Binary Algorithm C Programming

Question

So I have an algorithm designed to take in a 32 bit char array and translate it into its proper decimal equivalent.

double calculateDecimal(char *string){
    char *temp=string;
    double total=0;
    int i;
    int count=31;
    int x=pow(2,count);
    for(i=0;i<32;i++){
        if(!strncmp(temp,"1",1){
            printf("%d + %d\n",x,total);
            total+=x;
        }
        temp++;
        count--;
        x=pow(2,count);
    }
    printf("%d\n",total);
    return total;

}

1) My printf statements have proved that the proper 1's are being read and their proper powers are being calculated. What I have discovered is total and x keep equaling the same power which I'm confused about because I'm on the cusp.

2)My example calculation was 00000000100000010000000000000111 which if i typed in correctly should be the decimal equivalent of 84554151. Thanks for any contributions because I know I am close.

The problem is most certainly x = pow(2, count);, as x is an int, and pow() returns double. The integer equivalent of pow(2, count) is 1 << count. — jxh
– jxh, Commented Sep 21, 2013 at 0:10
And expanding on that, you probably want to just do x = 1 << count instead of x = pow(2,count). — Dennis Meng
– Dennis Meng, Commented Sep 21, 2013 at 0:11
I defined an array called char finalBinary[33] to make room for the null terminator. When I changed x to be a pow it opened up a whole new can of words with the proper pow's not being calculated. — Allie Steele
– Allie Steele, Commented Sep 21, 2013 at 0:13
You should declare x as unsigned. Otherwise, it will overflow when you try to set it to pow(2, 31). — Barmar
– Barmar, Commented Sep 21, 2013 at 0:15

Gene · Accepted Answer · 2013-09-21 00:16:53Z

2

If it's a 32-bit input, why not use a 32-bit integer to contain the result? The pow function is a floating point operation, which makes the task harder. Consider bit operations instead.

int toInt(char *str)
{
  int val = 0;
  while (*str) 
    val = (val << 1) | (*str++ == '1');
  return val;
}

Also note that by shifting the previous result left (multiplying by 2) each time a new character is found, this will work with any string up to 32 bits long.

answered Sep 21, 2013 at 0:16

Gene

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3 Comments

Barmar Over a year ago

I think your answer would be more helpful to the beginner if you split the calculation and increment up into separate statements, instead of trying to do it as tersely as C allows.

Gene Over a year ago

@Barmar I think not. Check again.

Gene Over a year ago

@Barmar He's doing fine. Looking up the idioms in K&R will be straightforward for him.

Tamás Zahola · Accepted Answer · 2013-09-21 00:26:24Z

Well, your code actually has a syntax error here (missing closing bracket):

if(!strncmp(temp,"1",1){

Regarding your problem of binary --> decimal conversion, I would recommend Horner's method of evaluating polynomials: http://en.wikipedia.org/wiki/Horner%27s_method

View the binary expansion as a polynomial where each coefficient is either 0 or 1. Evaluating this polynomial for x = 2 gives you the actual value, which you can print in decimal:

long calculateValue(char * string) {
    long result = 0;
    while(*string){
        result = ((*string) - '0') + result * 2;
        string++;
    }
    return result;
}

(And please don't use pow() and other floating-point functions for integer operations - especially for calculating powers of 2)

BTW, you can use this approach for evaluating numbers written in any base:

long calculateValue(char * string, int base) {
    long result = 0;
    while(*string){
        result = ((*string) - '0') + result * base;
        string++;
    }
    return result;
}

Of course this works for base 1-10, because '9' is followed by ':' in the ASCII table, but you get the idea.

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