我正在连接到Pervasive SQL数据库，该数据库在两个字段上分割一些数据。 DOUBLE字段实际上分为fieldName_1和fieldName_2，其中_1是2字节int，_2是4字节int。
我想获取这些值并使用PHP将它们转换为可用值。 我有一些示例代码来进行转换，但它是用Delphi编写的，我不明白：
{ Reconstitutes a SmallInt and LongInt that form }
{ a Real into a double. }
Function EntConvertInts (Const Int2 : SmallInt;
Const Int4 : LongInt) : Double; StdCall;
Var
TheRealArray : Array [1..6] Of Char;
TheReal : Real;
Begin
Move (Int2, TheRealArray[1], 2);
Move (Int4, TheRealArray[3], 4);
Move (TheRealArray[1], TheReal, 6);
Result := TheReal;
End;
一些数据[fieldName_1，fieldName_2]
[132,805306368] - ＆gt;这应该是11
[132,1073741824] - ＆gt;这应该是12
我不太了解逻辑，无法将其移植到PHP中。非常感激任何的帮助。感谢
EDIT。 这是他们提供的C代码，显示符号/指数：
double real_to_double (real r)
/* takes Pascal real, return C double */
{
union doublearray da;
unsigned x;
x = r[0] & 0x00FF; /* Real biased exponent in x */
/* when exponent is 0, value is 0.0 */
if (x == 0)
da.d = 0.0;
else {
da.a[3] = ((x + 894) << 4) | /* adjust exponent bias */
(r[2] & 0x8000) | /* sign bit */
((r[2] & 0x7800) >> 11); /* begin significand */
da.a[2] = (r[2] << 5) | /* continue shifting significand */
(r[1] >> 11);
da.a[1] = (r[1] << 5) |
(r[0] >> 11);
da.a[0] = (r[0] & 0xFF00) << 5; /* mask real's exponent */
}
return da.d;
}
答案 0 :(得分：4)
我已经在这个问题上工作了大约一个星期，现在试图为我们的组织整理它。
我们的财务部门使用IRIS Exchequer，我们需要降低成本。使用上面的PHP代码，我设法使用以下代码（包括依赖函数）在Excel VBA中使用它。如果下面没有正确归因，我从www.sulprobil.com获得了所有长期的功能。如果将以下代码块复制并粘贴到模块中，则可以从单元格中引用我的ExchequerDouble函数。
在继续之前，我必须在上面的C / PHP代码中指出一个错误。如果你看一下有效循环：
C/PHP: Significand = Significand + 2 ^ (-i)
VBA: Significand = Significand + 2 ^ (1 - i)
我在测试期间注意到答案非常接近但往往不正确。向下钻进我把它缩小到有意义。将代码从一种语言/方法转换为另一种语言/方法可能是一个问题，或者可能只是一个错字，但添加（1 - i）会产生重大影响。
Function ExchequerDouble(Val1 As Integer, Val2 As Long) As Double
Dim Int2 As String
Dim Int4 As String
Dim Real48 As String
Dim Exponent As String
Dim Sign As String
Dim Significand As String
'Convert each value to binary
Int2 = LongDec2Bin(Val1, 16, True)
Int4 = LongDec2Bin(Val2, 32, True)
'Concatenate the binary strings to produce a 48 bit "Real"
Real48 = Int4 & Int2
'Calculate the exponent
Exponent = LongBin2Dec(Right(Real48, 8)) - 129
'Calculate the sign
Sign = Left(Real48, 1)
'Begin calculation of Significand
Significand = "1.0"
For i = 2 To 40
If Mid(Real48, i, 1) = "1" Then
Significand = Significand + 2 ^ (1 - i)
End If
Next i
ExchequerDouble = CDbl(((-1) ^ Sign) * Significand * (2 ^ Exponent))
End Function
Function LongDec2Bin(ByVal sDecimal As String, Optional lBits As Long = 32, Optional blZeroize As Boolean = False) As String
'Transforms decimal number into binary number.
'Reverse("moc.LiborPlus.www") V0.3 P3 16-Jan-2011
Dim sDec As String
Dim sFrac As String
Dim sD As String 'Internal temp variable to represent decimal
Dim sB As String
Dim blNeg As Boolean
Dim i As Long
Dim lPosDec As Long
Dim lLenBinInt As Long
lPosDec = InStr(sDecimal, Application.DecimalSeparator)
If lPosDec > 0 Then
If Left(sDecimal, 1) = "-" Then 'negative fractions later..
