000FF00) SHR 8 ;
B2 := (N AND $00FF0000) SHR 16 ;
B3 := (N AND $FF000000) SHR 24 ;
Reverse := (B0 SHL 24) OR (B1 SHL 16) OR (B2 SHL 8) OR (B3 SHL 0) ;
End ;
There are certainly more efficient methods, some of which are quite machine and platform
dependent. Use what works best.
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