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# convert two bytes into an integer

Posted on 2000-04-11
1,502 Views
Is this right?

eg
dim z1 as byte
dim z2 as byte
dim MyInt as integer

MyInt = -1 * (z1 * 256 + z2)

0

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Expert Comment

ID: 2703980
get rid of the -1 and your fine

MYInt = (z1 * 256) + z2

Gordon

0

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Expert Comment

ID: 2704005
You must test and handle the case where z1 is greater than 127 (The MyInt is negative). If not you gen an Overflow error

If z1 And &H80 Then
MyInt = ((z1 And &H7F) * 256 + z2) Or &H8000
Else
MyInt = z1 * 256 + z2
End If
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Author Comment

ID: 2704056
Adjusted points from 5 to 10
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Author Comment

ID: 2704057
What does 'z1 And &H80' mean?
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Expert Comment

ID: 2704102
'z1 And &h80' maens logical binary operator. Select "And" word within VB environment and press F1 for more info.

Sincerely yours,

Crin
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Accepted Solution

Gordonp earned 10 total points
ID: 2704114
&H80 is a hex number = 128

(z1 And &H80) returns 128(&H80) is z1 is >= 128 and 0 if z1 < 128

Could rewrite it

IF z1 >= 128 then
MyInt = ((z1 - 128) * 256 + z2) * -1
else
MyInt = z1 * 256 + z2
End If

which is the same as lewys answers but using maths rather than bitwise manipulations.

Gordon
0

Expert Comment

ID: 2704138
Yes it is right
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