LongDec2Bin = CVErr(xlErrValue)
Exit Function
End If
sDec = Left(sDecimal, lPosDec - 1)
sFrac = Right(sDecimal, Len(sDecimal) - lPosDec)
lPosDec = Len(sFrac)
Else
sDec = sDecimal
sFrac = ""
End If
sB = ""
If Left(sDec, 1) = "-" Then
blNeg = True
sD = Right(sDec, Len(sDec) - 1)
Else
blNeg = False
sD = sDec
End If
Do While Len(sD) > 0
Select Case Right(sD, 1)
Case "0", "2", "4", "6", "8"
sB = "0" & sB
Case "1", "3", "5", "7", "9"
sB = "1" & sB
Case Else
LongDec2Bin = CVErr(xlErrValue)
Exit Function
End Select
sD = sbDivBy2(sD, True)
If sD = "0" Then
Exit Do
End If
Loop
If blNeg And sB <> "1" & String(lBits - 1, "0") Then
sB = sbBinNeg(sB, lBits)
End If
'Test whether string representation is in range and correct
'If not, the user has to increase lbits
lLenBinInt = Len(sB)
If lLenBinInt > lBits Then
LongDec2Bin = CVErr(x1ErrNum)
Exit Function
Else
If (Len(sB) = lBits) And (Left(sB, 1) <> -blNeg & "") Then
LongDec2Bin = CVErr(xlErrNum)
Exit Function
End If
End If
If blZeroize Then sB = Right(String(lBits, "0") & sB, lBits)
If lPosDec > 0 And lLenBinInt + 1 < lBits Then
sB = sB & Application.DecimalSeparator
i = 1
Do While i + lLenBinInt < lBits
sFrac = sbDecAdd(sFrac, sFrac) 'Double fractional part
If Len(sFrac) > lPosDec Then
sB = sB & "1"
sFrac = Right(sFrac, lPosDec)
If sFrac = String(lPosDec, "0") Then
Exit Do
End If
Else
sB = sB & "0"
End If
i = i + 1
Loop
LongDec2Bin = sB
Else
LongDec2Bin = sB
End If
End Function
Function LongBin2Dec(sBinary As String, Optional lBits As Long = 32) As String
'Transforms binary number into decimal number.
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim sBin As String
Dim sB As String
Dim sFrac As String
Dim sD As String
Dim sR As String
Dim blNeg As Boolean
Dim i As Long
Dim lPosDec As Long
lPosDec = InStr(sBinary, Application.DecimalSeparator)
If lPosDec > 0 Then
If (Left(sBinary, 1) = "1") And Len(sBin) >= lBits Then 'negative fractions later..
LongBin2Dec = CVErr(xlErrVa1ue)
Exit Function
End If
sBin = Left(sBinary, lPosDec - 1)
sFrac = Right(sBinary, Len(sBinary) - lPosDec)
lPosDec = Len(sFrac)
Else
sBin = sBinary
sFrac = ""
End If
Select Case Sgn(Len(sBin) - lBits)
Case 1
LongBin2Dec = CVErr(x1ErrNum)
Exit Function
Case 0
If Left(sBin, 1) = "1" Then
sB = sbBinNeg(sBin, lBits)
blNeg = True
Else
sB = sBin
blNeg = False
End If
Case -1
sB = sBin
blNeg = False
End Select
sD = "1"
sR = "0"
For i = Len(sB) To 1 Step -1
Select Case Mid(sB, i, 1)
Case "1"
sR = sbDecAdd(sR, sD)
Case "0"
'Do Nothing
Case Else
LongBin2Dec = CVErr(xlErrNum)
Exit Function
End Select
sD = sbDecAdd(sD, sD) 'Double sd
Next i
If lPosDec > 0 Then 'now the fraction
sD = "0.5"
For i = 1 To lPosDec
If Mid(sFrac, i, 1) = "1" Then
sR = sbDecAdd(sR, sD)
End If
sD = sbDivBy2(sD, False)
Next i
End If
If blNeg Then
LongBin2Dec = "-" & sR
Else
LongBin2Dec = sR
End If
End Function
Function sbDivBy2(sDecimal As String, blInt As Boolean) As String
