Path: Home =>
AVR-Overview =>
Binary calculations => Multiplication
Binary multiplication in AVR Assembler
In order to multiply two 8-bit-binaries we remind ourselves, how this is done
with decimal numbers:
1234 * 567 = ?
------------------------
1234 * 7 = 8638
+ 12340 * 6 = 74040
+ 123400 * 5 = 617000
------------------------
1234 * 567 = 699678
========================
In single steps decimal:
- We multiply the first number by the lowest significant digit of the second
number and add this to the result.
- We multiply the first number by 10 and then by the next higher digit of the
second number and add to the result.
- We multiply the first number by 100, then with the third-highest digit,
and add this to the result.
Now in binary. Multiplication with the single digits is not necessary, because there
are only the digits 1 (add the number) and 0 (don't add the number). Multiplication
by 10 in decimal goes to multiplication by 2 in binary mode. Multiplication
by 2 is done easily, either by adding the number with itself, or by shifting all
bits one position left and writing a 0 to the void position on the right. You see
that binary math is very much easier than decimal. Why didn't mankind use this from
the beginning?
The program for multiplication in AVR assembler language is demonstrated
in HTML following this link, the source code
is available at this link. This source code can
be assembled, the resulting object-file can be opened in the simulator studio
provided as free software by ATMEL. These are
the positions and registers used in the program:
| . | rmh = R1 = 0x00 | rm1 = R0 = 0xAA |
|---|
| Z1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| * | . | rm2 = R2 = 0x55 |
|---|
| Z2 | | 0 | 1 | 0 | 1 | 0 |
1 | 0 | 1 |
| = | reh = R4 = 0x00 | rel = R3 = 0x00 |
|---|
| Erg | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
For understanding the multiplication operation, it is necessary to understand the
binary rotation commands ROL and ROR. These instructions shift all bits of a
register one position left (ROL) resp. right (ROR). The void position in the register
is filled with the content of the carry bit of the status register, the bit that
rolls out of the register is shifted to the carry bit. This operation is demonstrated
using 0xAA as an example for ROL and 0x55 as an example for ROR:
 |
 |
The following screenshots show the multiplication program in the simulator.
- MULT8E1.gif: The object-code has been opened, the cursor
is placed on the first executable instruction. F11 does single steps.
- MULT8E2.gif: The registers R0 and R2 are set to 0xAA and
0x55, our test binaries, to be multiplied.
- MULT8E3.gif: R2 is rotated to the right, to roll the
least significant bit into the carry bit. 0x55 (0101.0101) yielded 0x2A (0010.1010).
- MULT8E4.gif: Because the carry bit had a one the content
of the registers R1:R0 is added to the (empty) register pair R4:R3, resulting in
0x00AA there.
- MULT8E5.gif: Now the register pair R1:R0 is rotated one
position left to multiply this binary by 2. From 0x00AA, multiplication by 2 yields
0x0154.
- The whole multiplication loop is repeated as long there is at least one binary 1
in register R2. These following loops are not shown here.
- MULT8E6.gif: Using key F5 of the studio we multi-stepped
over these loops to a breakpoint at the end of the multiplication routine. The result
register pair R4:R3 has the result of the multiplication of 0xAA by 0x55: 0x3872.
This wasn't that complicated, just remind yourself on the similiar decimal operations.
Binary multiplication is much easier than decimal.
To the top of that page
To the top of that page
©2002 by http://www.avr-asm-tutorial.net
