| Re: Reading 32-bit complex numbers in IDL (16-bit real / 16-bit imaginary) [message #77232 is a reply to message #77117] |
Mon, 15 August 2011 10:58  |
Waqas A. Qazi
Messages: 3
Registered: August 2011
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Junior Member |
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On Aug 13, 6:26 am, Wox <s...@nomail.com> wrote:
> When I say "convert integer to float" I don't mean that you would
> convert 10 to 10.0 for example. The value of the integer is not
> concidered at all, it's the "bit-content" that is used. The integer is
> just a bag of bits that represent a foating point number according to
> the IEEE754 standard. That's why the function is call BINARYTOFLOAT
> and not INTEGERTOFLOAT or something. In the example I gave:
>
> IDL> integer='3555'x
> IDL> f=binarytofloat(integer,precision=0)
> IDL> print,integer
> 13653
> IDL> print,f
> 0.333252
>
> You can see that the integer value 13653 has nothing to do with
> 0.33325... However both numbers have the same binary representation,
> namely
>
> IDL> print,integer,format='(b016)'
> 0011010101010101
>
> Note: Since IDL can't handle 16bit floats, the binary representation
> of f (32bit float) isn't the same anymore as that of 13653. The value
> of f is correct however: the value represented by 13653 under the
> half-precision IEEE 754 convention.
Great, thanks Wox, I understand now.
For the specific data I am working with, its not half-precision per
se, because the 2-byte binary numbers are defined as integers, not
floats. So after reading the 2-byte integers from the binary file, I
will not need to convert to any float. I can read each alternating 2-
byte integer into separate real and imaginary part arrays, and then
input these two arrays directly into the complex function (data type
converter) in IDL.
Thanks,
Waqas.
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