Question

就我所知,浮标可以准确反映14个数字。

So let s say we have

a = 564214623154
b = 54252

and we multiply this c=a*b and it should be 30609771735350808 but when compiled it shows me 3.0609771735351E+16 So as I understand it should lose some precision but when I divide c by a c/a I get 564214623154 exact result without any precision lost

另外一个例子让我们说:

c = 30609771735350808 
d = 30609761111111111

e=c-d应为10624239697,但在汇编后,显示我为1062423969696,因此失去精确度

So is precision lost only when I subtract or add two numbers?

If it matters I use php

Answer 1

It is possible to lose precision with multiplication and division also. PHP and JavaScript store numbers in IEEE-754 format with 52 bits of mantissa and 11 bits of exponent. Some integers are represented exactly and some are not.

Let s try these:

In Real Math (generated with Ruby):

45345657434523 * 9347287748322342 / 74387422372 = 5697991604786167788

PHP和Javad

45345657434523 * 9347287748322342 / 74387422372 = 5697991604786168000

因此,我们也在重复和分裂方面失去准确性。

EDIT:在重新审视被占领土上的问题时,似乎这不是一个很大的答案,因为结果包含超过15位精准数字。如果问题的意图是,每组数字在15位精度或更低的数位数上乘和分立,那么最终结果往往会保持相当的精确度(如果你不过度流入或流入)。因此,您可以乘以<代码>1.25E35* 2.5E7,并精确到3.125e+42。这是因为PHP和Java的成文将主要增加大量数字,并增加支出。然而,如果您的英文译文是英文版,则您的两种数值是1.25E35 + 2.5E7 = 1.25E35。这一权利,你们增加了2 500万,不会改变! 这是因为,正如《任择议定书》所述,你只获得14或15名精准数字。人工添加这两个数值如下:12000000000000000000000000000 + 25000。 14-15位数从左边开始计算,你可以 pick。

Bottom line is precision problems are more likely to arise with addition and subtraction. Good to be aware of.

Answer 2

In your first case you lose no precision, PHP is just formatting the larger number as a float. (Internally the number is kept as a float.) Try this go get the "precise" output:

$a = 564214623154;
$b = 54252;
$c = $a * $b;
printf("%u, %u
", $c, $c/$a);

其次,在<代码>c* d的情况下,您的两人个人人数已超过标准电子计算系统-64轨道浮动能力(即53比值,但至少需要55比值),因此,如果你储存这些数字,就已经失去精确性。

添加/减让期间失去精确性的问题称为“取消”: 您所花费的所有储存都被取消了。最后,你没有足够准确的比照来填补这些 man。 C est la vie.

想象你会再次坐在月球上,你对兄弟们在联合王国Worcester的耳光长度进行了两次测量。比较这两种测量,是因为你要求储存大量精确度。

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