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The "And" operator is faster than "Mod"
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| Total Hit (4040) |
Using the "And" operator instead of "Mod" may speed up your program under certain situations. Use the And operator instead of Mod when the divisor is a number in the form 2^N. For instance, there are two methods to extract the least significant byte in an Integer:
«Code LangId=1»
lowByte% = valu
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Avoid Integer Overflow
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| Total Hit (2871) |
When working with integer expressions there is often the risk of raising the "Overflow" error. More specifically, there can be two occasions when this frequently occurs:
When you multiply or add Integer values and the result exceeds 32,767.
When you create generic string routines that iterate
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Convert Hexadecimal numbers
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| Total Hit (4048) |
While Visual Basic offers the Hex$ function that converts a decimal value into its hexadecimal equivalent number, it seems that the inverse function is missing. Not true. Try out this one-liner:
«Code LangId=1»
Function HexToDec(HexValue As String) As Long
HexToDec = Val("&H" & HexValue)
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Counting Bits
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| Total Hit (3211) |
It seems that the only way to count the number of 1's or 0's in a binary value is creating a loop that iterates on all the 16 or 32 bits of the number. There is, however, a faster algorithm:
«Code LangId=1»
Function BitCount (ByVal number As Long) As Integer
Dim bits As Integer, temp As Long
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Determine whether the app is running on a flawed Pentium CPU
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| Total Hit (2645) |
Here's a simple test that you can use to determine whether the application is running on a system equipped with a Pentium CPU affected by the FDIV bug:
«Code LangId=1»
' return True if the CPU suffers from the FDIV bug
Function IsBuggedPentium() As Boolean
IsBuggedPentium = ((1 / 3221224
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Evaluate the integer equal or higher than a given value
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| Total Hit (2760) |
The VBA language offers the Int() function, which returns the integer equal or lower than a given value, but lacks a similar function that returns the integer equal or higher than a given value. You can remedy with the following function:
«Code LangId=1»
' Returns the integer equal or higher tha
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Understanding the "Allow Unrounded Floating Point Operations" option
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| Total Hit (2890) |
The Microsoft manuals preach that all the compiler options in the Advanced Optimization dialog box are to be considered unsafe, in that they might lead to incorrect results (or just program crashes!). This is true for most of them, but often one of such options - namely, the "Allow Unrounded Floatin
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Undocumented behavior of the CInt() function
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| Total Hit (3013) |
The CInt() function rounds to the nearest integer value. In other words, CInt(2.4) returns 2, and CInt(2.6) returns 3.
This function exhibits an under-documented behavior when the fractional part is equal to 0.5. In this case, this function rounds down if the integer portion of the argument is e
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Use integer division operator
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| Total Hit (2874) |
Use "\" instead of "/" when performing divisions between Integers. The "/" operator returns a Single value, therefore the seemingly efficient line
«Code LangId=1»
C% = A% / B%
«/Code»
actually requires three implicit conversions, two for converting the operands from Integer to Single (to pre
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Always use "$-typed" string functions
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| Total Hit (2784) |
VB official documentation seems to encourage you to use "$-less" string functions, e.g. Left, LTrim or UCase, instead of their time-honored Left$, LTrim$ and UCase$ counterparts. If you do so, however, you should be aware that the former ones return a variant value that, when used within string expr
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Avoid append operations with the Replace function
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| Total Hit (2861) |
I find the Replace function very useful when I want to avoid too many append operations just to insert non-printable characters. For example, the following statement:
«Code LangId=1»
MsgBox "Disk not ready." & vbCr & vbCr & _
"Please check that the diskette is in the drive" & vbCr & _
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Count distinct characters in a string
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| Total Hit (3785) |
When you need to count how many occurrences of a given character are in a string, you might be tempted to go along the "traditional" Visual Basic way:
«Code LangId=1»
' count spaces
For i = 1 To Len(text)
If Mid$(text, i, 1) = " " Then count = count + 1
Next
«/Code»
Ev
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Extract all quoted strings with the RegExp object
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| Total Hit (3617) |
When parsing a text file that contains quoted strings - such as a VB source file - you might want to quickly locate and extract all the quoted strings. Thanks to regular expressions and the RegExp object in the Microsoft VBScript Regular Expression type library, this task is surprisingly easy:
«co
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Extract words with the RegExp object
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| Total Hit (3731) |
The following routine extracts all the words from a source string and returns a collection. Optionally, the result contains only unique words.
