binary to decimal

This is a discussion on binary to decimal within the C Programming forums, part of the General Programming Boards category; Hi there, I am very new to c prog and most probly have thousands of questions to ask. As for ...

  1. #1
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    Question binary to decimal

    Hi there,

    I am very new to c prog and most probly have thousands of questions to ask.
    As for now, I have question on how do I make a conversion of Binary to Decimal. Having a standard in binary and standard out for decimal with an error msg where appropriate.

    1) Do I have to define the binary string (e.g. #define BSIZE 8) if I do that, does it mean that the standard in will have to be an exact of (e.g 10010001) and I can't input it as (e.g 101)

    2) How do I know when I can use a "char" or "int"? I know the function of it like char uses %d it is a single character value (e.g. ABC) and int uses %d (integer value)

    3) I know how to convert from binary to decimal in mathematical terms. However, I can't seem to have the logical ability (yet) to code it into a program.
    - in maths the binary will have the wt of 128, 64, 32, 16, 8, 4, 2,1
    - if I've to convert a binary of 101
    - it would be 101 = 4 + 1 = 5

    If I have to code it will it look like (will it?)
    - x=0 (binary standard in)
    - a=128 (the wt of the binary)
    - b=a/2 (the next wt)
    - x=b

    It is still a blunder to me.

    Really appreciate if anyone could help to clear this mud of mine. Being a novice is never a good feeling.

  2. #2
    and the hat of int overfl Salem's Avatar
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    Things like %d perform a conversion for you. So the only thing you can do for simple input is use %s

    char buff[100];
    scanf("%s",buff);

    Now you have "101" or "11001" or "11100011001101" typed in by the user

    Try converting "1234" in decimal
    base = 10
    result = 0
    loop
    - result = result * base
    - result = result + numeric value of char
    end loop
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  3. #3
    Registered User manofsteel972's Avatar
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    If you know the bitwise operators it can be very simple.

    You can rotate the bits and test for 1 or 0 using a mask if you AND 1 with the rotated number it will give you the bit then use a conditional statement to print either a 0 or a 1 to the screen.


    Code:
    int number=3;
    int number_copy=number;
    _rotr(number,1);  //rotates number right _rotl() goes left
    
    if(number & 1)  //if number & 1 is true
    {
    putchar('1');
    }
    putchar('0');
    you have to work with it a bit because if you shift it the wrong direction it prints it out in reverse but you get the idea

    [edit]
    OOPS wrong direction you wanted binary to decimal

    You could just do the reverse. For each 1 or 0 you can use the mask to set the bits and then rotate.
    Last edited by manofsteel972; 03-14-2004 at 07:04 AM.
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    Now I know what doesn't work.

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  4. #4
    Code Goddess Prelude's Avatar
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    >_rotr(number,1); //rotates number right _rotl() goes left
    My compiler must be broken. I just get undefined references concerning _rotr and _rotl.
    My best code is written with the delete key.

  5. #5
    Registered User manofsteel972's Avatar
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    Sorry am using Visual C++ compiler

    _rotl() and _rotr() are supposed to mimic the assembly language instructions for rotating bits.

    I suppose you could write your own using the bitwise shift operator>> and <<. Normally when you shift the bits left or right it just discards the bit and you eventually end up with 0. If you create a mask using the number for the bit you want to use (ie to test or set the 32nd bit set the mask=2^31) then you can use the following algorithm:

    for rotating right
    mask_lowbit=1; //2^0 the first bit set to 1
    mask_highbit=21474836848; //2^31 the 32nd bit set to 1

    test lowbit for 1 or 0 by ANDing (mask_lowbit & number)
    if result is 0 then simply shift number to the right
    if true then shift numbers to the right and set the 32nd bit by ORing(number | mask_highestbit)

    just reverse the process for shifting left
    Last edited by manofsteel972; 03-14-2004 at 05:26 PM.
    "Knowledge is proud that she knows so much; Wisdom is humble that she knows no more."
    -- Cowper

    Operating Systems=Slackware Linux 9.1,Windows 98/Xp
    Compilers=gcc 3.2.3, Visual C++ 6.0, DevC++(Mingw)

    You may teach a person from now until doom's day, but that person will only know what he learns himself.

    Now I know what doesn't work.

    A problem is understood by solving it, not by pondering it.

    For a bit of humor check out xkcd web comic http://xkcd.com/235/

  6. #6
    Code Goddess Prelude's Avatar
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    >_rotl() and _rotr() are supposed to mimic the assembly language instructions for rotating bits.
    I'm well aware of what they're supposed to do. My point was that the two functions/macros are not standard C. Maybe I missed the OP stating a platform and compiler where your suggestion would work, but you appear to be making far too many assumptions.
    My best code is written with the delete key.

  7. #7
    Registered User manofsteel972's Avatar
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    Thank you for setting me straight prelude.

    Your right. I just copied and paste one of my examples without thinking. I need to pay more attention to standards and compiler/op. Thank you. Anyway here is code example of home made functions.

    Code:
    unsigned int bit_rotate_right(unsigned int num, unsigned int numbits)
    {
        unsigned int lowbitmask=1;          // mask for 1st bit
        unsigned int highbitmask=2147483648;    //mask for 32nd bit 2^31
    
    	for(int i=0;i<numbits; i++)
    	    {
    		if(num & lowbitmask)
    		    {
    			num=num>>1;
    			num=(num|highbitmask); 
    		    }
    		else
    		    {
    			num=num>>1;
    		    }
    	    }
    return num;
    }
    
    unsigned int bit_rotate_left(unsigned int num, unsigned int numbits)
    {
        unsigned int lowbitmask=1;          // mask for 1st bit
        unsigned int highbitmask=2147483648;     //mask for 32nd bit 2^31
    
    	for(int i=0;i<numbits; i++)
    	    {
    		if(num &highbitmask )
    		    {
    			num=num<<1;
    			num=(num|lowbitmask); 
    		    }
    		else
    		    {
    			num=num<<1;
    		    }
    	    }
    return num;
    }
    "Knowledge is proud that she knows so much; Wisdom is humble that she knows no more."
    -- Cowper

    Operating Systems=Slackware Linux 9.1,Windows 98/Xp
    Compilers=gcc 3.2.3, Visual C++ 6.0, DevC++(Mingw)

    You may teach a person from now until doom's day, but that person will only know what he learns himself.

    Now I know what doesn't work.

    A problem is understood by solving it, not by pondering it.

    For a bit of humor check out xkcd web comic http://xkcd.com/235/

  8. #8
    Just Lurking Dave_Sinkula's Avatar
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    Code:
    unsigned int highbitmask=2147483648;
    For the high bit of an unsigned int, consider the following.
    Code:
    unsigned int highbitmask = (-1U >> 1) + 1;
    Also,
    Code:
    for(int i=0;i<numbits; i++)
    it may be preferable to conform to C89 until C99 implementations become more prevalent. Otherwise the distinction between C and C++ can be blurred.
    Last edited by Dave_Sinkula; 03-14-2004 at 08:41 PM.
    7. It is easier to write an incorrect program than understand a correct one.
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