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Float conversion to 4 bytes?

Posted on 1998-11-17
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Last Modified: 2013-11-20
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!  
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Question by:Surfer
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5 Comments
 

Author Comment

by:Surfer
ID: 1324810
Edited text of question
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LVL 2

Expert Comment

by:cyrilbdt
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

by:Surfer
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 ;
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LVL 4

Expert Comment

by:pagladasu
ID: 1324813
Your problem is due to the way floats are stored internally - IEEE format.
0
 
LVL 5

Accepted Solution

by:
scrapdog earned 200 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);

}

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