java – ArrayList.Remove第一次调用时不起作用

该集会成员的初始化如下：

private List<String> pkgs;

并且在类构造器中

pkgs = new ArrayList<>();

当我调用pkgs.remove(String)时,它不起作用,但是当我反复尝试删除时,它最终会起作用.

继承人我如何测试删除是否有效(我编辑了代码以便更容易阅读)

private void togglePackage(String selectedPackage,CheckBox chk_app){

    String m_pkg = selectedPackage.toString(); //redundant .toString()
    boolean checked = !chk_app.isChecked(); //checkbox boolean toggle

    if (checked && !pkgs.contains(m_pkg)) { //if not already in arraylist
        pkgs.add(m_pkg); //adding the newly checked package
    } else if (!checked && pkgs.contains(m_pkg)) { //if it needs to be removed
        pkgs.remove(m_pkg); //<-----------------------This works around the 3rd time i press the checkbox
    }
    //Here i check if the string was actually removed from the arrylist
    //This following code will not be in production,i just used it for testing

    if (pkgs.contains(m_pkg)) {
        if (checked) {
            chk_app.setChecked(checked);//Success
        } else {
            chk_app.setChecked(!checked);//Failure
        }
     } else {
         if (!checked) {
            chk_app.setChecked(checked);//Success
         } else {
            chk_app.setChecked(!checked);//Failure
         }
     }
 }

这是未经编辑的onclick事件

RelativeLayout rl_container = (RelativeLayout) child.findViewById(R.id.rl_container);
               rl_container.setTag(pkg);
               rl_container.setOnClickListener(new View.OnClickListener() {
                    @Override
                    public void onClick(View v) {
                        String m_pkg = v.getTag().toString();
                        System.out.println("Pkg = "+m_pkg);
                        boolean checked = !chk_app.isChecked();
                        if (checked && !pkgs.contains(m_pkg)) {
                            pkgs.add(m_pkg);
                            System.out.println("Adding " + m_pkg);
                        } else if (!checked && pkgs.contains(m_pkg)) {
                            pkgs.remove(m_pkg);
                            System.out.println("Removing " + m_pkg);
                        }


                        if (pkgs.contains(m_pkg)) {
                            if (checked) {
                                System.out.println("Success");
                                chk_app.setChecked(checked);
                            } else {
                                System.out.println("Fail");
                                chk_app.setChecked(!checked);
                            }

                        } else {
                            if (!checked) {
                                System.out.println("Success");
                                chk_app.setChecked(checked);
                            } else {
                                System.out.println("Fail");
                                chk_app.setChecked(!checked);
                            }
                        }


                    }
                });

作为一个例子,我已经包含了通过迭代列表获得的arraylists内容.
以及我的测试的日志输出

清单内容：

com.sbg.mobile.phone

com.google.android.youtube 

com.e8tracks 

com.vlingo.midas 

com.google.android.googlequicksearchbox 

com.truecaller

当我尝试取消选择“com.sbg.mobile.phone”时,未经编辑的代码输出Logcat

12-18 10:37:25 ViewPostImeInputStage ACTION_DOWN
12-18 10:37:25 Pkg = com.sbg.mobile.phone
12-18 10:37:25 Removing com.sbg.mobile.phone
12-18 10:37:25 Fail
12-18 10:37:28 ViewPostImeInputStage ACTION_DOWN
12-18 10:37:28 Pkg = com.sbg.mobile.phone
12-18 10:37:28 Removing com.sbg.mobile.phone
12-18 10:37:28 Fail
12-18 10:37:30 ViewPostImeInputStage ACTION_DOWN
12-18 10:37:31 Pkg = com.sbg.mobile.phone
12-18 10:37:31 Removing com.sbg.mobile.phone
12-18 10:37:31 Fail
12-18 10:37:32 ViewPostImeInputStage ACTION_DOWN
12-18 10:37:32 Pkg = com.sbg.mobile.phone
12-18 10:37:32 Removing com.sbg.mobile.phone
12-18 10:37:32 Fail
12-18 10:37:33 ViewPostImeInputStage ACTION_DOWN
12-18 10:37:33 Pkg = com.sbg.mobile.phone
12-18 10:37:33 Removing com.sbg.mobile.phone 
12-18 10:37:33 Success

PS.这是我的第一个问题,所以要温柔.
我也很感激有关如何改进提问的任何建议,我试图包含所有必需的信息,但如果您还需要其他信息,请告诉我.