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Making the most of a long

Posted on 2003-11-24
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Last Modified: 2010-04-01
An external library gives me a user-defined field of type long .

Trouble is I need to store flags in it AND i need to store an integer in it.

I have 6 flags and I don't need to store any number greater than 2^24 so I figured this is possible.

I can reserve 1 byte for the flags and use the other 3 to store the integer.

How can I do this?
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Question by:Sandra-24
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by:Karl Heinz Kremer
Comment Utility
Do you need to store negative values? If so, it gets a bit more complicated. If not, try the following:

        long l;

        long value = 10000;
        char flags = 0x3f;

        // store the value and the flags in the long variable

        l = flags << 24 | value;

        // extract the flags

        flags = l >> 24;

        // extract the value
        value = l & 0xffffff;

        printf("flags = %x, value = %u\n", flags, value);
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by:ashoooo
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You can also try storing a pointer to a struct which could contain whatever you want. But again, working with pointers is very implementation specific.
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bcladd earned 400 total points
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long stuffed;
const long flag0 = 0x01;
const long flag1 = 0x02;
const long flag2 = 0x04;
const long flag3 = 0x08;
//etc through 7

int i;

stuffed = i << 8; // moves integer over to the high end of the word

stuffed |= flag1; // set a flag with or
stuffed &= ~flag3; // unset with and and not

if (stuffed & flag2) {
}

i = stuffed >> 8; // shift right maintaining sign.

Ask if anything here is unclear.

Hope this helps, -bcl
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Expert Comment

by:bcladd
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Big-endian, little-endian.

You say to-MAY-to, I say to-MAH-to.

-bcl
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Assisted Solution

by:jkr
jkr earned 100 total points
Comment Utility
Use your own 'long':

struct flagged_long {

    typedef unsigned char flags;

    flagged_long () : m_long ( 0) {};
    flagged_long ( long l) : m_long ( l & 0x00ffffff) {};

    long operator long () const { return m_long & 0x00ffffff;}

    flags operator flags () const { return ( m_long & 0xff000000) >> 24;}

    flagged_long operator=(const long& l) { m_long = l & 0x00ffffff; return *this;}

    flagged_long operator=(const flags& l) { m_long |= l; return *this;}
    flagged_long operator=(const flagged_long& l) { m_long = l.m_long; return *this;}

protected:
    long m_long;
};

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by:bcladd
Comment Utility
jkr-

One possible downside of your implicit cast to long is that when the item is stored in the database (passed through some interface, whatever), the type of the provided element is most likely a long and your implicit cast will slice off the flag bits silently.  Might want to make the cast explicit OR use an accessor funciton for one or the other versions of the integer value.

-bcl
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by:jkr
Comment Utility
>>Might want to make the cast explicit OR use an accessor funciton

Argl, I knew I forgot something :o)

   long get_long () const { return m_long & 0x00ffffff;}

   flags get_flags () const { return ( m_long & 0xff000000) >> 24;}
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Expert Comment

by:bcladd
Comment Utility
One reason for storing the number in the upper 24 bits of the long is that the >> operator will replicate the sign bit when the number is right-shifted. This makes storing signed numbers straightforward. Anding with 0x00FFFFFF will simply chop off the upper eight bits without regard for the sign of the numeric field. You could always return  m_long << 8 >> 8 but that doesn't look very pretty.

Just thinking out loud here.
-bcl
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Expert Comment

by:itsmeandnobodyelse
Comment Utility
Try this:

struct MyLong
{
     long              value :24;   // 24 bits signed
     unsigned int   flag1 :1;     // 1 bit   0 or 1
     unsigned int   flag2 :1;
     unsigned int   flag3 :1;
     unsigned int   flag4 :1;
     unsigned int   flag5 :1;
     unsigned int   flag6 :1;
};



int main(int argc, char* argv[])
{
     MyLong  ml;

     ml.value = 1234567;
     ml.flag1 = 1;
     ml.flag2 = 0;
     ml.flag3 = 1;
     ml.flag4 = 0;
     ml.flag5 = 1;
     ml.flag6 = 0;

     int size = sizeof(MyLong);

     // copy to long
     long  l;
     memcpy(&l, &ml, sizeof(long));

     // copy to MyLong
     MyLong   ml1;
     memcpy(&ml, &l, sizeof(long));

     return 0;
}


Regards Alex
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Author Comment

by:Sandra-24
Comment Utility
Thanks for all the responses!

jkr - I couldn't get your code to work, here's how I used it:

flagged_long l;
flagged_long::flags f = 128;
long foo = 78452;
l = foo;
l = f;
std::cout << l.get_long() << std::endl;
std::cout << l.get_flags() << std::endl;

bcladd - Your code worked fine, but with one exception.

if you set some flags, then set the int, it erases the flags.

Is there an easy way around that?
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Author Comment

by:Sandra-24
Comment Utility
Sorry jkr, I meant to add that what was happening was that get_long returned 78452+128 = 78580 and get_flags did nothing.

I don't understand enough of this stuff to offer more than that.
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Assisted Solution

by:bcladd
bcladd earned 400 total points
Comment Utility
Yep, my mistake:

stuffed = i << 8 | (stuffed & 0xFF);

 
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Author Comment

by:Sandra-24
Comment Utility
Thanks, that did the trick. I rolled the code into a struct similar to jkr's to make things simple. An elegant solution if I do say so myself. Thanks guys.

-Sandra
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