Solved

# Float conversion to 4 bytes?

Posted on 1998-11-17
253 Views
I need to comvert 1 float into 4 bytes and then back to 1 float again.  Someone told me to try this:
FLOAT f1 = 123.456, f2 = 0.0 ;
BYTE byByte1 = HIBYTE( HIWORD( f1 ) ) ;
BYTE byByte2 = LOBYTE( HIWORD( f1 ) ) ;
BYTE byByte3 = HIBYTE( LOWORD( f1 ) ) ;
BYTE byByte4 = LOBYTE( LOWORD( f1 ) ) ;
//rebuild
f2  = byByte1 << 24 | byByte2 << 16 | byByte3 << 8 | byByte4 ;

The rebuild answer for this is f2 = 123.000 and not 123.456.  How can a get the rest of my float?  What am I doing wrong? Also what is required to break down to bytes, a double (8bytes) and rebuild the 8 bytes into one double? All sugestions welcome and sample code is greatly appreciated!
0
Question by:Surfer

Author Comment

ID: 1324810
Edited text of question
0

LVL 2

Expert Comment

ID: 1324811
I don't understand what you want to do.
variable (as you probably know) of type float takes 4 bytes in memory. So if you need access to each byte of them:

BYTE *p = (BYTE*)&f1;
p[0] - first byte and so on.

the code in your question doesn't work because bit operation OR can't return float type.

may be something like this
FLOAT f1 = 123.456, *f2;
int *pf, i2;
f2 = new FLOAT;
pf = (int*)&f1;
BYTE byByte1 = HIBYTE( HIWORD( *pf ) ) ;
BYTE byByte2 = LOBYTE( HIWORD( *pf ) ) ;
BYTE byByte3 = HIBYTE( LOWORD( *pf ) ) ;
BYTE byByte4 = LOBYTE( LOWORD( *pf ) ) ;
//rebuild
i2  = byByte1 << 24 | byByte2 << 16 | byByte3 << 8 | byByte4;
memcpy((void*)f2, (void*)&i2, 4);

0

Author Comment

ID: 1324812
I tried to modify your code sample that works with type float to work with type double and it dose not work.  All suggestions welcome.

double r1 = 123.456, *r2 ;
r2 = new double ;
_int64 i2 ;

BYTE *pr = (BYTE*)&r1 ;

BYTE byByte1 = pr[7] ;
BYTE byByte2 = pr[6] ;
BYTE byByte3 = pr[5] ;
BYTE byByte4 = pr[4] ;
BYTE byByte5 = pr[3] ;
BYTE byByte6 = pr[2] ;
BYTE byByte7 = pr[1] ;
BYTE byByte8 = pr[0] ;

i2 = byByte1 << 56 | byByte2 << 48 | byByte3 << 40 | byByte4  << 32 | byByte5 << 24 | byByte6 << 16 | byByte7 << 8  | byByte8 ;

memcpy((void*)r2,(void*)&i2,8) ;

delete r2 ;
0

LVL 4

Expert Comment

ID: 1324813
Your problem is due to the way floats are stored internally - IEEE format.
0

LVL 5

Accepted Solution

scrapdog earned 100 total points
ID: 1324814
#include <windows.h>
#include <stdio.h>

float f1 = 123.456;
float f2;
double d1 = 345.6789;
double d2;
BYTE *b;
BYTE *d;

void main()
{

b = (BYTE *)&f1;
printf("Float Bytes:   %i %i %i %i\n",b[0],b[1],b[2],b[3]);
f2 = *(float *)b;
printf("First float:  %f  \nSecond float:  %f\n\n",f1,f2);
d = (BYTE *)&d1;
printf("Double bytes:  %i %i %i %i %i %i %i %i\n",d[0],d[1],d[2],d[3],d[4],d[5],
d[6],d[7]);
d2 = *(double *)d;
printf("First double:  %f  \nSecond double:  %f   ",d1,d2);

}

0

## Featured Post

### Suggested Solutions

Introduction: Dynamic window placements and drawing on a form, simple usage of windows registry as a storage place for information. Continuing from the first article about sudoku.  There we have designed the application and put a lot of user int…
Introduction: Finishing the grid – keyboard support for arrow keys to manoeuvre, entering the numbers.  The PreTranslateMessage function is to be used to intercept and respond to keyboard events. Continuing from the fourth article about sudoku. …
This video will show you how to get GIT to work in Eclipse.   It will walk you through how to install the EGit plugin in eclipse and how to checkout an existing repository.
Here's a very brief overview of the methods PRTG Network Monitor (https://www.paessler.com/prtg) offers for monitoring bandwidth, to help you decide which methods you´d like to investigate in more detail.  The methods are covered in more detail in o…