为什么（长）9223372036854665200d给我9223372036854665216？ [英] Why is (long)9223372036854665200d giving me 9223372036854665216?

问题描述

我知道有精确错误的奇怪的东西，但我不能欺骗，

I know about weird stuff with precision errors, but I can't fathom,

为什么（长）9223372036854665200d 给我 9223372036854665216 ？

推荐答案

9223372036854665200d 是 double 类型的常量。但是， 9223372036854665200 不适用于 double ，且不会损失精度。 double 只有52位尾数，而有问题的数字需要63位来精确表示。

9223372036854665200d is a constant of type double. However, 9223372036854665200 does not fit in a double without loss of precision. A double only has 52 bits of mantissa, whereas the number in question requires 63 bits to be represented exactly.

最接近的 double 到 9223372036854665200d 是尾数等于的数字1.1111111111111111111111111111111111111111111110010100 ，其指数为63（十进制）。此数字不是 9223372036854665216 （称为 U ）。

The nearest double to 9223372036854665200d is the number whose mantissa equals 1.1111111111111111111111111111111111111111111110010100 in binary and whose exponent is 63 (decimal). This number is none other than 9223372036854665216 (call it U).

如果我们将尾数减去一个缺口到 1.1 ... 0011 ，我们得到 9223372036854664192 L ）。

If we decrease the mantissa one notch to 1.1...0011, we get 9223372036854664192 (call it L).

原始号码在和 U ，并且比 L U c>

The original number is between L and U and is much closer to U than it is to L

最后，如果你认为尾数的截断应该产生一个以一串零结尾的数字，你说得对。只有它发生在二进制，而不是十进制： U 在base-16是 0x7ffffffffffe5000 和 L 为 0x7ffffffffffe4c00 。

Finally, if you think that this truncation of the mantissa ought to result in a number that ends in a bunch of zeros, you're right. Only it happens in binary, not in decimal: U in base-16 is 0x7ffffffffffe5000 and L is 0x7ffffffffffe4c00.

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