* RESOLVED * How to declare the maximum value of a Double as a const

This code gives me an error. I'm not sure how to declare a const with an exponent. Does anyone know how to make this declaration? I want the maximum value that a double can hold as a constant for validation purposes.

VB Code:

public const dblMax as Double=1.79769313486232E308

I also need the smallest negative value. This doesn't give me an error, but I'm not sure that it's not subtracting 308 from the value which would make it incorrect.

VB Code:

Public Const dblMin As Double = 4.94065645841247E-324

Do you really need the absolute maximum and minimum?

I'm just curious, why would the user need to be able to enter in values in that entire range?

I just want to be able to validate user input before VB does so I can return a real error message instead of the default error that VB would raise.

I don't think it will be a problem for a double data type, but even so, I'd like to know how to do it.

basically, this is what I've got. The remmed out lines I haven't figured out how to make work yet.

VB Code:

Public Const MIN_BYTE As Byte = 0
Public Const MAX_BYTE As Byte = 255
Public Const MIN_CURRENCY As Currency = -922337203685477#
Public Const MAX_CURRENCY As Currency = 922337203685477#
'Public Const MIN_DOUBLE As Double = 4.94065645841247E-324
'public const MAX_DOUBLE as Double=1.79769313486232E+308
Public Const MIN_INTEGER As Integer = -32768
Public Const MAX_INTEGER As Integer = 32767
Public Const MIN_LONG As Long = -2147483648#
Public Const MAX_LONG As Long = 2147483647
'public const MIN_SINGLE as Single=same problem as double - but smaller value
'public const MAX_SINGLE as Single = Ditto

Just off the top of my head, maybe something like...

VB Code:

public const dblMax as Double=1.79769313486232^308

thanks. It likes that and therefore, so do I. Much appreciated. :)

So for the MIN_DOUBLE I'm declaring it not as a negative number, but as a negative exponent?

I'm confused now. That would mean a double can only be greater than 0? What am I missing here?

OK, the problem is that there are actually 4 limits to a single and a double so I can't just declare a min and max value. I need a min and max negative value and a min and max positive value.

Valid numbers are from -1.79769313486231E308 to -4.94065645841247E-324 for negative values;

4.94065645841247E-324 to 1.79769313486232E308 for positive values.

Also, what John said appears to work, so thank you for that information.

just curious... How do you know what the max value is??

you write a loop and print the value to a file until and error occurs?

From MSDN:

Quote:

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Visual Basic for Applications Reference

Data Type Summary


The following table shows the supported data types, including storage sizes and ranges.

Data type Storage size Range
Byte 1 byte 0 to 255
Boolean 2 bytes True or False
Integer 2 bytes -32,768 to 32,767
Long
(long integer) 4 bytes -2,147,483,648 to 2,147,483,647
Single
(single-precision floating-point) 4 bytes -3.402823E38 to -1.401298E-45 for negative values; 1.401298E-45 to 3.402823E38 for positive values
Double
(double-precision floating-point) 8 bytes -1.79769313486231E308 to
-4.94065645841247E-324 for negative values; 4.94065645841247E-324 to 1.79769313486232E308 for positive values
Currency
(scaled integer) 8 bytes -922,337,203,685,477.5808 to 922,337,203,685,477.5807
Decimal 14 bytes +/-79,228,162,514,264,337,593,543,950,335 with no decimal point;
+/-7.9228162514264337593543950335 with 28 places to the right of the decimal; smallest non-zero number is
+/-0.0000000000000000000000000001
Date 8 bytes January 1, 100 to December 31, 9999
Object 4 bytes Any Object reference
String
(variable-length) 10 bytes + string length 0 to approximately 2 billion
String
(fixed-length) Length of string 1 to approximately 65,400
Variant
(with numbers) 16 bytes Any numeric value up to the range of a Double
Variant
(with characters) 22 bytes + string length Same range as for variable-length String
User-defined
(using Type) Number required by elements The range of each element is the same as the range of its data type.


Note Arrays of any data type require 20 bytes of memory plus 4 bytes for each array dimension plus the number of bytes occupied by the data itself. The memory occupied by the data can be calculated by multiplying the number of data elements by the size of each element. For example, the data in a single-dimension array consisting of 4 Integer data elements of 2 bytes each occupies 8 bytes. The 8 bytes required for the data plus the 24 bytes of overhead brings the total memory requirement for the array to 32 bytes.

A Variant containing an array requires 12 bytes more than the array alone.

Note Use the StrConv function to convert one type of string data to another.


--------------------------------------------------------------------------------
Send feedback to MSDN.Look here for MSDN Online resources.

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Quote:

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Originally posted by cafeenman
I just want to be able to validate user input before VB does so I can return a real error message instead of the default error that VB would raise.

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Why not use VarType to validate whether the value can be passed to a certain data type?

Quote:

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Originally posted by WorkHorse
Why not use VarType to validate whether the value can be passed to a certain data type?

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Because I forgot about its existence. Thanks. :)

log on aim yo!

Quote:

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Originally posted by WorkHorse
Why not use VarType to validate whether the value can be passed to a certain data type?

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VarType doesn't tell you what kind of data a string holds. It tells you what kind of data is in a variable. You would use it like this:

VB Code:

Dim s As String
 
MsgBox VarType(s)
' Returns 8

Not like this (which is what I need)

VB Code:

MsgBox VarType("6.379")
' Also returns 8 because it's a string which is
' what a text box holds
 
' if I coerce the text to another variable before using vartype
' then I lose the opportunity to validate it before VB does

Quote:

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Originally posted by John McKernan
Just off the top of my head, maybe something like...

VB Code:

public const dblMax as Double=1.79769313486232^308

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Just a word of caution if someone else tries this, I dont think
Mathematicaly these two ARE the same:

1.79769313486232^308
1.79769313486232E +308

The first will be a very large Number, 2833..... 75 values to follow,
where as the second is 179769..... with 303 values to follow.

I could be wrong :)

Quote:

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Originally posted by Bruce Fox


Just a word of caution if someone else tries this, I dont think
Mathematicaly these two ARE the same:

1.79769313486232^308
1.79769313486232E +308

The first will be a very large Number, 2833..... 75 values to follow,
where as the second is 179769..... with 303 values to follow.

I could be wrong :)

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No Bruce, I think you are absolutely right. This validation scheme of mine is not well conceived at all. I need a better way to do it. VB will raise an error, and I could maybe implement that, but I'd really like to catch the error before VB so I can give the user specific information rather than just letting VB tell them that an error occurred somewhere. I.e. Type Mismatch. Most users say "Huh?"

Just a thought from left field, (may generate possibilities

Considering the High end for a moment, all it is a Large Number
1.7???E +308!
So, in theory, what if you were to divide it and find a suitable
number, so you can apply that as a "Number" mathematicaly
to a user input to test if the input is outa range....
(Yes, I have confussed myself too)

You know, ...you could write a program that incriments the smallest possible decimal value allowed in a double. Start it at 1.3E+308 and let it run like...all night...and trap the overflow error when it occurs. Voila your maximum value. Just a rough idea, considering its 4AM up here...

Quote:

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Originally posted by nishantp
You know, ...you could write a program that incriments the smallest possible decimal value allowed in a double. Start it at 1.3E+308 and let it run like...all night...and trap the overflow error when it occurs. Voila your maximum value. Just a rough idea, considering its 4AM up here...

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I already know what the number is. I just don't know how to code it as a constant.