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Jerry W. Lewis
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Hardly a bug, since both functions behave exactly as documented by their
respective Help's. VBA is a subset of VB, which is a completely separate entity from Excel. It is a mistake to think of them as the same language. Jerry "Don Wiss" wrote: > This is xl2002. VBA says that IsNumeric("3,45") is okay. There is a simple > workaround. WorksheetFunction.IsNumber("3,45") correctly notes that it is > not a number. > > Don't you just like the consistency of the function naming here? > > Don <www.donwiss.com> (e-mail link at home page bottom). |
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Rick Rothstein \(MVP - VB\)
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> This is xl2002. VBA says that IsNumeric("3,45") is okay. There is a simple
> workaround. WorksheetFunction.IsNumber("3,45") correctly notes that it is > not a number. The VBA IsNumeric function is actually a lot worse than you have yet discovered. Here is part of a canned response that I post (originally, over in the compiled VB newsgroups, but everything in my posting applies to VBA) whenever someone posts about (or recommends using) the VB/VBA IsNumeric function. Usually, the person having trouble with this function is trying to use it to determine if a user's entry is all digits or a "normally" formed number; hence the theme of the posting. I usually try and steer people away from using IsNumeric to "proof" supposedly numeric text. Consider this (also see note below): ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)") Most people would not expect THAT to return True. IsNumeric has some "flaws" in what it considers a proper number and what most programmers are looking for. I had a short tip published by Pinnacle Publishing in their Visual Basic Developer magazine that covered some of these flaws. Originally, the tip was free to view but is now viewable only by subscribers.. Basically, it said that IsNumeric returned True for things like -- currency symbols being located in front or in back of the number as shown in my example (also applies to plus, minus and blanks too); numbers surrounded by parentheses as shown in my example (some people use these to mark negative numbers); numbers containing any number of commas before a decimal point as shown in my example; numbers in scientific notation (a number followed by an upper or lower case "D" or "E", followed by a number equal to or less than 305 -- the maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for Hexadecimal, &O or just & in front of the number for Octal). NOTE: ====== In the above example and in the referenced tip, I refer to $ signs and commas and dots -- these were meant to refer to your currency, thousands separator and decimal point symbols as defined in your local settings -- substitute your local regional symbols for these if appropriate. As for your question about checking numbers, here are two functions that I have posted in the past for similar questions..... one is for digits only and the other is for "regular" numbers: Function IsDigitsOnly(Value As String) As Boolean IsDigitsOnly = Len(Value) > 0 And _ Not Value Like "*[!0-9]*" End Function Function IsNumber(ByVal Value As String) As Boolean ' Leave the next statement out if you don't ' want to provide for plus/minus signs If Value Like "[+-]*" Then Value = Mid$(Value, 2) IsNumber = Not Value Like "*[!0-9.]*" And _ Not Value Like "*.*.*" And _ Len(Value) > 0 And Value <> "." And _ Value <> vbNullString End Function Here are revisions to the above functions that deal with the local settings for decimal points (and thousand's separators) that are different than used in the US (this code works in the US too, of course). Function IsNumber(ByVal Value As String) As Boolean Dim DP As String ' Get local setting for decimal point DP = Format$(0, ".") ' Leave the next statement out if you don't ' want to provide for plus/minus signs If Value Like "[+-]*" Then Value = Mid$(Value, 2) IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _ Not Value Like "*" & DP & "*" & DP & "*" And _ Len(Value) > 0 And Value <> DP And _ Value <> vbNullString End Function I'm not as concerned by the rejection of entries that include one or more thousand's separators, but we can handle this if we don't insist on the thousand's separator being located in the correct positions (in other words, we'll allow the user to include them for their own purposes... we'll just tolerate their presence). Function IsNumber(ByVal Value As String) As Boolean Dim DP As String Dim TS As String ' Get local setting for decimal point DP = Format$(0, ".") ' Get local setting for thousand's separator ' and eliminate them. Remove the next two lines ' if you don't want your users being able to ' type in the thousands separator at all. TS = Mid$(Format$(1000, "#,###"), 2, 1) Value = Replace$(Value, TS, "") ' Leave the next statement out if you don't ' want to provide for plus/minus signs If Value Like "[+-]*" Then Value = Mid$(Value, 2) IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _ Not Value Like "*" & DP & "*" & DP & "*" And _ Len(Value) > 0 And Value <> DP And _ Value <> vbNullString End Function Rick |
