Copy bit to bit

This is a discussion on Copy bit to bit within the C Programming forums, part of the General Programming Boards category; Dear list! I have 2 pointers, let's call them source_ptr and destination_ptr. I want to copy 1 bit from the ...

  1. #1
    Coder2Die4
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    Question Copy bit to bit

    Dear list!

    I have 2 pointers, let's call them source_ptr and destination_ptr. I want to copy 1 bit from the source byte to another bit of the destination byte. Are there a simple instruction to do that?

    Example: source bit 2 to destination bit 5

  2. #2
    Pursuing knowledge confuted's Avatar
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    Code:
    int source,dest,temp;
    
    source=43;
    
    dest=source&&2;  //2 is 0000 0010 in binary
    dest<<3;  //shift it over to 3 bits so it's in the right spot
    That should work, but it involves a logical AND and bit shifting, which are likely to confuse you. I know of no *simple* ways to do it, and if you're trying to deal specifically with just bit 5 in dest and not affect the other bits, my code there won't work. You might want to check into bitfields, but chances are very good that whatever you're trying to accomplish can be done without manipulating bits like that. What are you trying to do, btw?
    Away.

  3. #3
    Super Moderator VirtualAce's Avatar
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    I agree and cannot see why you would need to do this.


    Code:
    typedef unsigned long DWORD;
    typedef unsigned int WORD;
    
    DWORD ReplaceBit(DWORD source,DWORD target,WORD sourcebit,WORD destbit)
    {
      WORD bitmask=(WORD)pow(2,sourcebit);
      WORD bitvalue=source&bitmask;
      if (bitvalue) target+=(WORD)pow(2,destbit);
      return target;
    }

    I've not tested this yet. I think it should work. It tests the sourcebit to see if it is 1 or not. If it is, it sets the destbit to 1 by adding pow(2,destbit) to the target value, if not then target does not need to be altered since the sourcebit evaluated to 0.

    This function should work on any 8,16, or 32-bit non-floating point value. It does not check to make sure that the destbit and sourcebit are valid nor does it check for overflow condition in the target. Bit positions start at 0.

    This function will not cause target to lose any data because it does not use shifting - unless target overflows.


    Try it out. I'm not at my house right now so I cannot compile it and see if it works.
    Last edited by VirtualAce; 06-25-2003 at 09:07 AM.

  4. #4
    Registered User Casey's Avatar
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    This works
    Code:
    #define bit(n) (1UL << (n))
    
    unsigned copybit(unsigned a, unsigned b, int abit, int bbit)
    {
        return (a & bit(abit)) ? b | bit(bbit) : b & ~bit(bbit);
    }
    It checks to see if the bit to be copied is set, if it is then the bit to be copied to is set, otherwise it's cleared. Not as easy as a direct assignment, but it does the trick. The operations look complicated, but they're totally idiomatic, just remember what they look like.

    a & bit(abit) tests a bit to see if it's set and returns zero if it isn't set and nonzero otherwise.

    b | bit(bbit) sets the selected bit to 1

    b & ~bit(bbit) clears the selected bit to 0

  5. #5
    Coder2Die4
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    Originally posted by blackrat364
    Code:
    int source,dest,temp;
    
    source=43;
    
    dest=source&&2;  //2 is 0000 0010 in binary
    dest<<3;  //shift it over to 3 bits so it's in the right spot
    That should work, but it involves a logical AND and bit shifting, which are likely to confuse you. I know of no *simple* ways to do it, and if you're trying to deal specifically with just bit 5 in dest and not affect the other bits, my code there won't work. You might want to check into bitfields, but chances are very good that whatever you're trying to accomplish can be done without manipulating bits like that. What are you trying to do, btw?
    Well, I just need some code that copy 1 bit from source to destination and overwrites the destination bit whatever the value is and at the same time do not affect the other bits in source or destination. Maybe I can test the destination bit and set the source to the same state? Any code example for that?

  6. #6
    Super Moderator VirtualAce's Avatar
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    Good job Casey. Much nicer and cleaner than my code.


  7. #7
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    Look at follwing Function:

    unsigned char copybit(unsigned char a, unsigned char b, int abit, int bbit)
    {
    if(((1<<(bbit-1))&b) == 0)
    return a & (~(1<<(abit-1)));
    else
    return a | (1<<(abit-1));
    }

    ===
    Copybit function replaces abit of a with bbit of b.

    Consider bit number as follows:

    Value : 1 1 1 1 1 1 1 1
    Bit Num : 8 7 6 5 4 3 2 1
    Chintan R Naik

  8. #8
    ATH0 quzah's Avatar
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    Originally posted by blackrat364
    Code:
    dest=source&&2;  //2 is 0000 0010 in binary
    dest<<3;  //shift it over to 3 bits so it's in the right spot
    This isn't really the way to do it. Why? Because you have your AND operators mixed up. You need a bitwise and, not a logical. (Assuming that you didn't just typo that.)

    The above expression evaluates to this:

    dest = (source is not zero and 2 is not zero).

    In which case, dest is now set to 1. If either of those two are not true, then dest is set to zero.

    If you are just trying to copy a single bit, rather, if you are just trying to make dest equal to that one specific bit, you'd do:
    Code:
    dest = source & bit;
    So here you'd have:
    Code:
    dest = source & 2;
    This would set dest to 2 if the bit for 2 is set. If it isn't, it sets it to zero.

    Quzah.
    Hope is the first step on the road to disappointment.

  9. #9
    Pursuing knowledge confuted's Avatar
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    It wasn't exactly a typo, but I know better. It was just an early morning lack of sleep stupid thing. Thanks for catching that Quzah.
    Away.