'Divide sDecimal by two, blInt = TRUE returns integer only
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim i As Long
Dim lPosDec As Long
Dim sDec As String
Dim sD As String
Dim lCarry As Long
If Not blInt Then
lPosDec = InStr(sDecimal, Application.DecimalSeparator)
If lPosDec > 0 Then
'Without decimal point lPosDec already defines location of decimal point
sDec = Left(sDecimal, lPosDec - 1) & Right(sDecimal, Len(sDecimal) - lPosDec)
Else
sDec = sDecimal
lPosDec = Len(sDec) + 1 'Location of decimal point
End If
If ((1 * Right(sDec, 1)) Mod 2) = 1 Then
sDec = sDec & "0" 'Append zero so that integer algorithm calculates division exactly
End If
Else
sDec = sDecimal
End If
lCarry = 0
For i = 1 To Len(sDec)
sD = sD & Int((lCarry * 10 + Mid(sDec, i, 1)) / 2)
lCarry = (lCarry * 10 + Mid(sDec, i, 1)) Mod 2
Next i
If Not blInt Then
If Right(sD, Len(sD) - lPosDec + 1) <> String(Len(sD) - lPosDec + 1, "0") Then
'frac part Is non - zero
i = Len(sD)
Do While Mid(sD, i, 1) = "0"
i = i - 1 'Skip trailing zeros
Loop
'Insert decimal point again
sD = Left(sD, lPosDec - 1) _
& Application.DecimalSeparator & Mid(sD, lPosDec, i - lPosDec + 1)
End If
End If
i = 1
Do While i < Len(sD)
If Mid(sD, i, 1) = "0" Then
i = i + 1
Else
Exit Do
End If
Loop
If Mid(sD, i, 1) = Application.DecimalSeparator Then
i = i - 1
End If
sbDivBy2 = Right(sD, Len(sD) - i + 1)
End Function
Function sbBinNeg(sBin As String, Optional lBits As Long = 32) As String
'Negate sBin: take the 2's-complement, then add one
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim i As Long
Dim sB As String
If Len(sBin) > lBits Or sBin = "1" & String(lBits - 1, "0") Then
sbBinNeg = CVErr(xlErrValue)
Exit Function
End If
'Calculate 2 's-complement
For i = Len(sBin) To 1 Step -1
Select Case Mid(sBin, i, 1)
Case "1"
sB = "0" & sB
Case "0"
sB = "1" & sB
Case Else
sbBinNeg = CVErr(xlErrValue)
Exit Function
End Select
Next i
sB = String(lBits - Len(sBin), "1") & sB
'Now add 1
i = lBits
Do While i > 0
If Mid(sB, i, 1) = "1" Then
Mid(sB, i, 1) = "0"
i = i - 1
Else
Mid(sB, i, 1) = "1"
i = 0
End If
Loop
'Finally strip leading zeros
i = InStr(sB, "1")
If i = 0 Then
sbBinNeg = "0"
Else
sbBinNeg = Right(sB, Len(sB) - i + 1)
End If
End Function
Function sbDecAdd(sOne As String, sTwo As String) As String
'Sum up two string decimals.
'Reverse("moc.LiborPlus.www") V0.3 PB 16-Jan-2011
Dim lStrLen As Long
Dim s1 As String
Dim s2 As String
Dim sA As String
Dim sB As String
Dim sR As String
Dim d As Long
Dim lCarry As Long
Dim lPosDec1 As Long
Dim lPosDec2 As Long
Dim sF1 As String
Dim sF2 As String
lPosDec1 = InStr(sOne, Application.DecimalSeparator)
If lPosDec1 > 0 Then
s1 = Left(sOne, lPosDec1 - 1)
sF1 = Right(sOne, Len(sOne) - lPosDec1)
lPosDec1 = Len(sF1)
Else
s1 = sOne
sF1 = ""
End If
lPosDec2 = InStr(sTwo, Application.DecimalSeparator)
If lPosDec2 > 0 Then
s2 = Left(sTwo, lPosDec2 - 1)
sF2 = Right(sTwo, Len(sTwo) - lPosDec2)
