Extracting digits from a float C++

Question

Given double x, and assuming that it lies in [0,1] . Assume for example that x=0.3 In binary, (keeping 10 digits after the decimal point), it is represented as

x=0.0100110011...

I want to write some C++ code which will extract the 10 digits shown after the decimal point. In other words I want to extract the integer (0100110011)_2.

Now I am quite new to bit shifting and the (naive) solution which I have for the problem is the following

int temp= (int) (x*(1<<10))

Then temp in binary will have the necesary 10 digits.

Is this a safe way to perform the above process? OR are there safer / more correct ways to do this?

Note: I don't want the digits extracted in the form of a character array. I specifically want an integer (OR unsigned integer) for this. The reason for doing this is that in generation of octrees, points in space are given hash keys based on their position named as Morton Keys. These keys are usually stored as integers. After getting the integr keys for all the points they are then sorted. Theoretically these keys can be obtained by scaling the coordinates to [0,1], extracting the bits , and interleaving them.

Why not print the *entire* mantissa, with as many zeros prepended as necessary? — Kerrek SB, Dec 16 '11 at 17:42
@ Oli Charlesworth: Please see edit. I have clarified my question. — smilingbuddha, Dec 16 '11 at 17:46
For a positive value in the range `[0,1)` your method should work just fine. I've got to question your ability to count though, your example is only 9 bits. — Mark Ransom, Dec 16 '11 at 17:50
Is it what you want? http://stackoverflow.com/questions/5672960/how-can-i-extract-the-mantissa-of-a-double — kan, Dec 16 '11 at 17:55
@Mark Ransom Oops! :D Thanks for pointing that out. Corrected it now. — smilingbuddha, Dec 16 '11 at 17:56
You should also be aware that you won't get an even distribution due to Benford's Law: http://en.wikipedia.org/wiki/Benford%27s_law — Mark Ransom, Dec 16 '11 at 19:19
@TonyK,smilingbuddha: I'm sorry, I have no idea what I was thinking. I'm sure it made sense at the time, and someone agreed with me. :/ I looked at the edit history, but I still don't know what I was thinking. — Mooing Duck, May 11 '12 at 16:28

score 1 · Accepted Answer · answered Dec 16 '11 at 20:49

1

Use memcpy to copy double into an array of 32-bit numbers, like this:

unsigned int b[2]; // assume int is 32-bits
memcpy(b, &x, 8);

The most 10 significant binary digits are in b[0] or b[1], depending on whether your machine is big- or little-endian.

EDIT: The same can be achieved by some casting instead of memcpy, but that would violate strict aliasing rules. An alternative is to use a union.

answered Dec 16 '11 at 20:49

zvrba

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The OP asked for the 10 digits after the decimal point, not the 10 most significant digits. So the exponent also has to be taken into account. (But the OP's proposed solution is perfectly sufficient for the task, without having to twiddle bits.) – TonyK May 11 '12 at 16:26

score 0 · Answer 2 · answered Dec 16 '11 at 19:25

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Read this: http://chrishecker.com/images/f/fb/Gdmfp.pdf

If you can grok what that article is telling you, you can derive the algorithm you're looking for. Just remember the bias factor in the exponent and the implicit leading one in the mantissa and the rest should fall into place.

answered Dec 16 '11 at 19:25

mcmcc

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ack, reading comprehension failure. Your naive solution should work just as well. The article above is good info anyway. – mcmcc Dec 16 '11 at 19:49

Extracting digits from a float C++

2 Answers2