Java集合的remove异常

Question

ArrayList<Integer>list=new ArrayList<>();//与变量名无关
//省略add方法节省空间
for(Integer a:list){
    if(a==30){
        list.remove(a);
    }
}

//这段代码就没有异常
ArrayList<String>arrayList=new ArrayList<>();
for(String s:arrayList){
    if("3".equals(s)){
        arrayList.remove(s);
    } 
}

为什么同样的形式上面会报并发修改异常、而下面的就不会
反编译源码形式也差不多

        Iterator var2 = arrayList.iterator();

while(var2.hasNext()) {
            String s = (String)var2.next();
            if (s == "3") {
                arrayList.remove(s);
            }
        }


 Iterator var2 = arrayList.iterator();

        while(var2.hasNext()) {
            Integer a = (Integer)var2.next();
            if (a == 30) {
                arrayList.remove(a);
            }
        }

Answer 1

ArrayList<String>arrayList=new ArrayList<>();
for(String s:arrayList){}

这一段代码反编译后

ArrayList<String> arrayList = new ArrayList();

String var3;
for(Iterator var2 = arrayList.iterator(); var2.hasNext(); var3 = (String)var2.next()) {
}

看上面，一开始调用了next方法。找到next原码如下

 public E next() {
            checkForComodification();//
            int i = cursor;
            if (i >= size)
                throw new NoSuchElementException();
            Object[] elementData = ArrayList.this.elementData;
            if (i >= elementData.length)
                throw new ConcurrentModificationException();
            cursor = i + 1;
            return (E) elementData[lastRet = i];
        }

关注标注释处、ArrayList的remove不会更正修改次数、故报异常===》并发访问异常。