lPosDec2 = Len(sF2)
Else
s2 = sTwo
sF2 = ""
End If
If lPosDec1 + lPosDec2 > 0 Then
If lPosDecl > lPosDec2 Then
sF2 = sF2 & String(lPosDec1 - lPosDec2, "0")
Else
sF1 = sFl & String(lPosDec2 - lPosDec1, "0")
lPosDec1 = lPosDec2
End If
sF1 = sbDecAdd(sF1, sF2) 'Add fractions as integer numbers
If Len(sF1) > lPosDecl Then
lCarry = 1
sF1 = Right(sF1, lPosDec1)
Else
lCarry = 0
End If
Do While lPosDec1 > 0
If Mid(sF1, lPosDec1, 1) <> "0" Then
Exit Do
End If
lPosDec1 = lPosDec1 - 1
Loop
sF1 = Left(sF1, lPosDec1)
Else
lCarry = 0
End If
lStrLen = Len(sl)
If lStrLen < Len(s2) Then
lStrLen = Len(s2)
sA = String(lStrLen - Len(s1), "0") & s1
sB = s2
Else
sA = s1
sB = String(lStrLen - Len(s2), "0") & s2
End If
Do While lStrLen > 0
d = 0 + Mid(sA, lStrLen, 1) + Mid(sB, lStrLen, 1) + lCarry
If d > 9 Then
sR = (d - 10) & sR
lCarry = 1
Else
sR = d & sR
lCarry = 0
End If
lStrLen = lStrLen - 1
Loop
If lCarry > 0 Then
sR = lCarry & sR
End If
If lPosDec1 > 0 Then
sbDecAdd = sR & Application.DecimalSeparator & sF1
Else
sbDecAdd = sR
End If
End Function
此代码有效，但有时（大约1％的测试数据）与Excel Addin中的Iris'EntDouble函数相比，最终只有几便士。除非有人能搞清楚，否则我会将其归结为精确度。
最终让这个在VBA工作是我的概念验证，检查一切是否有效。此功能的目标平台是SQL Server。如果您将Exchequer DB链接到SQL Server，则应该能够直接针对Pervasive DB中的数据运行此功能。在我的例子中，我们将把最近2。5年的交易数据转储到SQL Server上的静态表中，但我们每年只处理一次这样的数据，所以这不是问题。以下两个函数应该排除你。在精度方面，它们相当于上面的VBA代码，有时候有些便宜，但有99％的时间看起来完全一样。我们使用SQL Server 2000，因此对于较新的版本，可能会对某些内容进行优化（Varchar（MAX）），但据我所知，最终这应该可以正常工作。
CREATE FUNCTION dbo.FUNCTION_Exchequer_Double
(
@Val1 AS SmallInt,
@Val2 AS BigInt
)
RETURNS Decimal(38, 10)
AS
BEGIN
-- Declare and set decoy variables
DECLARE @Val1_Decoy AS SmallInt
DECLARE @Val2_Decoy AS BigInt
SELECT @Val1_Decoy = @Val1,
@Val2_Decoy = @Val2
-- Declare other variables
DECLARE @Val1_Binary AS Varchar(16)
DECLARE @Val2_Binary AS Varchar(32)
DECLARE @Real48_Binary AS Varchar(48)
DECLARE @Real48_Decimal AS BigInt
DECLARE @Exponent AS Int
DECLARE @Sign AS Bit
DECLARE @Significand AS Decimal(19, 10)
DECLARE @BitCounter AS Int
DECLARE @Two As Decimal(38, 10) -- Saves us casting inline in the code
DECLARE @Output AS Decimal(38, 10)
-- Convert values into two binary strings of the correct length (Val1 = 16 bits, Val2 = 32 bits)
SELECT @Val1_Binary = Replicate(0, 16 - Len(dbo.FUNCTION_Convert_To_Base(Cast(@Val1_Decoy AS Binary(2)), 2)))
+ dbo.FUNCTION_Convert_To_Base(Cast(@Val1_Decoy AS Binary(2)), 2),
@Val2_Binary = Replicate(0, 32 - Len(dbo.FUNCTION_Convert_To_Base(Cast(@Val2_Decoy AS Binary(4)), 2)))
+ dbo.FUNCTION_Convert_To_Base(Cast(@Val2_Decoy AS Binary(4)), 2)
-- Find the decimal value of the new 48 bit number and its binary value
SELECT @Real48_Decimal = @Val2_Decoy * Power(2, 16) + @Val1_Decoy
SELECT @Real48_Binary = @Val2_Binary + @Val1_Binary
-- Determine the Exponent (takes the first 8 bits and subtracts 129)
SELECT @Exponent = Cast(@Real48_Decimal AS Binary(1)) - 129
-- Determine the Sign