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Peter T
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Don,
There's a big difference between IsNumeric and IsNumber. Surely the text "3.45" is indeed numeric. If you want to test the 'type' of a variable look as VarType. Rick, > ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)") Wow ! Actually, as my currency is not USD I did this - s = "($1,23,,3.4,,,5,,E67$)" s = Replace(s, "$", Application.International(xlCurrencyCode)) Raised my curiosity - Dim d as double on error resume next d = s ' try to coerce If err.number then err.clear else blnIsNumeric = true end if With the your string, Internationalized if necessary, it can be coerced and somehow returns -1.23345E+70 However - v = application.Evaluate(s) ' Error 2015 Which leads me to wonder if the following in help - "Returns a Boolean value indicating whether an expression can be evaluated as a number" might be better written as "- can be COERCED as a number" Not sure why a Date is not considered as numeric although it can 'sometimes' if not typically be coerced to a double. Regards, Peter T "Rick Rothstein (MVP - VB)" <(E-Mail Removed)> wrote in message news:(E-Mail Removed)... > > This is xl2002. VBA says that IsNumeric("3,45") is okay. There is a simple > > workaround. WorksheetFunction.IsNumber("3,45") correctly notes that it is > > not a number. > > The VBA IsNumeric function is actually a lot worse than you have yet > discovered. Here is part of a canned response that I post (originally, over > in the compiled VB newsgroups, but everything in my posting applies to VBA) > whenever someone posts about (or recommends using) the VB/VBA IsNumeric > function. Usually, the person having trouble with this function is trying to > use it to determine if a user's entry is all digits or a "normally" formed > number; hence the theme of the posting. > > I usually try and steer people away from using IsNumeric to "proof" > supposedly numeric text. Consider this (also see note below): > > ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)") > > Most people would not expect THAT to return True. IsNumeric has some "flaws" > in what it considers a proper number and what most programmers are looking > for. > > I had a short tip published by Pinnacle Publishing in their Visual Basic > Developer magazine that covered some of these flaws. Originally, the tip was > free to view but is now viewable only by subscribers.. Basically, it said > that IsNumeric returned True for things like -- currency symbols being > located in front or in back of the number as shown in my example (also > applies to plus, minus and blanks too); numbers surrounded by parentheses as > shown in my example (some people use these to mark negative numbers); > numbers containing any number of commas before a decimal point as shown in > my example; numbers in scientific notation (a number followed by an upper or > lower case "D" or "E", followed by a number equal to or less than 305 -- the > maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for > Hexadecimal, &O or just & in front of the number for Octal). > > NOTE: > ====== > In the above example and in the referenced tip, I refer to $ signs and > commas and dots -- these were meant to refer to your currency, thousands > separator and decimal point symbols as defined in your local settings -- > substitute your local regional symbols for these if appropriate. > > As for your question about checking numbers, here are two functions that I > have posted in the past for similar questions..... one is for digits only > and the other is for "regular" numbers: > > Function IsDigitsOnly(Value As String) As Boolean > IsDigitsOnly = Len(Value) > 0 And _ > Not Value Like "*[!0-9]*" > End Function > > Function IsNumber(ByVal Value As String) As Boolean > ' Leave the next statement out if you don't > ' want to provide for plus/minus signs > If Value Like "[+-]*" Then Value = Mid$(Value, 2) > IsNumber = Not Value Like "*[!0-9.]*" And _ > Not Value Like "*.*.