any hints to solving this?

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  1. #1
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    any hints to solving this?

    Convert (10.1101) binary to decimal

  2. #2
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    Code:
    #include <stdio.h>
    #include <math.h>
    #include <string.h>
    
    int main( void ) {
        char *bitstring = "10.1101";
        int decimalPoint = 0;        /* Location of decimal */ 
        int i = 0;                    /* Counter1 */ 
        int j = 0;                    /* Counter2 */
       double value = 0.00;
         
        /* Locate decimal point */ 
        for( i = 0; i <  strlen(bitstring); i++ ) { 
            if( bitstring[i] == '.' ) { 
                decimalPoint = i; 
                break; 
            } 
        } 
         
        /* Calculate exponent */ 
        for( i += 1, j = 1; i < strlen(bitstring); i++, j++ ) { 
            if( bitstring[i] == '1' ) { 
                value += (double)(1.00 / (double)pow(2.00,(double)j)); 
            } 
        } 
    
        /* Calculate Mantissa */
        for( i = decimalPoint-1, j = 0; i >= 0; i--, j++ ) { 
            if( bitstring[i] == '1' ) { 
                value += (double)pow(2.00,(double)j); 
            } 
        }
       
       printf("%s = %f\n", bitstring, value);
       return 0;
    }
    EDIT:
    Sorry I'll explain it a little..

    The bitstring is a series of 1's and 0's, the delimiter between the mantissa and exponent being the decimal point.
    Step 1: You need to find the location of the decimal point, and you also need to know the location of the LSB of the mantiisa - so you do this.
    Step 2: You've found the decimal, set i to decimalPoint and decrement i to point to the LSB of the mantissa and begin calculating powers. You start with i = 0 and (2^i)*bit and as you move towards the MSB you increment t i by 1. So you have the summation of( 2^i)*bit where i = (decimalPoint-1) to 0.
    Step 3: Now you need to exponent side of the bitstring, so you point to the next bit after the decimal. You start with i = 1 and (1.00 / (2^i))*bit to get the actual decimal value. As you move towards the LSB of the mantissa you need to increment i by 1. So you have the summation of (1.00 / (2^i)*bit, where i = 0 to strlen(bitstring)-decimalPoint.

    The summation of the value obtained from step 2 and the value obtained from step 3 is your decimal value.
    Last edited by saeculum; 03-12-2009 at 08:53 AM.

  3. #3
    spurious conceit MK27's Avatar
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    This converts binary numbers too:
    Code:
    #include <stdio.h>
    #include <string.h>
    
    int main (int argc, char *argv[]) {
    	int i, bit=1, result=0, len=strlen(argv[1]);
    	for (i=len-1;i>=0;i--) {
    		if (argv[1][i]!='1') { bit*=2; continue; }
    		result|=bit;
    		bit*=2;
    	}
    	printf("Binary %s is decimal %d\n",argv[1],result);
    	return 0;
    }
    It doesn't deal with decimals tho, and you can't convert the right side the same way I wouldn't think.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  4. #4
    Registered User hk_mp5kpdw's Avatar
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    Values to the left of the decimal represent powers of 2 (starting at 2 raised to the 0th power). Values to the right of the decimal represent powers of (starting at raised to the 1st power).

    Example: 1010.0101

    The values to the left of the decimal are 1010. These represent the value (1 * 2 ^ 3 + 0 * 2 ^ 2 + 1 * 2 ^ 1 + 0 * 2 ^ 0). Simplifying this you get (1 * 8 + 0 * 4 + 1 * 2 + 0 * 1) which is (8 + 2) or 10 (decimal).

    The values to the right of the decimal are 0101. These represent the value (0 * ^ 1 + 1 * ^ 2 + 0 * ^ 3 + 1 * ^ 4). Simplifying this you get ( 0 * 0.5 + 1 * 0.25 + 0 * 0.125 + 1 * 0.0625) which is (0.25 + 0.0625) or 0.3125.

    Combining these two you get 10.3125 as the final answer.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
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  5. #5
    spurious conceit MK27's Avatar
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    Thanks hk_mp5kpdw for that. So I think this is an optimized solution:
    Code:
    #include <stdio.h>
    #include <string.h>
    
    int getdot (char *string, int len) {
    	int i;
    	for (i=0;i<len;i++) if (string[i]=='.') return i;
    	return len;  /* no decimal places */
    }
    
    int main (int argc, char *argv[]) {
    	float result,right=0.0f, tmp;
    	int i, ii, left=0, len=strlen(argv[1]), dot=getdot(argv[1],len), bit=1;
    	
    		/* integer */	
    	for (i=dot-1;i>=0;i--) {
    		if (argv[1][i]!='1') { bit*=2; continue; }
    		left|=bit;
    		bit*=2;
    	}
    	result=(float)left;
    		/* integer fraction */
    	for (i=dot+1;i<len;i++) {
    		if (argv[1][i]!='1') continue;
    		tmp=0.5f;
    		for (ii=0;ii<i-dot-1;ii++) tmp/=2;
    		right+=tmp;
    	}
    	result+=right;	
    	printf("Binary %s is decimal %f\n",argv[1],result);
    	return 0;
    }
    
    Example output:
    [root~/C] ./a.out 1010.0101
    Binary 1010.0101 is decimal 10.312500
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  6. #6
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by hk_mp5kpdw View Post
    Values to the left of the decimal represent powers of 2 (starting at 2 raised to the 0th power). Values to the right of the decimal represent powers of (starting at raised to the 1st power).
    Or you could basically take it that every digit before or after the decimal point is a power of two. Those that are to the right of the decimal point are simply to a negative power of two.
    0.1 ==> 2^-1
    0.01 ==> 2^-2
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