如何解决'＆LT;＆GT;'运营商不允许低于1.6 1.7源代码级？ [英] How to fix '<>' operator is not allowed for source level below 1.7 in 1.6?

问题描述

我有一个Java 1.6的Andr​​oid项目。我有一个第三方的code，这并不编译：

I have a Java 1.6 Android project. I have a third-party code that does not compile:

import org.springframework.http.HttpEntity;
//...
HttpHeaders requestHeaders = new HttpHeaders();
//...
new HttpEntity<>(requestHeaders);

它说：'＆LT;>'经营者不得低于1.7源代码级

It says: '<>' operator is not allowed for source level below 1.7

我不希望我的项目切换到1.7。我已经改变了该行

I do not want to switch my project to 1.7. I have changed that line to

new HttpEntity<Object>(requestHeaders);

和现在编译罚款。

不过是我的修正是否正确？什么是Java的1.7跟空方括号？

But is my fix correct? What does Java 1.7 do with empty brackets?

更新

这是新的对象传递给接受 HttpEntity＆LT功能;＆GT; 参数。我明白了类型推断的想法，但我不明白什么呢从给定的code线1.7编译器推断。

That new object is passed to function that accepts HttpEntity<?> argument. I understand the idea of type inference, but I do not understand what does 1.7 compiler infer from the given code line.

推荐答案

您修复几乎是正确的，反正没有危险。

Your fix is almost correct and anyway not dangerous.

对象是层次结构和＆lt根;>表示让编译器推断类型，这样会在1.7被推断任何类型将是一个对象的专业化反正。

Object is the root of the hierarchy and <> means "let the compiler infer the type", so any type that would have been inferred in 1.7 would be a specialization of Object anyway.

在看到您的更新后：<？code>＆LT;＆GT; 其实就是通配符的（见这里），所以对象的罚款。

After having seen your update: <?> actually means "wildcard" (see here), so Object is fine.

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