DoubleToHEX!!!Urgent

[vittal]

Hi,

Is there any built-in function which can convert a double value into a hexadecimal value and back.

OR is there any other way of doing it.


Please help me ASAP, its very urgent.


Thanks in Advance

Vittal

[Rail Jon Rogut]

How about something simple like:

CString DecimalToHex(DWORD dwValue)

{

BOOL bFlag = TRUE;

CString szReturn;

CString szLetter;

float fRemainder;

unsigned long lTemp, lTemp2;

lTemp = dwValue;

while (lTemp)

{

lTemp2 = lTemp;

lTemp = lTemp2/16;

fRemainder = ((float)lTemp2/16.0F) - (float)lTemp;

lTemp2 = (long)(fRemainder * 16.0F);

if (lTemp2 > 9)

{

switch (lTemp2)

{

case 10: szLetter = "A";

break;

case 11: szLetter = "B";

break;

case 12: szLetter = "C";

break;

case 13: szLetter = "D";

break;

case 14: szLetter = "E";

break;

case 15: szLetter = "F";

break;

}

}

else

szLetter.Format("%d", lTemp2);

szReturn += szLetter;

}

szReturn.MakeReverse();

return szReturn;

}

Please excuse the formatting... the BBS has been changing my formatting tonite.

Rail

Recording Engineer/Software Developer

Rail Jon Rogut Software


Rail Jon Rogut Software

[Rail Jon Rogut]

Of course, besides that other method, there's always:


CString szTemp;


DWORD dwTemp = 4290384468;


szTemp.Format("%lX", dwTemp);


which sets szTemp equal to: FFBA1254


Rail

Recording Engineer/Software Developer

Rail Jon Rogut Software


Rail Jon Rogut Software

[Orlando]

Hi,


Type double is a real number and has a fractional part.

What happens if there is anything after the decimal point??

[Bore]

double myDouble;

CString myString;


myString.Format("%x",(long)myDouble);

[Bore]

Sorry, I missed the "and back" part of your question. From the original example:

double myDouble;

CString myString;

char *dummy;

myString.Format("%x",(long)double);

myDouble = (double)strtol((LPCTSTR)myString,&dummy,16);

Of course, you'll never get the fractional part back. And I didn't mention it before, but of course, you'll also lose

data if the value of the double is greater than 2^32.

[Gomez Addams]

This really made me curious. I decided to try a back-door approach.

The key is using a union to share the eight bytes of a double with

a 64-bit integer so sscanf and sprintf can take of the conversions.

I actually compiled this and it runs.

typedef union

{

__int64 intval;

double dblval;

} DIUNION;

int main( int ac, char *av[] )

{

// should verify that there is an argument

double dval = atof( av[1] );

printf( "original is : %.6f\n", dval );

DIUNION diu1;

diu1.dblval = dval;

char buffer[32];

sprintf( buffer, "%016I64X", diu1.intval );

printf( "hex string is %s\n", buffer );

DIUNION diu2;

sscanf( buffer, "%I64X", &diu2.intval );

printf( "double val is %.6f\n", diu2.dblval );


return 0;

}

[Paul McKenzie]

The trick is to convert floating point decimal to hex is to do the following:


0) Convert the integer portion of the number to hex. Append a decimal point.

Store this result in "sub-answer".


1) Take the decimal portion of the original number (the digits after the

decimal point). Include the decimal point (this is very important!). Call

this number the "decimal portion".


2) multiply this decimal portion by 16.


3) strip the integer portion of the answer you got for 2) and append to "sub-

answer in 1).


4) Exhausted? Stop.


5) Use the stripped integer answer you got for step 3 and that is your new

decimal portion. Go to step 2


Example:


25.48 to hex


Step 0) 25 = 0x19.

Step 1) .48

Step 2) .48 x 16 = 7.68

Step 3) 0x19. --> 0x19.7 (got the 7 from step 2)

Step 4) Not tired. keep going

Step 5) Use the .68 and do step 2

Step 2) .68 x 16 = 10.88

Step 3) 0x19.7 --> 0x19.7A (I converted the 10 to the hex A)

Step 4) Not tired.

Step 5) Use the .88 and goto step 2

Step 2) .88 x 16 = 14.08

Step 3) 0x19.7A --> 0x19.7AE

Step 4) Tired. Stop


Answer 25.48 is approximately 0x19.7AE


I hope I didn't make any errors :-()


Regards,


Paul

[Rail Jon Rogut]

Oh, then how about:

CString szTemp;

double dwTemp = 25.48;

int iVal = (int)dwTemp;

double iResolution = 4.0; // How many positions after the point

int iVal2 = (int)((dwTemp - (double)iVal) * pow(16.0, iResolution));

szTemp.Format("%X.%X", iVal, iVal2);


Rail


Recording Engineer/Software Developer

Rail Jon Rogut Software


Rail Jon Rogut Software

[Paul McKenzie]

Ok. You beat me :-()


Hey wait a minute...my method can't overflow!


Regards,


Paul

[Rail Jon Rogut]

Hey... didn't know this was a competition... :^)


Actually your method is perfect, similar to the way I posted 2 methods initially

last nite (or this morning)... this was simply another way to get the same results.


As for the overflow... yup, you're right :^)


All the best.


Rail

Recording Engineer/Software Developer

Rail Jon Rogut Software


Rail Jon Rogut Software

[Gomez Addams]

Just to play devil's advocate, what if it's a really

huge number that needs an exponent to display correctly ?

