clock cycles per instruction?

Goorman · Joined: 12 Jul 2017 Posts: 13

'Bascom AVR 2.0.7.3

$regfile = "m32def.dat"
$crystal = 16000000
$hwstack = 32 ' default use 32 for the hardware stack
$swstack = 10 ' default use 10 for the SW stack
$framesize = 40 ' default use 40 for the frame space
Dim A As Word , B As Word , C As Word
Dim E As Byte, F As Byte, G As Byte
Do
A = B / 100 '255 Cycles !
A = B / 1000 '255 Cycles!
A = B / C '31 Cycles
A = A / C '31 Cycles
A = A / A '31 Cycles

E = E / 100 '112 Cycles!!!!!!
E = F / G '111 Cycles !!!!!
E = E / F '111 Cycles !!!!
E = E / E '111 Cycles !!!!
Loop

End

The question is. Why Bascom needs 111 cycles to divide eight bits register?

Evert :-) · Joined: 18 Feb 2005 Posts: 2156

Thanks for the numbers.

Goorman · Joined: 12 Jul 2017 Posts: 13

Evert :-) · Joined: 18 Feb 2005 Posts: 2156

Hi Goorman,

Some reading to do: http://www.atmel.com/Images/doc0936.pdf

Goorman · Joined: 12 Jul 2017 Posts: 13

Per Svensson · Joined: 03 Oct 2004 Posts: 235 Location: Gothenburg, Sweden

The time and nr of cycles of a division depend strongly upon the data. If the reminder fast becomes zero as in 12/2 or 15/3 then few cycles are needed. If you divide by a constant and need fast and/or predictable divide time then try signed rightshift operator or combo of multiplication and shift. Also use as short data types as possible. Use word instead if integer for unsigned numbers, etc.
N*5 shifted 2 times right (N*5/4) is faster than N*1.25. etc.
There are more tricks for faster math, like truncated series and cordix.
Use google for more...