for loop with step increment of 0.1
 

Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

for loop with step increment of 0.1
 

2.1


--
Don Guillett
Microsoft MVP Excel
SalesAid Software

"choi4u" wrote in message
...
Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

for loop with step increment of 0.1
 

Sub test()
For i = 1.5 To 2.01 Step 0.1
Debug.Print i
Next i
End Sub


--
Gary''s Student - gsnu2007h

for loop with step increment of 0.1
 

To add to thos ecomments try this

Sub test()
Dim i As Single
For i = 1.5 To 2 Step 0.1
p = i + 0.1 2
Debug.Print i
Debug.Print p
Next i
End Sub

As you will note as soon as i = 1.9 then i + 0.1 is greater than 2 (True) so
it never executes.

Mike
"choi4u" wrote:

Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

for loop with step increment of 0.1
 

I'd guess that there was an internal rounding problem that you got hit with.

I'd use:

Sub test()
Dim i As Long
For i = 15 To 20 Step 1
Debug.Print i / 10
Next i
End Sub



choi4u wrote:

Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

--

Dave Peterson

for loop with step increment of 0.1
 

The problem has to do with floating point representation and the inability
of VB (actually, any language) of being able to store all fractions in a
given integer data type (the problem is akin to the decimal representation
of 2/3, a decimal point followed by an unending string of 6's, invariably
the number is written as a fixed number of 6's followed by an ending 7)...
whatever has accumulated in the loop variable "i" at the 1.9 stage, adding
0.1 to it gives a number ever so slightly greater than 2 which is outside of
the ending loop limit, hence the loop ends at the 1.9 value. The way around
this problem is to not use floating point numbers. For example...

Sub test()
For i = 15 To 20
Debug.Print i / 10
Next i
End Sub

(note the modified loop limits) where whenever you would have used "i" in
your original loop code, you would use "i/10" instead.

Rick


"choi4u" wrote in message
...
Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

for loop with step increment of 0.1
 

Here are some links to WebPages that will show you the problem in more
detail...

INFO: Visual Basic and Arithmetic Precision
http://support.microsoft.com/default...NoWebContent=1

(Complete) Tutorial to Understand IEEE Floating-Point Errors
http://support.microsoft.com/default...NoWebContent=1

What Every Computer Scientist Should Know About Floating Point
http://docs.sun.com/source/806-3568/ncg_goldberg.html

Rick


"Rick Rothstein (MVP - VB)" wrote in
message ...
The problem has to do with floating point representation and the inability
of VB (actually, any language) of being able to store all fractions in a
given integer data type (the problem is akin to the decimal representation
of 2/3, a decimal point followed by an unending string of 6's, invariably
the number is written as a fixed number of 6's followed by an ending 7)...
whatever has accumulated in the loop variable "i" at the 1.9 stage, adding
0.1 to it gives a number ever so slightly greater than 2 which is outside
of the ending loop limit, hence the loop ends at the 1.9 value. The way
around this problem is to not use floating point numbers. For example...

Sub test()
For i = 15 To 20
Debug.Print i / 10
Next i
End Sub

(note the modified loop limits) where whenever you would have used "i" in
your original loop code, you would use "i/10" instead.

Rick


"choi4u" wrote in message
...
Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

for loop with step increment of 0.1
 

On May 16, 9:33 am, choi4u wrote:
Running this code
Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
Next i
End Sub

gives this output
1.5
1.6
1.7
1.8
1.9 .

Why does the output not include 2.0?
and what do I have to do to make the output include 2.0?

Thanks in advance.

It's due to the age old binary to decimal conversion problem.
Computers use binary representations of real numbers that often (most
often) do not convert exactly to their decimal equivalents without
some remainder. This leads to the commonly occurring problem that you
see here.

As Dan Guillett suggested, one workaround is simply increase the end
value to compensate for the error, though I might use 2.01 to keep
from confusing the issue more - that is, "Why isn't 2.1 in the list?"
Another approach might be to convert the loop from a FOR to a Do Until
with a Round function to correct for the error ...

Sub test()
step = 0.1
i = 1.5
Do Until i 2
Debug.Print i
i = Round(i + step, 1)
Loop
End Sub

Another way that would work is to round the index in the FOR loop ...

Sub test()
For i = 1.5 To 2 Step 0.1
Debug.Print i
i = Round(i, 1)
Next i
End Sub

Tom Lavedas
===========
http://members.cox.net/tglbatch/wsh/