GPT generates an infinite loop error when outputting Java generics

ChatGPT Bugs
1

The issue of GPT outputting generics and then being unable to continue, causing a loop, has persisted for several days on both the web version of ChatGPT 3.5 and 4. I encountered this problem around 7:00 PM Pacific Time on April 25th, and it still exists. This problem persists even when using other computers.

image
image698×787 11.3 KB

3 Likes
2

question：

优化这段代码
public void filterEncounterData(List list, UIEncounterBean uiEncounterBean, RTMKitCallback callback) {
UIEncounterFilterBean filterBean = new UIEncounterFilterBean();
int index = findDuplicateIndex(list, uiEncounterBean);
if (index != -1) {
UIEncounterBean encounterBean = list.get(index);
uiEncounterBean.setContact(encounterBean.isContact());
}
filterBean.setUiEncounterBean(uiEncounterBean);
filterBean.setDuplicateIndex(index);
callback.onSuccess(filterBean);
}

private int findDuplicateIndex(List<UIEncounterBean> list, UIEncounterBean uiEncounterBean) {
    for (int i = 0; i < list.size(); i++) {
        if (list.get(i).getUserId().equals(uiEncounterBean.getUserId())) {
            return i;
        }
    }
    return -1;
}

answer

这段代码的目的是过滤 UIEncounterBean 列表中的数据，并将结果通过回调返回。这里有几个可以优化的地方：

  1. 避免不必要的对象创建：每次调用 list.get(index) 都会创建一个新的对象，可以直接在列表中获取对应的元素进行操作，避免了不必要的对象创建。
  2. 使用增强的 for 循环：使用增强的 for 循环可以简化代码，使其更易读。
  3. 命名规范：使用更具描述性的变量名和方法名。
  4. 空值检查：考虑添加对空列表和空元素的检查，以避免可能的空指针异常。

下面是优化后的代码：

public void filterEncounterData(List



```java
public void filterEncounterData(List

```java
public void filterEncounterData(List

```java
public void filterEncounterData(List

```java
public void filterEncounterData(List

```java


```java
public void filterEncounterData(List

```java


```java
public void filterEncounterData(List

```java


```java


```java
public void filterEncounterData(List

```java
public void filterEncounterData(List