This code is remarkably simpler than an equivalent "pure" VB solution because it takes advantage of the RegExp object in the Microsoft VBScript Regular Expr
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Faster string appending with Mid$ command
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| Total Hit (3146) |
As you probably know, the "&" operator is rather slow, especially with long strings. When you have to repeatedly append chucks of characters to the same variable, you can speed up your code using a simple trick based on the Mid$ command. The idea is that you pre-allocate a buffer long enough to acco
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Fixed-length strings allocate and deallocate faster
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| Total Hit (2692) |
Fixed-length strings are generally slower than conventional strings, because all VB string functions and operators only recognize conventional strings, thus all fixed-length strings must be transparently converted to and from conventional strings.
However, arrays of fixed-length strings occupy a
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Fixed-length strings allocate and deallocate faster
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| Total Hit (2822) |
Fixed-length strings are generally slower than conventional strings, because all VB string functions and operators only recognize conventional strings, thus all fixed-length strings must be transparently converted to and from conventional strings.
However, arrays of fixed-length strings occupy a
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Non obvious uses for the LIKE operator
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| Total Hit (2991) |
LIKE is probably the VB operator is most underutilized if compared to its potential. The most immediate use for this operator is to check that a string begins or ends with a given pattern. For example, it is simple to check that a Product ID is made up by a alphabetic character followed by three dig
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Process string characters using Byte arrays
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| Total Hit (2843) |
When you need to process each character in a string, you can assign the string to a byte array and iterate on its elements (remember: each Unicode character corresponds to two bytes). This approach is usually much faster because it saves a lot of Mid$ functions and, hence, a lot of temporary strings
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Replace the last occurrence of a string
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| Total Hit (3138) |
Here is a one-liner that replaces the last occurrence of a substring in a string. It is slightly less efficient than using the InstrRev plus a Replace statement, but at least you can use it in-line inside another expression. And it is also one of the few occasions to use the StrReverse function, und
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Replicate a string of any length
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| Total Hit (2962) |
The String$ function can replicate only 1-char strings, so it seems that you need a loop to duplicate strings that contain 2 or more characters. However, this is a one-liner that does the trick:
«Code LangId=1»
Function ReplicateString(Source As String, Times As Long) As String
' build a s
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Retrieve the currency symbol
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| Total Hit (3043) |
It is very simple to retrieve the correct currency symbol, using just plain VBA statements:
«Code LangId=1»
currSymbol = Trim$(Replace(FormatCurrency(0,0), "0", ""))
«/Code»
If you can also determine if the symbol precedes or follows the currency amount using this method: «Code LangId=1»
If
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Search multiple substrings with the RegExp object
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| Total Hit (3720) |
The RegExp object in the Microsoft VBScript Regular Expression type library supports regular expression patterns containing the | (or) operator, which lets you search for multiple substrings at the same time. For example, the following piece of code lets you search for a month name in a source text:
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Swap strings the fast way
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| Total Hit (3236) |
Consider the usual way of swapping two strings:
«Code LangId=1»
Dim s1 As String, s2 As String
Dim tmp As String
' initialize the strings
s1 = String$(1000, "-")
s2 = String$(1500, "+")
' do the swap, through a temporary variable
tmp = s1
s1 = s2
s2 = tmp
«/Code»
If you put the abo
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Write concise code with the IIf function
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| Total Hit (2134) |
Use IIf when setting a value based upon an expression. This works only in situations where you have two choices, based upon the result of a Boolean expression.
«Code LangId=1»
'long way
If Var1 > Var2 Then
Var3 = Var1 / Var2
Else
Var3 = Var2 / Var1
End If
'this is more concise
Var
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Write concise code with the InStr function
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| Total Hit (2043) |
You can often use the Instr function in an unorthodox way to write more concise code. A typical example is when you need to test a single character:
«Code LangId=1»
' test whether CHAR contains a vowel
' the standard way
If UCase$(char) = "A" Or UCase$(char) = "E" Or UCase$(char) = "I" Or UCas
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Write concise code with the Switch function
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| Total Hit (1717) |
Many VB developers don't realize that the Switch built-in function can often save a lot of code. This function takes any number of (expr, value) pairs, it evaluates all expressions and return the value related to the first expression that is found to be True. Typically you can use a Switch function
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Write concise code: a collection of simple tips
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| Total Hit (2114) |
If you fully understand how VB and VBA work, you can often save some statements. This makes your listings more concise and more readable, and indirectly optimizes your program. Here is a short list of the many tricks you can use with this goal. You don't need to initialize a numeric variable to 0 or
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A simple expression evaluator
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| Total Hit (3301) |
While any Windows user could pop up the Calculator accessory to perform any type of math calculations, it would be great if you could offer him or her the capability to do simple math from within your application. This is a very simple expression evaluator function that does it:
This evaluator i
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