Multiply vs Divide, and use of Equates

[Daniel Raymer]

I'm still working on my super-big program. There are a lot of places where I do units conversions twice - first I multiply by 1.689, then later I must divide by 1.689. These are peppered throughout a highly iterative routine so I may hit them 50,000 times in a single analysis.

Two Expert's Questions:

1) Is it really much faster to multiply by 0.592 than to divide by 1.689?

2) Should I create equates for 1.689 and 0.592 and use them rather than repeat those constant values wherever needed in the code? Ie., is it (a) faster and/or (b) smaller executable?

Thanks, mighty online gurus!

Dan

[Davide Vecchi]

About the 1 I don't know, you might try measuring that with a loop with many iterations.

About the 2, that should make no difference at all because equates work by text substitution when compiling. But equates are available only for integers.

[Mark Hunter]

Dividing takes longer than multiplying. The following is in PBCC, not PBDOS:

%count = 100000     'number of iterations

'From winAPI32
Declare Function QueryPerformanceFrequency Lib "KERNEL32.DLL" Alias "QueryPerformanceFrequency" (lpFrequency As Quad) As Long
Declare Function QueryPerformanceCounter   Lib "KERNEL32.DLL" Alias "QueryPerformanceCounter" (lpPerformanceCount As Quad) As Long

Function PBMain
  Local Ctr1 As Quad
  Local Ctr2 As Quad
  Local Freq As Quad
  Local Overhead As Quad
  Local Time1 As Ext, Time2 As Ext
  Local i As Long, x As Single

  QueryPerformanceFrequency Freq
  QueryPerformanceCounter Ctr1      'find API overhead
  QueryPerformanceCounter Ctr2
  Overhead = Ctr2 - Ctr1

'------------
  Print "Multiply:"
  QueryPerformanceCounter Ctr1
  For i = 1 To %count
   x = i*0.592
  Next
  QueryPerformanceCounter Ctr2
  Time1 = (Ctr2 - Ctr1 - overhead) / Freq
  Print "(" + Format$(Ctr2) + " - " + Format$(Ctr1) + " - " + Format$(Overhead) + ") / " + Format$(Freq)
  Print "= " + Format$(Time1, "##.0000") + " seconds"
  Print
'------------
  Print "Divide:"
  QueryPerformanceCounter Ctr1
  For i = 1 To %count
   x = i/1.689
  Next
  QueryPerformanceCounter Ctr2
  Time2 = (Ctr2 - Ctr1 - overhead) / Freq
  Print "(" + Format$(Ctr2) + " - " + Format$(Ctr1) + " - " + Format$(Overhead) + ") / " + Format$(Freq)
  Print "= " + Format$(Time2, "##.0000") + " seconds"
  Print
'------------
  Print "Divide time / Multiply time:"
  Print Time2/Time1

  WaitKey$
End Function

[Mark Hunter]

Here's similar code for PBDOS

%count = 30000
dim x as single
dim timer1 as single
dim timer2 as single

cls

print "Mutliply"
mtimer
for i = 1 to %count
 x = i*0.592
next
time1 = mtimer
print time1;" microseconds"
print

print "Divide"
mtimer
for i = 1 to %count
 x = i/1.689
next
time2 = mtimer
print time2;" microseconds"
print

print "Divide / Multiply"
print time2/time1

[Michael Mattias]

I'm not really a math type, but if you first have to multiply by 1.69, then divide by 1.69 later, do you really need to do either?

[John Gleason]

At the cpu level, it is almost exactly three times slower to divide than multiply for a pentium 4.

[Daniel Raymer]

Why am I multiplying then dividing

FYI, it's my big aircraft design program. Users like to input in knots (nautical miles per hour). All the equations use feet-per-second. So, gotta do units conversions.

For users who like to use metric, all their inputs first get multiplied by that conversion factor.

Per a previous thread of mine, sacraficing goats would be easier.