SELECT @Sign = Left(@Real48_Binary, 1)
-- A bit of setup for determining the Significand
SELECT @Significand = 1,
@Two = 2,
@BitCounter = 2
-- Determine the Significand
WHILE @BitCounter <= 40
BEGIN
IF Substring(@Real48_Binary, @BitCounter, 1) Like '1'
BEGIN
SELECT @Significand = @Significand + Power(@Two, 1 - @BitCounter)
END
SELECT @BitCounter = @BitCounter + 1
END
SELECT @Output = Power(-1, @Sign) * @Significand * Power(@Two, @Exponent)
-- Return the output
RETURN @Output
END
CREATE FUNCTION dbo.FUNCTION_Convert_To_Base
(
@value AS BigInt,
@base AS Int
)
RETURNS Varchar(8000)
AS
BEGIN
-- Code from http://dpatrickcaldwell.blogspot.co.uk/2009/05/converting-decimal-to-hexadecimal-with.html
-- some variables
DECLARE @characters Char(36)
DECLARE @result Varchar(8000)
-- the encoding string and the default result
SELECT @characters = '0123456789abcdefghijklmnopqrstuvwxyz',
@result = ''
-- make sure it's something we can encode. you can't have
-- base 1, but if we extended the length of our @character
-- string, we could have greater than base 36
IF @value < 0 Or @base < 2 Or @base > 36
RETURN Null
-- until the value is completely converted, get the modulus
-- of the value and prepend it to the result string. then
-- devide the value by the base and truncate the remainder
WHILE @value > 0
SELECT @result = Substring(@characters, @value % @base + 1, 1) + @result,
@value = @value / @base
-- return our results
RETURN @result
END
随意使用我的VBA或SQL代码。真正的努力工作是由谁将其转换为PHP以上。如果有人找到任何改进方法，请告诉我，这样我们就可以使这段代码尽可能完美。
谢谢！
答案 1 :(得分：2)
Delphi的Move命令用于将内存块从一个地方移动到另一个地方。这看起来像旧的Delphi代码 - Real类型已过时，替换为Double（编辑 Real48替换6字节Real），以及使用Byte类型可能比使用Char更好。两者都是字节，但Char更适用于单字节字符（ascii）。这段代码的作用是：
1）声明一个Char数组（这里可以使用Byte），长度为6个字节。同时声明Real（编辑现在Real48类型）以存储转换后的值。
TheRealArray : Array [1..6] Of Char;
TheReal : Real;
2）将双字节Int值移动到TheRealArray - 从索引1开始并移动2个字节的数据（即：所有Int2，一个SmallInt（16位））。对Int4执行相同操作并在索引[3]处启动它，长度为4个字节。
Move (Int2, TheRealArray[1], 2);
Move (Int4, TheRealArray[3], 4);
如果你开始（图片，而不是代码）
Int2 = [2_byte0][2_byte1]
Int4 = [4_byte0][4_byte1][4_byte2][4_byte3]
TheRealArray = [2_byte0][2_byte1][4_byte0][4_byte1][4_byte2][4_byte3]
最终移动命令将此数组复制到TheReal的内存位置，这是一个真实的（6字节浮点）类型。它从数组的索引1开始，将其复制到TheReal，并复制总共六个字节（即：整个事件）。
Move (TheRealArray[1], TheReal, 6);
假设存储在Int2和Int4中的数据，当这样连接时，产生一个格式正确的Real48，那么你最终得到的TheReal以正确的格式保存数据。
PHP字符串中的基本上是字节数组（如Delphi中Char的Array [1..6]）所以你可以使用unpack()做类似的事情来转换为float。
答案 2 :(得分：2)
将此添加为另一个答案，因为我终于弄明白了。这是PHP代码，它将转换值。它必须手动计算，因为PHP不知道如何解包Real48（非标准）。以下评论中的解释。
function BiIntToReal48($f1, $f2){
$x = str_pad(decbin($f1), 16, "0", STR_PAD_LEFT);
$y = str_pad(decbin($f2), 32, "0", STR_PAD_LEFT);
//full Real48 binary string
$real48 = $y . $x;
//Real48 format is V = (-1)^s * 1.f * 2^(exp-129)
// rightmost eight bits are the exponent (bits 40-->47)
// subtract 129 to get the final value
$exp = (bindec(substr($real48, -8)) - 129);
//Sign bit is leftmost bit (bit[0])
$sign =$real48[0];
