List of shiftLeft() Examples

Examples of shiftLeft()

com.hp.hpl.jena.tdb.index.bplustree.BPTreeRecords.shiftLeft()
com.sun.enterprise.admin.jmx.remote.internal.Shifter.shiftLeft()
java.math.BigInteger.shiftLeft()
Returns a BigInteger whose value is {@code (this << n)}. The shift distance, {@code n}, may be negative, in which case this method performs a right shift. (Computes floor(this * 2ⁿ).) @param n shift distance, in bits. @return {@code this << n} @see #shiftRight
org.eclipse.ui.internal.texteditor.quickdiff.compare.rangedifferencer.RangeDifference.shiftLeft()
Shifts the offset into the left document of the receiver. @param shift the number of elements to shift
org.rabinfingerprint.polynomial.Polynomial.shiftLeft()
Computes (this << shift) in GF(2^k)

Examples of java.math.BigInteger.shiftLeft()

        int result = -1;
        BigInteger p = ONE;
        while (p.compareTo(a) <= 0)
        {
            result++;
            p = p.shiftLeft(1);
        }
        return result;
    }


    /**

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Examples of java.math.BigInteger.shiftLeft()

                mhi = BigInteger.valueOf(1<<Log2P);
            }


            b = b.shiftLeft(e[0] + s2);
            BigInteger s = BigInteger.valueOf(1);
            s = s.shiftLeft(s2);
            /* At this point we have the following:
             *   s = 2^s2;
             *   1 > df = b/2^s2 > 0;
             *   (d - prevDouble(d))/2 = mlo/2^s2;
             *   (nextDouble(d) - d)/2 = mhi/2^s2. */

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Examples of java.math.BigInteger.shiftLeft()

            m2 += i;
            s2 += i;
        }
        /* Now S*2^s2 has exactly four leading zero bits in its most significant word. */
        if (b2 > 0)
            b = b.shiftLeft(b2);
        if (s2 > 0)
            S = S.shiftLeft(s2);
        /* Now we have d/10^k = b/S and
           (mhi * 2^m2) / S = maximum acceptable error, divided by 10^k. */
        if (k_check) {

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Examples of java.math.BigInteger.shiftLeft()

                            && ((word1(d) & 1) == 0)
                    )) {
                    if (j1 > 0) {
                        /* Either dig or dig+1 would work here as the least significant decimal digit.
                           Use whichever would produce a decimal value closer to d. */
                        b = b.shiftLeft(1);
                        j1 = b.compareTo(S);
                        if (((j1 > 0) || (j1 == 0 && (((dig & 1) == 1) || biasUp)))
                            && (dig++ == '9')) {
                                buf.append('9');
                                if (roundOff(buf)) {

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Examples of java.math.BigInteger.shiftLeft()

                b = b.multiply(BigInteger.valueOf(10));
            }


        /* Round off last digit */


        b = b.shiftLeft(1);
        j = b.compareTo(S);
        if ((j > 0) || (j == 0 && (((dig & 1) == 1) || biasUp))) {
//        roundoff:
//            while(*--s == '9')
//                if (s == buf) {

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Examples of java.math.BigInteger.shiftLeft()

                mantissa = -mantissa;
            }
            exp -= 1075;
            BigInteger x = BigInteger.valueOf(mantissa);
            if (exp > 0) {
                x = x.shiftLeft(exp);
            } else if (exp < 0) {
                x = x.shiftRight(-exp);
            }
            intDigits = x.toString(base);
        }

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Examples of java.math.BigInteger.shiftLeft()

                mhi = BigInteger.valueOf(1<<Log2P);
            }


            b = b.shiftLeft(e[0] + s2);
            BigInteger s = BigInteger.valueOf(1);
            s = s.shiftLeft(s2);
            /* At this point we have the following:
             *   s = 2^s2;
             *   1 > df = b/2^s2 > 0;
             *   (d - prevDouble(d))/2 = mlo/2^s2;
             *   (nextDouble(d) - d)/2 = mhi/2^s2. */

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Examples of java.math.BigInteger.shiftLeft()

            m2 += i;
            s2 += i;
        }
        /* Now S*2^s2 has exactly four leading zero bits in its most significant word. */
        if (b2 > 0)
            b = b.shiftLeft(b2);
        if (s2 > 0)
            S = S.shiftLeft(s2);
        /* Now we have d/10^k = b/S and
           (mhi * 2^m2) / S = maximum acceptable error, divided by 10^k. */
        if (k_check) {

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Examples of java.math.BigInteger.shiftLeft()

                            && ((word1(d) & 1) == 0)
                    )) {
                    if (j1 > 0) {
                        /* Either dig or dig+1 would work here as the least significant decimal digit.
                           Use whichever would produce a decimal value closer to d. */
                        b = b.shiftLeft(1);
                        j1 = b.compareTo(S);
                        if (((j1 > 0) || (j1 == 0 && (((dig & 1) == 1) || biasUp)))
                            && (dig++ == '9')) {
                                buf.append('9');
                                if (roundOff(buf)) {

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Examples of java.math.BigInteger.shiftLeft()

                b = b.multiply(BigInteger.valueOf(10));
            }


        /* Round off last digit */


        b = b.shiftLeft(1);
        j = b.compareTo(S);
        if ((j > 0) || (j == 0 && (((dig & 1) == 1) || biasUp))) {
//        roundoff:
//            while(*--s == '9')
//                if (s == buf) {

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