  10. #10
    ATH0 quzah's Avatar
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    Originally posted by blackrat364
    It wasn't exactly a typo, but I know better. It was just an early morning lack of sleep stupid thing. Thanks for catching that Quzah.
    Not a problem. I know how that goes. You should have seen me yesterday...

    Quzah.
    Hope is the first step on the road to disappointment.

  11. #11
    Toaster Zach L.'s Avatar
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    One other small thing: dest<<3;
    Ya just lost your shift there... How 'bout <<=.
    The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.

  12. #12
    Coder2Die4
    Guest
    Originally posted by Casey
    This works
    Code:
    #define bit(n) (1UL << (n))
    
    unsigned copybit(unsigned a, unsigned b, int abit, int bbit)
    {
        return (a & bit(abit)) ? b | bit(bbit) : b & ~bit(bbit);
    }
    It checks to see if the bit to be copied is set, if it is then the bit to be copied to is set, otherwise it's cleared. Not as easy as a direct assignment, but it does the trick. The operations look complicated, but they're totally idiomatic, just remember what they look like.

    a & bit(abit) tests a bit to see if it's set and returns zero if it isn't set and nonzero otherwise.

    b | bit(bbit) sets the selected bit to 1

    b & ~bit(bbit) clears the selected bit to 0
    Absolutely fantastic solution!! I have tested the code now in my app and it does the job without asking for trouble :-))



    Thanks for helping :-)

  13. #13
    ATH0 quzah's Avatar
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    Originally posted by Casey
    It checks to see if the bit to be copied is set, if it is then the bit to be copied to is set, otherwise it's cleared.
    Wouldn't XORing be easier? Because basicly that's what you're doing.
    Code:
    unsigned int foo( unsigned int a, unsigned int b )
    {
        return (a^(1<<b));
    }
    Your code basicly just toggles the specified bit, so you may as well just do that.

    Although, to be "safe", you'd probably want to "%(sizeof(b)<<3)" on B first to make sure it's actually in range...

    Quzah.
    Hope is the first step on the road to disappointment.

  14. #14
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    Your code basicly just toggles the specified bit, so you may as well just do that.
    no it doesn't. the code sets or resets a single bit in 'a' dependent on the value of a single bit in another int.
    DavT
    -----------------------------------------------

  15. #15
    ATH0 quzah's Avatar
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    Originally posted by DavT
    no it doesn't. the code sets or resets a single bit in 'a' dependent on the value of a single bit in another int.
    No, it actually doesn't even do that. Honestly, I am at a loss of how to use said function. Look at it again (I had to):
    Code:
    #define bit(n) (1UL << (n))
    
    unsigned copybit(unsigned a, unsigned b, int abit, int bbit)
    {
        return (a & bit(abit)) ? b | bit(bbit) : b & ~bit(bbit);
    }
    No, it's damn messed up. Took me the longest time to actually figure out what it's doing. Here, let me illustrate:
    Code:
    #include <stdio.h>
    
    #define bit(n) (1UL << (n))
    
    unsigned copybit(unsigned a, unsigned b, int abit, int bbit)
    {
        return (a & bit(abit)) ? b | bit(bbit) : b & ~bit(bbit);
    }
    
    void drawbit( unsigned int a )
    {
    	int b = 0;
    	for( b = 0; b < 32; b++ )
    		printf("%d", !!(a & (1<<b)) );
    	printf("\n");
    }
    
    void dump( unsigned int a, unsigned int b )
    {
    	int x,y;
    
    	for( x = y = 0; y < 16; y++ )
    	{
    		printf("x = %d, y is %2d: ", x, y );
    		drawbit( copybit( a, b, x, y ) );
    	}
    	printf("Hit enter.\n");
    	getchar( );
    }
    
    int main ( void )
    {
    	unsigned int a = 0xFFFFFFFF, b = 0xFFFFFFFF;
    
    	dump( a, b );
    
    	a = 0x00000000;
    	dump( a, b );
    
    	b = 0x00000000;
    	dump( a, b );
    	
    	a = 0xFFFFFFFF;	
    	dump( a, b );
    
    	return 0;
    }
    The code only sets the bit on b, if the bit in question on a is not set. If they are both set, nothing at all happens.

    As for practical application of a four argument bit toggle, I have no idea. This isn't meant as an insult to the author, I just am boggled by it. Maybe I'm tired. That's always a good excuse...

    On a side note, you are correct in saying I was wrong.

    [edit]
    Test output is way too huge to list.
    Picture 16 lines of 1s, followed by 16 lines where there is a zero that moves down the row of ones:
    011...
    101...
    110...
    Followed by 16 lines of 0s, followed by 16 lines of a traveling one.
    [/edit]

    [edit #2]
    Ok, nevermind. You and I were both wrong. (I finally wrapped my mind around it as I was about to drop into bead.) It performas as it should. (At least that's what I'm convincing myself of now.)

    The example output from the text function should confirm it.

    If both bits are the same, then it is indeed copying the value to that location, (in b), which is why we get a solid output of 1s. If they are different, it does in fact copy that result to said bit, which is why we get the "traveling 1" or "traveling 0" effect.

    We were both incorrect. The reason for the confusion was the way they worded what it does:
    It checks to see if the bit to be copied is set, if it is then the bit to be copied to is set, otherwise it's cleared.
    It should read:

    "If the source bit and the destination bit are both set, nothing changes. If one is set and the other isn't, the bit is 'toggled'."
    [/edit #2]

    Quzah.
    Last edited by quzah; 06-26-2003 at 05:12 AM.
    Hope is the first step on the road to disappointment.

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