//Now work through the significand - bits are fractional binary
//(1/2s place, 1/4s place, 1/8ths place, etc)
// bits 1-->39
// significand is always 1.fffffffff... etc so start with 1.0
$sgf = "1.0";
for ($i = 1; $i <= 39; $i++){
if ($real48[$i] == 1){
$sgf = $sgf + pow(2,-$i);
}
}
//final calculation
$final = pow(-1, $sign) * $sgf * pow(2,$exp);
return($final);
}
$field_1 = 132;
$field_2 = 805306368;
$ConvVal = BiIntToReal48($field_1, $field_2);
// ^ gives $ConvVal = 11, qed
答案 3 :(得分：1)
只是转动J ...的回答。 利用变体记录，代码有所简化：
Function EntConvertInts (Const Int2 : SmallInt;
Const Int4 : LongInt) : Double; StdCall;
Type
TReal48PlaceHolder = record
case boolean of
true : (theRealArray : array [1..6] of byte);
false : (r48 : Real48);
end;
Var
R48Rec : TReal48PlaceHolder;
Begin
Move (Int2, R48Rec.theRealArray[1], 2);
Move (Int4, R48Rec.theRealArray[3], 4);
Result := R48Rec.r48;
End;
var
r : Double;
begin
r:= EntConvertInts(132,805306368);
WriteLn(r); // Should be 11
r:= EntConvertInts(141,1163395072);
WriteLn(r); // Should be 6315
ReadLn;
end.
答案 4 :(得分：0)
这也不是“PHP代码”意义上的答案。我只想警告任何可能通过Delphi标签找到此代码的人。
那不是DELPHI !!!
这是旧的Turbo Pascal代码。好吧，也许是16位Delphi 1，它确实是类固醇的TP。
不要在32位Delphi上尝试此代码，至少在替换更改的Char和Real类型之前不要。这两种类型都是从Turbo Pascal时代改变的，特别是6字节的Real，从来都不是硬件FPU兼容的！
如果设置为正确的模式，可能FreePascal可以承载vanilla TurboPascal代码，但最好还是使用Delphi模式和更新的代码。
还应确保SmallInt类型为16位整数（int16），LongInt为32位（int32）。这似乎适用于16位，32位和64位Delphi编译器，但在其他Pascal实现中可能会有所改变。
下面我尝试修改与现代Delphi兼容的代码。我虽然无法测试它。
希望有一天可能会帮助某人将一些类似的旧类型TurboPascal代码转换为更新的代码。
此代码直接遵循原始代码，更兼容，简洁，快速。
{ Reconstitutes a SmallInt and LongInt that form }
{ a Real into a double. }
Function EntConvertInts (Const Int2 : SmallInt;
Const Int4 : LongInt) : Double;
(* StdCall; - only needed for non-Pascal DLLs *)
Var
TheRealArray : Packed Array [1..6] Of Byte; //AnsiChar may suffice too
TheReal : Real48 absolute TheRealArray;
TheInt2 : SmallInt absolute TheRealArray[1];
TheInt4 : LongInt absolute TheRealArray[3];
Begin
Assert(SizeOf(TheInt2) = 2);
Assert(SizeOf(TheInt4) = 2);
Assert(SizeOf(TheReal) = 6);
TheInt2 := Int2; (* Move (Int2, TheRealArray[1], 2); *)
TheInt4 := Int4; (* Move (Int4, TheRealArray[3], 4); *)
(* Move (TheRealArray[1], TheReal, 6); *)
Result := TheReal;
End;
此代码直接使用原生Turbo Pascal功能tagless variant record
{ Reconstitutes a SmallInt and LongInt that form }
{ a Real into a double. }
Function EntConvertInts (Const Int2 : SmallInt;
Const Int4 : LongInt) : Double;
(* StdCall; - only needed for non-Pascal DLLs *)
Var
Value : Packed Record
Case Byte of
0: (TheReal: Real48);
1: (Packed Record TheInt2: SmallInt;
TheInt4: LongInt; end; );
end;
Begin
Assert(SizeOf(Value.TheInt2) = 2);
Assert(SizeOf(Value.TheInt4) = 2);
Assert(SizeOf(Value.TheReal) = 6);
Value.TheInt2 := Int2; (* Move (Int2, TheRealArray[1], 2); *)
Value.TheInt4 := Int4; (* Move (Int4, TheRealArray[3], 4); *)
(* Move (TheRealArray[1], TheReal, 6); *)
Result := Value.TheReal;
End;