如何判断数字是否可以精确地用浮点数表示？

Question

由于浮点数是基于 2 的数字系统，因此不可能0.24F直接表示为相同的数字，因此不可能在没有重复小数点的1/3十进制系统中表示，即或。1/3=0.3333...0.(3)

0.24F因此，当打印回十进制表示形式时，浮点数会0.23因四舍五入而发生变化：

println(0.24F) => 0.23999999463558197021484375

while0.25F可以直接显示：

println(0.25F) => 0.25

但是我如何确定一个数字可以完全表示呢？

isExactFloat(0.25F) ==> true
isExactFloat(0.24F) ==> false

也许 Java API 已经有一些函数可以做到这一点？

UPD 这是一个代码，显示 [-4, 4] 范围内的浮点数及其内部表示：

public class FloatDestructure {
    public static void main(String[] args) {
        BigDecimal dec = BigDecimal.valueOf(-4000L, 3);
        BigDecimal incr = BigDecimal.valueOf(1L, 3);
        for (int i = 0; i <= 8000; i++) {
            double dbl = dec.doubleValue();
            floatDestuct(dbl, dec);
            dec = dec.add(incr);
        }

    }
    static boolean isExactFloat(double d) { return d == (float) d; }

    static void floatDestuct(double val, BigDecimal dec) {
        float value = (float) val;
        int bits = Float.floatToIntBits(value);
        int sign = bits >>> 31;
        int exp = (bits >>> 23 & ((1 << 8) - 1)) - ((1 << 7) - 1);
        int mantissa = bits & ((1 << 23) - 1);
        float backToFloat = Float.intBitsToFloat((sign << 31) | (exp + ((1 << 7) - 1)) << 23 | mantissa);
        boolean exactFloat = isExactFloat(val);
        boolean exactFloatStr = Double.toString(value).length() <= 7;
        System.out.println(dec.toString() + " " + (double) val + " " + (double) value + " sign: " + sign + " exp: " + exp + " mantissa: " + mantissa + " " + Integer.toBinaryString(mantissa) + " " + (double) backToFloat + " " + exactFloat + " " + exactFloatStr);
    }
}

当尾数为零时，浮点数绝对是精确的。但在其他情况下，比如-0.375或 则-1.625不太清楚。

Answer 1

我想在这里分享这个功能。

// Determine whether number is exactly representable in double.
// i.e., No rounding to an approximation during the conversion.
// Results are valid for numbers in the range [2^-24, 2^52].

public static boolean isExactFloat(double val) {

    int exp2 = Math.getExponent(val);
    int exp10 = (int) Math.floor(Math.log10(Math.abs(val)));

    // check for any mismatch between the exact decimal and
    // the round-trip representation.
    int rightmost_bits = (52 - exp2) - (16 - exp10);

    // create bitmask for rightmost bits
    long mask = (1L << rightmost_bits) - 1;

    // test if all rightmost bits are 0's (i.e., no rounding)
    return (Double.doubleToLongBits(val) & mask) == 0;
}

编辑：上面的函数可以更短

public static boolean isExactFloat(double val) {

    int exp2 = Math.getExponent(val);
    int exp10 = (int) Math.floor(Math.log10(Math.abs(val)));

    long bits = Double.doubleToLongBits(val);

    // test if at least n rightmost bits are 0's (i.e., no rounding)
    return Long.numberOfTrailingZeros(bits) >= 36 - exp2 + exp10;        
}

演示