How to extract the low byte of a float variable and store it in a char variable?

hi
i need to extract the low byte of a float variable and store it in a char variable

any suggestions
thanx

Re: qustion for C expert

It depends byte ordering in the machine. i.e. the machine is little endian or big endian.
Here is a C code for little endian

float a; // float variable whose low byte is needed
char *ptr, b;

ptr = (char *)&a;

b = ptr[0]; // b contains the low byte of a now.

.. Hope this is what u wanted.

qustion for C expert

#define LOBYTE(a) (*((unsigned char*)&(a)))
#define HIBYTE(a) (*((unsigned char*)&(a) + 1))

Re: qustion for C expert

booklog said:

It depends byte ordering in the machine. i.e. the machine is little endian or big endian.
Here is a C code for little endian

float a; // float variable whose low byte is needed
char *ptr, b;

ptr = (char *)&a;

b = ptr[0]; // b contains the low byte of a now.

.. Hope this is what u wanted.

Click to expand...

Hi,

Why did you say b = ptr[0]; instead of b = *ptr;

Thanks alot

Re: qustion for C expert

Here's a way to do it that's independent of endianness:


The above was modified per booklog's suggestion (see next post) because I got too clever and tried to take the address of a constant.

Re: qustion for C expert

Hi Lambtron,
Whatever u have posted, need some modification.
The correct code is as below
char GetLsb( float *f )
{
int a = 5 ;
return ((char *)f)[ (((char*)&a)[0] == 5)) ? 0 : (sizeof(float)-1) ];
}

qustion for C expert

sorry i tried it but too many errors

please i want to convert a log routine from basic to c and i know that there is a log function but i need it detailed as it in the basic code

can anyone convert this for me please
---------------------------------------------------------------
' Create some variables for use with LN and LOG

Dim LOG_VALUE as Float
Dim LOG_POWER as Float
Dim LOG_TEMP as Float
Dim LOG_X as Float
Dim LOG_FACTOR as Float
Dim LOG_XSQR as Float
Dim LOG_N as Byte
Dim LOG_TEMP2 as Byte



LNc:

' We can't have ln(1) so we must return a zero if it is
If LOG_VALUE.Byte0 = 0 Then LOG_VALUE = 0: Return

' The difference between LOG_N.BYTE0 and $7E will be the
' amount of 2^LOG_N's that we want to multiply times ln(2)
If LOG_VALUE.Byte0 <= $7E Then
LOG_N = $7E - LOG_VALUE.Byte0
LOG_FACTOR = -0.69314718 * LOG_N
Else
LOG_N = LOG_VALUE.Byte0 - $7E
LOG_FACTOR = 0.69314718 * LOG_N
Endif
LOG_VALUE.Byte0 = $7E

' Begin the taylor series expansion
' ln(1+LOG_X) = LOG_X - (LOG_X^2/2) + (LOG_X^3/3) -+...

LOG_VALUE = LOG_VALUE - 1
LOG_X = LOG_VALUE
LOG_XSQR = LOG_VALUE
LOG_N = 2
Repeat
LOG_XSQR = LOG_XSQR * LOG_X
LOG_VALUE = LOG_VALUE - (LOG_XSQR / LOG_N)
LOG_XSQR = LOG_XSQR * LOG_X
LOG_TEMP2 = (LOG_N + 1)
LOG_VALUE = LOG_VALUE + (LOG_XSQR / LOG_TEMP2)
LOG_N = LOG_N + 2
Until LOG_N > 12
LOG_VALUE = LOG_VALUE + LOG_FACTOR
Return
-----------------------------------------------------------
thanx alot

Re: qustion for C expert

I think structure and union is best

union
{
float value ;

struct
{
unsigned char value_a0 ;
unsigned char value_a1 ;
unsigned char value_a2 ;
unsigned char value_a3 ;
}
split
}
floatcombo ;

floatcombo.value = 323947.24742424 ;

floatcombo.split.value_aX is now contais 4 bytes of float variable

qustion for C expert

atmelAVR91
i dont know how to use it in my code
could you please write my basic code using your method?
thankx

Re: qustion for C expert

hi

consult the following book