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- NAME
- #include <aros/libcall.h>
#include <proto/mathieeesingbas.h>
#include <proto/mathieeesingtrans.h>
#include <proto/exec.h>
#include <exec/types.h>
#include <libraries/mathieeesp.h>
float IEEESPSqrt ()
- SYNOPSIS
- float y
- FUNCTION
- Calculate square root of IEEE single precision number
- INPUTS
- y
- IEEE single precision floating point number
- RESULT
- IEEE single precision number
flags:
zero : result is zero
negative : 0
overflow : square root could not be calculated
- NOTES
-
- EXAMPLE
- BUGS
-
- SEE ALSO
- MathIEEESingleTrans
- INTERNALS
- ALGORITHM:
First check for a zero and a negative argument and take
appropriate action.
y = M * 2^E
Exponent is an odd number:
y = ( M*2 ) * 2^ (E-1)
Now E' = E-1 is an even number and
-> sqrt(y) = sqrt(M) * sqrt(2) * sqrt (2^E')
= sqrt(M) * sqrt(2) * 2^(E'/2)
(with sqrt(M*2)>1)
= sqrt(M) * sqrt(2) * 2^(E'/2)
= sqrt(M) * 1/sqrt(2) * 2^(1+(E'/2))
= sqrt(M/2) * 2^(1+(E'/2))
Exponent is an even number:
-> sqrt(y) = sqrt(M) * sqrt (2^E) =
= sqrt(M) * 2^(E/2)
Now calculate the square root of the mantisse.
The following algorithm calculates the square of a number + delta
and compares it to the mantisse. If the square of that number +
delta is less than the mantisse then keep that number + delta.
Otherwise calculate a lower offset and try again.
Start out with number = 0;
Exponent = -1;
Root = 0;
repeat
{
if ( (Root + 2^Exponent)^2 < Mantisse)
Root += 2^Exponent
Exponent --;
}
until you`re happy with the accuracy
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