truncating least n significant bits in a double precison number

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assembly code will do.I want to integrate it into my embarcadero c++ polygon clipping algorithm.
I need a c++ function which takes a double precision number and strips of the last n lsb bits from
a double.
I am only a occasional c++ programmer.

Posted 24-Aug-12 3:40am

Shunya

Updated 24-Aug-12 4:54am

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[no name] 24-Aug-12 9:44am

Okay go right ahead and do that. Perhaps you meant to ask some sort of a question?

Christian Graus 24-Aug-12 9:46am

Surely someone who can write a 'polygon clipping algorithm' can write a bit of code to strip bits from a number ?

[no name] 24-Aug-12 9:48am

You would think that someone that has been a member of CP for the past 8+ years would know how to use the site by now.

Shunya 24-Aug-12 10:48am

sorry,i come to the site rarely.
My need was how to write a c++ function like round which returns a value
back after stripping the last n lsb bits.Input is a double precision number.

Shunya 24-Aug-12 10:48am

You think so?.

Christian Graus 24-Aug-12 10:51am

Yes. At least you should try and post some code, instead of just asking us to do it. You're just stripping bytes, which is done with boolean operators.

Shunya 24-Aug-12 11:01am

Yes,Yes!, I will try with some C code and post the actual here.

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JackDingler · Accepted Answer · 2012-08-24T05:53:00

I suspect that the reason you think you want this, is because you're getting comparison errors from artifacts in the double precision math.

Numbers that should be mathematically equivalent according to the formulas they were produced from, are not equivalent in practice.

Stripping bits is a shortcut that won't solve the problem. It'll introduce it's own errors.

Applying error ranges to your operations, is the only reliable way to make this work.

Take a look at this function to see how it's done.

C++

bool AlmostEqualDoubles2(double nVal1, double nVal2, double nEpsilon)
{
	bool bRet = (((nVal2 - nEpsilon) < nVal1) && (nVal1 < (nVal2 + nEpsilon)));
	return bRet;
}

Reliable Floating Point Equality Comparison[^]

Or possibly..

C++

#define AlmostEqualDoubles2(nVal1, nVal2, nEpsilon) \
    ((((nVal2) - (nEpsilon)) < (nVal1)) && ((nVal1) < ((nVal2) + (nEpsilon))));

JackDingler · Accepted Answer · 2012-08-24T05:19:00

Something like this?

I'm not so sure you'll find it useful.

C++

// StripDoubleBits.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"


int _tmain(int argc, _TCHAR* argv[])
{
    int nBits = 4;
    double OriginalValue = 1.0 / 9.0;
    double StrippedValue = OriginalValue;
    ((unsigned __int32 *) &StrippedValue)[1] &= (0xFFFFFFFF << nBits);

    printf("Original Hex Value = %08X%08X\n", ((unsigned __int32 *) &OriginalValue)[0], ((unsigned __int32 *) &OriginalValue)[1]);
    printf("Stripped Hex Value = %08X%08X\n", ((unsigned __int32 *) &StrippedValue)[0], ((unsigned __int32 *) &StrippedValue)[1]);

    printf("Original Value = %0.17f\n", OriginalValue);
    printf("Stripped Value = %0.17f\n", StrippedValue);

    return 0;
}

lewax00 · Accepted Answer · 2012-08-24T05:21:00

I'm assuming you're using the MS compiler, but if not you may have to adjust the types (long long is the 64-bit integer type in the MS 32-bit compiler, which is why I used it here to match the size of double, but the size of primitives can vary between compilers).

C++

double stripBits(int n, double value) {
    long long intVal = *((long long*)&value); // Convert the value to the binary-equivalent integer type

    long long mask = ~0; // start mask at all 1's

    // set n bits to 0, starting at the end
    for (long long i = 0; i < n; ++i)
    {
        mask -= (1 << i);
    }

    intVal &= mask; // apply mask

    return *((double*)&intVal); // return value with mask applied as a double
}

truncating least n significant bits in a double precison number

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