Google javaparser IfStmt不计算后续< else-if> [英] Google javaparser IfStmt not counting consequent <else-if>
问题描述
我正在使用Google javaparser来解析java文件,当我尝试计算If语句时,似乎我无法获得else-if语句的数量。
I am using Google javaparser to parse the java file, when I try to count the "If" statement, it seems like I can not get the number of "else-if" statement.
例如,我想解析以下代码:
For example, I want to parse the following code:
if(i>1){
i++;
}else if(i>2){
i++;
}else if(i>3){
i++;
}else{
i++;
}
我想获得Cyclomatic的复杂性,以便我需要计算数量if和else-if。
当我使用访客模式时,我只能访问API中定义的IfStmt,代码如下:
I want to get the Cyclomatic complexity so that I need to count the number of "if" and "else-if". When I use Visitor pattern, I can only visit the "IfStmt" defined in the API, the code looks like:
private static class IfStmtVisitor extends VoidVisitorAdapter<Void> {
int i = 0;
@Override
public void visit(IfStmt n, Void arg) {
//visit a if statement, add 1
i++;
if (n.getElseStmt() != null) {
i++;
}
}
public int getNumber() {
return i;
}
}
无法获得else-if但是带有IfStmt的访问者模式将整个代码块视为一个ifStatement.So,我希望该数字为4,但它是2。
There is no way to get "else-if" but the Visitor pattern with "IfStmt" treats the whole code block as one "if" Statement.So, I expect the number to be 4, but it is 2.
任何人都有一些想法?
推荐答案
if语句只包含一个then Statement和一个else Statement。 else语句可以是隐藏的if语句。所以有一个递归。为了跟踪您所需的复杂性,以下递归方法可能有所帮助:
A if-statement only contains one "then Statement" and one "else Statement". The else Statement can be a hidden if statement. So there is a recursivity. To track your needed complexity the following recursive method may help:
private static class IfStmtVisitor extends VoidVisitorAdapter<Void> {
int i = 0;
@Override
public void visit(IfStmt n, Void arg)
{
cyclomaticCount(n);
}
private void cyclomaticCount(IfStmt n)
{
// one for the if-then
i++;
Statement elseStmt = n.getElseStmt();
if (elseStmt != null)
{
if ( IfStmt.class.isAssignableFrom(elseStmt.getClass()))
{
cyclomaticCount((IfStmt) elseStmt);
}
else
{
// another for the else
i++;
}
}
}
public int getNumber() {
return i;
}
}
希望有所帮助。
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