*" And _ > Len(Value) > 0 And Value <> "." And _ > Value <> vbNullString > End Function > > Here are revisions to the above functions that deal with the local settings > for decimal points (and thousand's separators) that are different than used > in the US (this code works in the US too, of course). > > Function IsNumber(ByVal Value As String) As Boolean > Dim DP As String > ' Get local setting for decimal point > DP = Format$(0, ".") > ' Leave the next statement out if you don't > ' want to provide for plus/minus signs > If Value Like "[+-]*" Then Value = Mid$(Value, 2) > IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _ > Not Value Like "*" & DP & "*" & DP & "*" And _ > Len(Value) > 0 And Value <> DP And _ > Value <> vbNullString > End Function > > I'm not as concerned by the rejection of entries that include one or more > thousand's separators, but we can handle this if we don't insist on the > thousand's separator being located in the correct positions (in other words, > we'll allow the user to include them for their own purposes... we'll just > tolerate their presence). > > Function IsNumber(ByVal Value As String) As Boolean > Dim DP As String > Dim TS As String > ' Get local setting for decimal point > DP = Format$(0, ".") > ' Get local setting for thousand's separator > ' and eliminate them. Remove the next two lines > ' if you don't want your users being able to > ' type in the thousands separator at all. > TS = Mid$(Format$(1000, "#,###"), 2, 1) > Value = Replace$(Value, TS, "") > ' Leave the next statement out if you don't > ' want to provide for plus/minus signs > If Value Like "[+-]*" Then Value = Mid$(Value, 2) > IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _ > Not Value Like "*" & DP & "*" & DP & "*" And _ > Len(Value) > 0 And Value <> DP And _ > Value <> vbNullString > End Function > > Rick > |
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Rick Rothstein \(MVP - VB\)
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See all my inline comments....
>> ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)") > > Wow ! Yeah, that was pretty much my reaction when it first dawned on me that IsNumeric didn't do what I thought it did. There are a lot of answers that I posted in my early days of volunteering in the compiled VB newsgroups that include the IsNumeric function as a "number proofer". Then, one day, by chance, I typed in something like 12e3 into a TextBox in order to test the error handling section of some code I wrote and, lo-and-behold, no error was generated. It took a few seconds for it to dawn on me that the 4-character garbage "number" I thought I typed in was really an actual 5-digit number (12e3=12000) to the IsNumeric function. That set me off on my investigation of the IsNumeric function which led to my publish article that I cited in my last post. While we are talking about surprises like what IsNumeric considers numbers, VB/VBA contains lots of such surprises. Let me give you another one that I had misunderstood at first... Integer Division (where you use the backward slash to for the division sign). In the beginning, I had thought (like I have found many people do) that A\B was a sort of short-hand for Int(A/B). IT IS NOT! In Integer Division, if A and/or B are floating point numbers, those numbers are rounded first, BEFORE the integer division takes place. On top of that, the rounding method used is the one known as Banker's Rounding (where numbers ending in 5 that are being rounded up to the previous digit position round to the nearest even digit). Here is the problem... in my beginnings with VB, I had figured (as most people still do) that 4.5\1.5 was equivalent to Int(4.5/1.5) and that the answer would be 3; but IT IS NOT, rather, the answer it 2! The 4.5 is rounded to 4 before the division and the 1.5 is rounded to 2 before the division... only then are they divided producing 2 (from this...4/2) as the answer. > Actually, as my currency is not USD I did this - > > s = "($1,23,,3.4,,,5,,E67$)" > s = Replace(s, "$", Application.International(xlCurrencyCode)) > > Raised my curiosity - > > Dim d as double > on error resume next > d = s ' try to coerce > If err.number then > err.clear > else > blnIsNumeric = true > end if > > With the your string, Internationalized if necessary, it can be coerced > and somehow returns -1.23345E+70 The commas are completely ignored... the string becomes -123345E+67 which, in scientic notation, is simplified to -1.23345E+70. > However - > v = application.Evaluate(s) ' Error 2015 > > Which leads me to wonder if the following in help - > "Returns a Boolean value indicating whether an expression can be > evaluated as a number" > > might be better written as "- can be COERCED as a number" I think the "evaluated" part refers to the Cxxx converter such as