这段代码对我来说很复杂。
$sql5 = "SELECT id, value FROM pc_taxes WHERE id = :taxId AND client_id = :clientID LIMIT 1";
$stmt5 = getaccessdata::getInstance()->prepare($sql5);
$stmt5->bindValue(':clientID', $client_id, PDO::PARAM_INT);
$stmt5->bindValue(':taxId', $taxID, PDO::PARAM_INT);
try {
$stmt5->execute();
$req5 = $stmt5->fetchAll(PDO::FETCH_ASSOC);
$countTaxVal = 0;
foreach ($req5 as $r5) {
$countTaxVal++;
}
} catch (Exception $e) {
}
if($countTaxVal>0){
$rowExist=true;
}
我如何用C ++跟踪这段代码,我不知道哪个算子先行动。
答案 0 :(得分:2)
参见例如
除法和乘法首先从左到右完成:
n = ((n/100) * 100) + ((n%100) / 10) + ((n%10) * 10)
^-------------^ ^------------^ ^-----------^
然后从左到右添加:
n = ( ( ((n/100) * 100) + ((n%100) / 10) ) + ((n%10) * 10) )
| ^-----------------------------------^ |
+--------------------------------------------------------+
最后是作业。
答案 1 :(得分:1)
结果是总和3部分:
temp1 = n / 100; # it is integer division (decimal part is throwing out)
temp1 = temp1 * 100;
temp2 = n % 100; # remainder after division by 100, i. e. last 2 digits
temp2 = temp2 / 10;
temp3 = n % 10; # remainder after division by 10, i. e. last digit
temp3 = temp3 * 10;
n = temp1 + temp2 + temp3;