1. Recently, during an online assessment, I encountered an interview question: design a high-precision addition function that supports adding long numbers without losing precision and ignores negative numbers (do not use built-in functions like bcadd).
2. Checking the PHP official website, bcadd — Adds two arbitrary precision numbers. As shown in Figure 1.
3. The final implementation code is as follows. I plan to refine it further when I have time later. Currently, the fault tolerance for request parameters is insufficient, and negative numbers are not supported.
PHP
<?php
function highPrecisionAdd(string $number1, string $number2, $scale = null) {
// 提取出整数与小数
$numArr1 = explode('.', $number1);
$numArr2 = explode('.', $number2);
$result = [];
// 计算小数的最长长度
$decimalMaxLen = max(strlen($numArr1[1]), strlen($numArr2[1]));
// 判断 $scale 是否为 null，如果是，则设置其为结果中小数点后的小数位数
if ($scale === null) {
$scale = $decimalMaxLen;
}
// 删除前导零并反转整数，然后在末尾连接小数
$number1 = strrev(ltrim($numArr1[0], '0') . str_pad($numArr1[1], $decimalMaxLen, '0'));
$number2 = strrev(ltrim($numArr2[0], '0') . str_pad($numArr2[1], $decimalMaxLen, '0'));
// 计算需要处理的最长长度
$maxLen = max(strlen($number1), strlen($number2));
// 填充两个数字，使它们的长度相等(都等于 $maxLen)
$number1 = str_pad($number1, $maxLen, '0');
$number2 = str_pad($number2, $maxLen, '0');
// 处理每个数字，保留个位，携带十位(余数)
for($i = 0; $i < $maxLen; $i++) {
$sum = ((int) $number1[$i] + (int) $number2[$i]);
if (isset($result[$i])) {
$sum += $result[$i];
}
$result[$i] = $sum % 10;
if ($sum > 9) {
$result[$i + 1] = 1;
}
}
// 将数组转换为字符串并将其反转
$result = strrev(implode($result));
// 计算出小数
$decimal = str_pad(substr($result, -$decimalMaxLen, $scale), $scale, '0');
// 计算出整数
$can = (($maxLen - $decimalMaxLen < 1) ? '0' : substr($result, 0, -$decimalMaxLen));
// 拼接整数与小数
$result = $can . (($scale > 0) ? '.' . $decimal : '');
return $result;
}
echo highPrecisionAdd('64474.47265', '24562.45124');
?>4. Adding 64474.47265 and 24562.45124 yields a result of 89036.92389, which meets expectations.