CLng, CDbl, etc. Perhaps you didn't know... try printing out (in the Immediate window) that same string inside a CDbl function call.... s = "($1,23,,3.4,,,5,,E67$)" s = Replace(s, "$", Application.International(xlCurrencyCode)) Print CDbl(s) Yep, it prints out -1.23345E+70. If IsNumeric returns True for a string expression, then CInt, CLng, CCur, CSng and CDbl will accept the string as a valid argument and convert it to a number (assuming that number is small enough to fit in the indicated data type). Cute, huh? > Not sure why a Date is not considered as numeric although it > can 'sometimes' if not typically be coerced to a double. I never thought to try Date before. I think the answer is in what the date represents. As a String, something like "6/7/08" (or even "6/7/2008") is not really a number of any sort, so IsNumeric reports False for it and CDbl will error out on it. Odd though, if you use the date delimiter symbols (#) around it instead (that is, #6/7/08# or #6/7/2008#), CDbl convert it, but IsNumeric will still report False for it. Very odd indeed. Rick |
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Peter T
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OK we agree that
cDbl("($1,23,,3.4,,,5,,E67$)") ' change '$' to local currency symbol returns a value and also the string will coerce to a number if assigned to a variable declared as a double. Therefore surely we should expect IsNumeric to return true when testing that string, which it does however surprising at first it may seem. So I'm wondering after all that, is there anything actually unreliable or buggy about IsNumeric. However I still suspect Help might be better written along the lines I suggested using the word 'coerced' vs 'evaluated', as further suggested in the following Sub test() Dim s$, v Dim dbl As Double Dim dt As Date s = "3,4.5,,,6,7" s = "3.45" v = Application.Evaluate(s) ' Error 2015 dbl = s ' 34.567 Debug.Print v, dbl, IsNumeric(s) ' can coerce but can't evaluate ''' Date Debug.Print IsNumeric(Date) ' false as per help s$ = Format(Date, "mmm dd yyyy") dt = s ' OK dbl = CDate(s) ' OK' dbl = s ' error 'can only coerce to variable declared 'As Date' End Sub btw, behind the scenes is there any difference between doing dbl = s vs dbl = cDbl(s) ? I guess for consistency the authors thought best to exclude dates from being considered IsNumeric, ie depending on the string the date may or may not simply coerce to a number. Concerning the Integer divisor: I had been aware of it's Banker's Rounding nature for a long time but was surprised once to be hit with something like this: n = 4 \ 0.5 ' error div by zero Regards, Peter T "Rick Rothstein (MVP - VB)" <(E-Mail Removed)> wrote in message news:%(E-Mail Removed)... > See all my inline comments.... > > >> ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)") > > > > Wow ! > > Yeah, that was pretty much my reaction when it first dawned on me that > IsNumeric didn't do what I thought it did. There are a lot of answers that I > posted in my early days of volunteering in the compiled VB newsgroups that > include the IsNumeric function as a "number proofer". Then, one day, by > chance, I typed in something like 12e3 into a TextBox in order to test the > error handling section of some code I wrote and, lo-and-behold, no error was > generated. It took a few seconds for it to dawn on me that the 4-character > garbage "number" I thought I typed in was really an actual 5-digit number > (12e3=12000) to the IsNumeric function. That set me off on my investigation > of the IsNumeric function which led to my publish article that I cited in my > last post. > > While we are talking about surprises like what IsNumeric considers numbers, > VB/VBA contains lots of such surprises. Let me give you another one that I > had misunderstood at first... Integer Division (where you use the backward > slash to for the division sign). In the beginning, I had thought (like I > have found many people do) that A\B was a sort of short-hand for Int(A/B). > IT IS NOT! In Integer Division, if A and/or B are floating point numbers, > those numbers are rounded first, BEFORE the integer division takes place. On > top of that, the rounding method used is the one known as Banker's Rounding > (where numbers ending in 5 that are being rounded up to the previous digit > position round to the nearest even digit). Here is the problem... in my > beginnings with VB, I had figured (as most people still do) that 4.5\1.5 was > equivalent to Int(4.5/1.5) and that the answer would be 3; but IT IS NOT, > rather, the answer it 2! The 4.5 is rounded to 4 before the division and the > 1.5 is rounded to 2 before the division... only then are they divided > producing 2 (from this...4/2) as the answer. > > > > Actually, as my currency is not USD I did this - > > > > s = "($1,23,,3.4,,,5,,E67$)" > > s = Replace(s, "$", Application.International(xlCurrencyCode)) > > > > Raised my curiosity - > > > > Dim d as double > > on error resume next > > d = s ' try to coerce > > If err.number then > > err.clear > > else > > blnIsNumeric = true > > end if > > > > With the your string, Internationalized if necessary, it can be coerced > > and somehow returns -1.23345E+70 > > The commas are completely ignored... the string becomes -123345E+67 which, > in scientic notation, is simplified to -1.23345E+70. > > > > However - > > v = application.Evaluate(s) ' Error 2015 > > > > Which leads me to wonder if the following in help - > > "Returns a Boolean value indicating whether an expression can be > > evaluated as a number" > > > > might be better written as "- can be COERCED as a number" > > I think the "evaluated" part refers to the Cxxx converter such as CLng, > CDbl, etc. Perhaps you didn't know... try printing out (in the Immediate > window) that same string inside a CDbl function call.... > > s = "($1,23,,3.4,,,5,,E67$)" > s = Replace(s, "$", Application.International(xlCurrencyCode)) > Print CDbl(s) > > Yep, it prints out -1.23345E+70. If IsNumeric returns True for a string > expression, then CInt, CLng, CCur, CSng and CDbl will accept the string as a > valid argument and convert it to a number (assuming that number is small > enough to fit in the indicated data type). Cute, huh? > > > > Not sure why a Date is not considered as numeric although it > > can 'sometimes' if not typically be coerced to a double. > > I never thought to try Date before. I think the answer is in what the date > represents. As a String, something like "6/7/08" (or even "6/7/2008") is not > really a number of any sort, so IsNumeric reports False for it and CDbl will > error out on it. Odd though, if you use the date delimiter symbols (#) > around it instead (that is, #6/7/08# or #6/7/2008#), CDbl convert it, but > IsNumeric will still report False for it. Very odd indeed. > > > Rick > |
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Peter T
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> However I still suspect Help might be better written along the lines I
> suggested using the word 'coerced' vs 'evaluated' Forgot about this bit you mentioned previously: "I think the "evaluated" part refers to the Cxxx converter such as CLng, CDbl, etc. Perhaps you didn't know... try printing out (in the Immediate window) that same string inside a CDbl function call...." Yes, with that interpretation of 'evaluated' vs app.evaluate it makes sense, with the exception of cDate Regards, Peter T |
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Rick Rothstein \(MVP - VB\)
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> OK we agree that
> cDbl("($1,23,,3.4,,,5,,E67$)") ' change '$' to local currency symbol > returns a value and also the string will coerce to a number if assigned to > a > variable declared as a double. > > Therefore surely we should expect IsNumeric to return true when testing > that > string, which it does however surprising at first it may seem. So I'm > wondering after all that, is there anything actually unreliable or buggy > about IsNumeric. Well, I think both the Cxxx functions and IsNumeric are to "generous" in what they consider numbers. Why is a number with a minus sign in both the front **and** back considered a valid number? > However I still suspect Help might be better written along the lines I > suggested using the word 'coerced' vs 'evaluated', Agreed > ''' Date > Debug.Print IsNumeric(Date) ' false as per help > s$ = Format(Date, "mmm dd yyyy") > dt = s ' OK > dbl = CDate(s) ' OK' > dbl = s ' error > 'can only coerce to variable declared 'As Date' In your last statement, the variable 's' is a Variant of sub-type String, but that String value is no different to the assignment than a/b/c would be, so I understand that type mismatch error. > btw, behind the scenes is there any difference between doing > dbl = s vs dbl = cDbl(s) ? Well, technically, I think yes. Using CDbl specifically casts 's' as a Double before the assignment is made. Just assigning 's' directly to 'dbl' means VB's type coercion mechanism has to decide what to do... while I know it doesn't do this, it could theoretically cast it internally to a Single (losing precision) first before deciding it should be a Double. As I said, I'm positive it doesn't do this, but why leave the job to chance when you know what it should be? > I guess for consistency the authors thought best to exclude dates from > being > considered