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java-round函数源码

时间:2018-03-02 01:19:57      阅读:497      评论:0      收藏:0      [点我收藏+]

标签:else   integer   return   ini   expr   abs   number   let   ati   

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public static long round(double a) {
        long longBits = Double.doubleToRawLongBits(a);
        long biasedExp = (longBits & DoubleConsts.EXP_BIT_MASK)
                >> (DoubleConsts.SIGNIFICAND_WIDTH - 1);
        long shift = (DoubleConsts.SIGNIFICAND_WIDTH - 2
                + DoubleConsts.EXP_BIAS) - biasedExp;
        if ((shift & -64) == 0) { // shift >= 0 && shift < 64
            // a is a finite number such that pow(2,-64) <= ulp(a) < 1
            long r = ((longBits & DoubleConsts.SIGNIF_BIT_MASK)
                    | (DoubleConsts.SIGNIF_BIT_MASK + 1));
            if (longBits < 0) {
                r = -r;
            }
            // In the comments below each Java expression evaluates to the value
            // the corresponding mathematical expression:
            // (r) evaluates to a / ulp(a)
            // (r >> shift) evaluates to floor(a * 2)
            // ((r >> shift) + 1) evaluates to floor((a + 1/2) * 2)
            // (((r >> shift) + 1) >> 1) evaluates to floor(a + 1/2)
            return ((r >> shift) + 1) >> 1;
        } else {
            // a is either
            // - a finite number with abs(a) < exp(2,DoubleConsts.SIGNIFICAND_WIDTH-64) < 1/2
            // - a finite number with ulp(a) >= 1 and hence a is a mathematical integer
            // - an infinity or NaN
            return (long) a;
        }
    }

  

java-round函数源码

标签:else   integer   return   ini   expr   abs   number   let   ati   

原文地址:https://www.cnblogs.com/zhizhiyin/p/8491135.html

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