How can I perform multiplication without the '*' operator?

Question

I was just going through some basic stuff as I am learning C. I came upon a question to multiply a number by 7 without using the * operator. Basically it\'s like this

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Answer 1

If you can use the log function:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    //Find the 2^b which is larger than "a" which turns out to be the 
    //ceiling of (Log base 2 of b) == numbers of digits to shift
    double logBase2 = Math.log(absB) / Math.log(2);
    long bits = (long)Math.ceil(logBase2);

    //Get the value of 2^bits
    long biggerInteger = (int)Math.pow(2, bits);

    //Find the difference of the bigger integer and "b"
    long difference = biggerInteger - absB;

    //Shift "bits" places to the left
    long result = absA<0) {
        result -= difference;
        diffLoop--;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

If you cannot use the log function:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    //Get the number of bits for a 2^(b+1) larger number
    int bits = 0;
    int bitInteger = absB;
    while (bitInteger>0) {
        bitInteger /= 2;
        bits++;
    }

    //Get the value of 2^bit
    long biggerInteger = (int)Math.pow(2, bits);

    //Find the difference of the bigger integer and "b"
    long difference = biggerInteger - absB;

    //Shift "bits" places to the left
    long result = absA<0) {
        result -= difference;
        diffLoop--;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

I found this to be more efficient:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    long result = 0L;
    while (absA>0) {
        if ((absA&1)>0) result += absB; //Is odd
        absA >>= 1;
        absB <<= 1;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

and yet another way.

public static final long multiplyUsingLogs(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);
    long result = Math.round(Math.pow(10, (Math.log10(absA)+Math.log10(absB))));
    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}