Mega Code Archive
Converting a double into a string representation
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Title: Converting a double into a string representation
Question: How do I take a double-precision number and return it in textual form? For example, 101.50 becomes:
ONE HUNDRED ONE AND 50/100
Answer:
I wrote this code a couple of years ago for a Check Printing routine. I've learned a lot since then, and I'm certain it could be improved upon. However, it works right up to one trillion - 1
function TCheckPrinter.DescAmount(amount: Double): String;
var
sAmount : string; // String value of the amount.
sResult : string; // Working string to return.
iLenAmt : integer; // Length of the amount string.
iExponent : integer; // Order of magnitude of the amount
// (hundred, thousand, etc.)
iCounter : integer; // Generic Counter.
pStringStart: pchar; // Pointer to read sAmount from left to
// right.
pStringEnd : pchar; // Pointer to read sAmount from right
// to left.
procedure Swap( var C1: char; var C2: char );
var
cTemp : char;
begin
cTemp := C1;
C1 := C2;
C2 := cTemp;
end;
function ExpText( sValue: string ): string;
var
sResult : string;
iLenVal : integer;
iChrVal : integer;
begin
iLenVal := Length( sValue );
sResult := '';
// Calculate hundreds amount
if ( iLenVal 2 ) then
begin
iChrVal := StrToInt( sValue[ 3 ] );
if ( iChrVal 0 ) then
sResult := sResult + sOnes[ iChrVal ] + ' HUNDRED ';
end;
// Calculate tens and ones.
if ( iLenVal 1 ) then
begin
iChrVal := StrToInt( sValue[ 2 ] );
if ( iChrVal 1 ) then
begin
sResult := sResult + sTens[ iChrVal ];
iChrVal := StrToInt( sValue[ 1 ] );
if ( iChrVal 0 ) then
sResult := sResult + '-' + sOnes[ iChrVal ] + ' '
else
sResult := sResult + ' ';
end
else if ( iChrVal = 1 ) then
begin
iChrVal := StrToInt( sValue[ 1 ] );
if ( iChrVal 0 ) then
sResult := sResult + sTeen[ iChrVal ] + ' '
else
sResult := sResult + sTens[ 1 ] + ' ';
end
else
begin
iChrVal := StrToInt( sValue[ 1 ] );
if ( iChrVal 0 ) then
sResult := sResult + sOnes[ iChrVal ] + ' ';
end;
end;
// Calculate ones.
if ( iLenVal = 1 ) then
begin
iChrVal := StrToInt( sValue[ 1 ] );
if ( iChrVal 0 ) then
sResult := sResult + sOnes[ iChrVal ] + ' ';
end;
Result := sResult;
end;
begin
sAmount := Format( '%-1.2f', [ Amount ] );
iLenAmt := Length( sAmount );
iExponent := 0;
// Determine the Magnitude (power of 100)
case ( iLenAmt ) of
4..6 : iExponent := 1;
7..9 : iExponent := 2;
10..12: iExponent := 3;
13..15: iExponent := 4;
end;
// Reverse the order of sAmount.
pStringStart := pchar( sAmount );
pStringEnd := pchar( pStringStart + iLenAmt - 1 );
iCounter := 1;
while not( iCounter ( iLenAmt div 2 ) ) do
begin
Swap( pStringStart^, pStringEnd^ );
inc( pStringStart );
dec( pStringEnd );
inc( iCounter );
end;
// Step through the groups of three integer values ( 100's,
// 1000's, etc. )
iCounter := iExponent;
while not( iCounter = 0 ) do
begin
case ( iCounter ) of
4: sResult := sResult + ExpText( Copy( sAmount, 13, 3 ) ) + 'BILLION ';
3: sResult := sResult + ExpText( Copy( sAmount, 10, 3 ) ) + 'MILLION ';
2: sResult := sResult + ExpText( Copy( sAmount, 7, 3 ) ) + 'THOUSAND ';
1: sResult := sResult + ExpText( Copy( sAmount, 4, 3 ) );
end;
dec( iCounter );
end;
Result := sResult + 'AND ' + sAmount[ 2 ] + sAmount[ 1 ] + '/100';
end;
Ghannodahn Kirrel
http://www.icitadel.com
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