bcadding 非常小的浮点数给出 0

Question

感谢堆栈，我了解了浮点不精度，所以我转到了 bc 函数。

这在“普通”浮点数上效果很好，但对于非常小的浮点数，比如10^-10 类型，bcadd 总是给出0。

谁能告诉我我做错了什么才能使这些小浮点数精确地相加？

提前非常感谢！

---

PHP

$numerator = 1;
$denominator = 1000000000;
$quotientOne = $numerator / $denominator;

$numerator = 1;
$denominator = 1000000000000000;
$quotientTwo = $numerator / $denominator;

$smallSum = bcadd($quotientOne, $quotientTwo, 100);

echo $quotientOne . "<br>";
echo $quotientTwo . "<br>";
echo $smallSum . "<br>";

给予

1.0E-9
1.0E-15
0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Answer 1

The / operator 返回浮点数或整数。它干扰了您正确使用 bcadd() 的尝试。在 bcadd() 有机会展示它的东西之前，浮点算术的局限性就会出现。

echo gettype($numerator / $denominator);
double

请改用bcdiv()。请注意，bc*() 函数将字符串作为参数，并且它们也返回一个字符串。

$ cat code/php/test.php
<?php
$num_1 = 1;
$denom_1 = 1000000000;

# Cast values to string type to be crystal clear about what you're doing.
$q_1 = bcdiv((string)$num_1, (string)$denom_1, strlen($denom_1));
printf("q_1: %s\n", $q_1);

$num_2 = 1;
$denom_2 = 1000000000000000;

# php will do an implicit conversion to string anyway, though.
$q_2 = bcdiv($num_2, $denom_2, strlen($denom_2));
printf("q_2: %s\n", $q_2);

printf("sum: %s\n", bcadd($q_1, $q_2, strlen($denom_2)));
?>

$ php code/php/test.php
q_1: 0.0000000010
q_2: 0.0000000000000010
sum: 0.0000000010000010

任意精度算术本质上比浮点算术慢。这就是您为获得数十、数百或数千位精度所付出的代价。