POW（）似乎是由一个在这里被淘汰 [英] pow() seems to be out by one here

问题描述

这是怎么回事：

#include <stdio.h>
#include <math.h>
int main(void) {
    printf("17^12 = %lf\n", pow(17, 12));
    printf("17^13 = %lf\n", pow(17, 13));
    printf("17^14 = %lf\n", pow(17, 14));
}

我得到这样的输出：

I get this output:

17^12 = 582622237229761.000000
17^13 = 9904578032905936.000000
17^14 = 168377826559400928.000000

13和14不匹配钨ALPA CF：

12: 582622237229761.000000
    582622237229761

13: 9904578032905936.000000
    9904578032905937

14: 168377826559400928.000000
    168377826559400929

此外，这不是错了一些奇怪的分数 - 这正好跟一个错

Moreover, it's not wrong by some strange fraction - it's wrong by exactly one!

如果这倒给我什么深远的限制POW（）能为我做的，就是在那里，可以计算出该选择？我需要一个可以计算 X ^是，其中 X ^是总是小于ULLONG_MAX的功能。

If this is down to me reaching the limits of what pow() can do for me, is there an alternative that can calculate this? I need a function that can calculate x^y, where x^y is always less than ULLONG_MAX.

推荐答案

POW 适用于双击号码。这些重形式S * 2 ^电子商务present号码其中s是一个53位整数。因此，双击可存储所有整数低于2 ^ 53，但只的部分的上述2 ^ 53的整数。特别是，它只能重新present偶数> 2 ^ 53，由于E> 0的值始终是2的倍数。

pow works with double numbers. These represent numbers of the form s * 2^e where s is a 53 bit integer. Therefore double can store all integers below 2^53, but only some integers above 2^53. In particular, it can only represent even numbers > 2^53, since for e > 0 the value is always a multiple of 2.

17 ^ 13需要54位来准确重新present，所以E设定为1，从而计算出的值变成偶数。正确的值是奇数，所以这并不奇怪它由一个人的了。同样，17 ^ 14需要58位重新present。这也推后一个是机缘巧合（只要你不施加太大的数论），这恰好是一次性从32的多个的，这是粒度在该双击如此规模的数字四舍五入。

17^13 needs 54 bits to represent exactly, so e is set to 1 and hence the calculated value becomes even number. The correct value is odd, so it's not surprising it's off by one. Likewise, 17^14 takes 58 bits to represent. That it too is off by one is a lucky coincidence (as long as you don't apply too much number theory), it just happens to be one off from a multiple of 32, which is the granularity at which double numbers of that magnitude are rounded.

有关详细的整数幂，你应该使用整数一路。自己写的双击 - 免费幂程序。通过平方，如果是可以大量使用幂，但我相信它总是小于64，使这个问题没有实际意义。

For exact integer exponentiation, you should use integers all the way. Write your own double-free exponentiation routine. Use exponentiation by squaring if y can be large, but I assume it's always less than 64, making this issue moot.

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