IsNumeric, ie depending on the string the date may or may not > simply coerce to a number. What I don't understand is why IsNumeric(Now) returns False while CDbl(Now) evaluates to a value (and, of course, the example I gave earlier). If IsNumeric is a "proofer" for the Cxxx functions, why not handle this also? > Concerning the Integer divisor: I had been aware of it's Banker's Rounding > nature for a long time but was surprised once to be hit with something > like > this: > n = 4 \ 0.5 ' error div by zero Yeah, I got caught with that one too. As a point of information, VB/VBA uses Banker's Rounding in every situation where rounding is necessary EXCEPT inside the Format function. The Format function uses what I consider to be normal rounding rules (5's are always rounded upward).... Print Format(5/2, "0") ====> 3 Rick |
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Peter T
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Hi again,
Several interesting observations. FWIW cell's numberformat works similarly to your Format example, ie an exact .5 always rounds up. If anything I would have expected it to truncate down. > Well, I think both the Cxxx functions and IsNumeric are to "generous" in > what they consider numbers. Why is a number with a minus sign in both the > front **and** back considered a valid number? I don't know about 'too generous', why not 'helpfully accommodating' <g> As for "both front **and** back", I take it you mean either/or and not both together. "-123-" fails, but why shouldn't "123-" be allowed to evaluate, or rather coerce to a valid number. If one accepts the IsNumeric function is consistent with 'can cDbl without error' and 'is not a date' (as documented), perhaps it should not be considered as buggy! Regards, Peter T "Rick Rothstein (MVP - VB)" <(E-Mail Removed)> wrote in message news:(E-Mail Removed)... > > OK we agree that > > cDbl("($1,23,,3.4,,,5,,E67$)") ' change '$' to local currency symbol > > returns a value and also the string will coerce to a number if assigned to > > a > > variable declared as a double. > > > > Therefore surely we should expect IsNumeric to return true when testing > > that > > string, which it does however surprising at first it may seem. So I'm > > wondering after all that, is there anything actually unreliable or buggy > > about IsNumeric. > > Well, I think both the Cxxx functions and IsNumeric are to "generous" in > what they consider numbers. Why is a number with a minus sign in both the > front **and** back considered a valid number? > > > However I still suspect Help might be better written along the lines I > > suggested using the word 'coerced' vs 'evaluated', > > Agreed > > > ''' Date > > Debug.Print IsNumeric(Date) ' false as per help > > s$ = Format(Date, "mmm dd yyyy") > > dt = s ' OK > > dbl = CDate(s) ' OK' > > dbl = s ' error > > 'can only coerce to variable declared 'As Date' > > In your last statement, the variable 's' is a Variant of sub-type String, > but that String value is no different to the assignment than a/b/c would be, > so I understand that type mismatch error. > > > btw, behind the scenes is there any difference between doing > > dbl = s vs dbl = cDbl(s) ? > > Well, technically, I think yes. Using CDbl specifically casts 's' as a > Double before the assignment is made. Just assigning 's' directly to 'dbl' > means VB's type coercion mechanism has to decide what to do... while I know > it doesn't do this, it could theoretically cast it internally to a Single > (losing precision) first before deciding it should be a Double. As I said, > I'm positive it doesn't do this, but why leave the job to chance when you > know what it should be? > > > I guess for consistency the authors thought best to exclude dates from > > being > > considered IsNumeric, ie depending on the string the date may or may not > > simply coerce to a number. > > What I don't understand is why IsNumeric(Now) returns False while CDbl(Now) > evaluates to a value (and, of course, the example I gave earlier). If > IsNumeric is a "proofer" for the Cxxx functions, why not handle this also? > > > Concerning the Integer divisor: I had been aware of it's Banker's Rounding > > nature for a long time but was surprised once to be hit with something > > like > > this: > > n = 4 \ 0.5 ' error div by zero > > Yeah, I got caught with that one too. As a point of information, VB/VBA uses > Banker's Rounding in every situation where rounding is necessary EXCEPT > inside the Format function. The Format function uses what I consider to be > normal rounding rules (5's are always rounded upward).... > > Print Format(5/2, "0") ====> 3 > > Rick > |
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