Or, for that matter, a really really small one ?


Personally, I kind of like my method but it is far from

portable since it relies on an __int64. I guess an array

of eight bytes could accomplish the same thing and then

there is only the LSB-MSB issue to worry about.

That is, if anyone really cares about portability.

Who knows, if Linux picks up steam portability could

become much more of an issue.

[Gomez Addams]

I started thinking about this some more and decided the union

with an __int64 wasn't the best option. I revised it so that

there is no union and this will work with any data type. This

should be fairly portable also in case you care. The one caveat

I can think of about this approach is that the hex data will

probably not be completely transportable between different types

machines. This is an issue of little-endian versus big-endian.

Again, this is an 'if you care' issue.

typedef unsigned char uchar;

void ConvertToHex( char *dstbuf, void *pvalue, size_t valsize )

{

*dstbuf = '\0'; // initialize string

// step through each byte of the value

uchar * pchval = (uchar *)pvalue;

for( size_t x = 0; x < valsize; x += 1 )

{

// obtain current byte

uchar itembyte = *pchval;

// load a string with hex value of byte

char tmpbuf[4];

sprintf( tmpbuf, "%02X", itembyte );

// concatenate to destination buffer

strcat( dstbuf, tmpbuf );

pchval += 1;

}

}

void ConvertFromHex( char *srcbuf, void *pvalue, size_t valsize )

{

uchar * pchval = (uchar *)pvalue; // start a first byte

for( size_t x = 0; x < valsize; x += 1 )

{

// determine string index - hex has 2 chars per byte

int chindex = x * 2;

// grab hex string for 1 byte

char tmpbuf[4];

strncpy( tmpbuf, srcbuf + chindex, 2 );

tmpbuf[2] = '\0';

// convert hex string to a byte value

uchar byteval = (uchar)strtol( tmpbuf, NULL, 16 );

// load byte into the destination

*pchval = byteval;

pchval += 1;

}

}

// test the algorithms - what goes in should come out

void main()

{

char buffer[32];

double dblval = 3.1415926539;

printf( "\nvalue is : %.9f\n", dblval );

ConvertToHex( buffer, &dblval, sizeof( double ) );

printf( "hex value : %s\n", buffer );

ConvertFromHex( buffer, &dblval, sizeof( double ) );

printf( "dbl value : %.9f\n", dblval );

int intval = 3563428;

printf( "\nvalue is : %.d\n", intval );

ConvertToHex( buffer, &intval, sizeof( int ) );

printf( "hex value : %s\n", buffer );

ConvertFromHex( buffer, &intval, sizeof( int ) );

printf( "int value : %d\n", intval );

}



PS. Sorry to spoil your fun - I like an occasional challenge.

[Rail Jon Rogut]

Hi Gomez


Your algorithm converts to and from Hex correctly, but you need to shift the bytes around for the display of the Hex value...


Remember when you store the value in memory, it's not gonna be displayed in the same way we humans would want to look at it


For example... You test function sets the value


int intval = 3563428;


and you algorithm converts that to Hex:


A45F3600


which is wrong because of byte ordering...


switch them around and you get the right answer:


A4 5F 36 00 => 00 36 5F A4 .... 0x00365FA4


Your other example:


double dblval = 3.1415926539;


creates the hex value:


B4D54E54FB210940 => 040921FB544ED5B4 (from the above logic)


But that doesn't fall into the format which we expect to see displayed...


(integer).(fraction)


which each digit:


X(n), X(n-1),....X(2), X(1), X(0).Y(-1), Y(-2), ...Y(z+1), Y(z)


and we multiply each digit X or Y by the base to the power of the subscript and sum them all together:


Value = X(0)*pow(Base, 0) + X(1)*pow(Base, 1)+...+X(n)*pow(Base,n) + Y(-1)*pow(Base, -1) + Y(-2)*pow(Base, -2)+...+Y(z)*pow(Base, z)


for instance:


Decimal 25.48


= 2*pow(10, 1) + 5*pow(10, 0) + 4*pow(10, -1) + 8*pow(10, -2)


= 20 + 5 + .4 + .08


= 25.48


convert that to HEX:


Hex 0x19.7AE


= 1*pow(16, 1) + 9*pow(16, 0) + 7*pow(16, -1) + A*pow(16, -2) + E*pow(16, -3)


= 16 + 9 + .0625 + .0390 + .0034


= 25.4799


Anyway...


All the best.


Rail

Recording Engineer/Software Developer

Rail Jon Rogut Software





Rail Jon Rogut Software

[Rail Jon Rogut]

Sorry, I made a typeo:


convert that to HEX:


Hex 0x19.7AE


= 1*pow(16, 1) + 9*pow(16, 0) + 7*pow(16, -1) + A*pow(16, -2) + E*pow(16, -3)


= 16 + 9 + .0625 + .0390 + .0034


= 25.4799


should of course read:


convert that to HEX:


Hex 0x19.7AE


= 1*pow(16, 1) + 9*pow(16, 0) + 7*pow(16, -1) + A*pow(16, -2) + E*pow(16, -3)


= 16 + 9 + .4375 + .0390 + .0034


= 25.4799


I had written .0625 instead of 7 * .0625 which equals .4375


Rail

Recording Engineer/Software Developer

Rail Jon Rogut Software


Rail Jon Rogut Software