blocks|key|2808137|text|从逻辑上讲，请考虑以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2808138|SandBox<String,+String>+sandBox+=+new+SandBox<String,+String>();
sandBox.put("foo");+//+which+put+is+invoked?|code-block|syntax|javascript|2808139|(尽管我应该承认，可以生成完全有效的代码来产生类似的情况-然后在“随机”中选择一种方法。)|2808140|从形式上讲，我认为this+section+of+the+JLS是相关的。两个版本的put都有相同的参数类型--如果我没看错的话。|offset|length|2808141|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/docs/books/jls/third_edition/html/classes.html#38649^0|0|0|0|9|N|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Logically, consider the following code:

<pre><code>SandBox&lt;String, String&gt; sandBox = new SandBox&lt;String, String&gt;();
sandBox.put("foo"); // which put is invoked?
</code></pre>

(Although I should admit that it's possible to produce perfectly valid code that produces similar situations - and then a method is chosen at "random".)

Formally, I think <a href="http://java.sun.com/docs/books/jls/third_edition/html/classes.html#38649" rel="noreferrer">this section of the JLS</a> is relevant. Both versions of put have the same argument types - if I read that section correctly.

blocks|key|4217129|text|Java中的泛型仅在代码级可用，在运行时不可用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4217130|因此，当编译器翻译你的源码时，你为A和B指定的类型就会被“擦除”。这意味着两种类型都设置为java.lang.Object。|offset|length|style|CODE|4217131|所以你最终得到了两个相同的方法签名。|4217132|entityMap^0|0|19|G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

Generics in Java are only available at code level and not at runtime.

So when the compiler translates your sourcecode, the types you specified for A and B are "erased". This means both types are set to <code>java.lang.Object</code>.

So you end up with two identical method signatures.

blocks|key|923480|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|923481|两个不同的泛型如何拥有相同的签名？那是完全分开的！|blockquote|923482|923483|923484|Java泛型是基于type+erasure的。这意味着你的代码基本上是这样编译的：|offset|length|923485|public+class+Sandbox{
+public+void+put(Object+b)+{
+}
+public+void+put(Object+a)+{
+}
}|code-block|syntax|javascript|923486|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/tutorial/java/generics/erasure.html^0|0|0|0|0|9|C|0|0|0^^$0|@$1|2|3|-4|4|5|6|W|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|X|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|Y|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|10|7|@]|8|@$H|11|I|12|1|13]]|9|$]]|$1|J|3|K|4|L|6|14|7|@]|8|@]|9|$M|N]]|$1|O|3|-4|4|5|6|15|7|@]|8|@]|9|$]]]|P|$Q|$4|R|S|T|9|$U|V]]]]

<blockquote>
 How do two different generic types
 have the same signature? There
 completely separate!
</blockquote>

Java generics are based on <a href="http://download.oracle.com/javase/tutorial/java/generics/erasure.html" rel="nofollow noreferrer">type erasure</a>. That means your code basically compiles to this:

<pre><code>public class Sandbox{
 public void put(Object b) {
 }
 public void put(Object a) {
 }
}
</code></pre>

blocks|key|923507|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|923508|是吧？两个不同的泛型如何拥有相同的签名？那是完全分开的！|blockquote|923509|923510|923511|问题的根源似乎是您没有很好地掌握Java术语。|923512|编译器是而不是，它告诉你两个泛型类型具有相同的签名：|offset|length|style|BOLD|923513|923514|类型(泛型或其他类型)没有正式意义上的签名。具有signatures.|unordered-list-item|923515|A和B的方法不是泛型类型。它们是类型和东西，编译器告诉你有相同的签名是put(A)和put(B)。这些是方法，不是类型。|CODE|923516|923517|编辑|923518|好的..。所以你确实理解这个术语。|923519|如果你不正确地使用技术术语，在以下情况下不要感到惊讶：|923520|923521|有些人曲解了你的问题(和答案)，|923522|有些人因为你不正确地使用术语而指责你，|923523|有些人认为你问的/说的是不知情的。|923524|923525|但我最担心的是这个案例(也是我花时间回答您问题的真正原因！)是当一些可怜的困惑的Java新手在谷歌上搜索时，发现你的SO问题充满了草率的术语，并且变得更加困惑。|923526|entityMap^0|0|0|0|0|0|4|3|0|0|0|0|1|2|1|Z|6|16|6|0|0|0|2|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1D|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1E|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1F|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|1G|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|1H|7|@]|8|@]|9|$]]|$1|H|3|I|4|5|6|1I|7|@$J|1J|K|1K|L|M]]|8|@]|9|$]]|$1|N|3|-4|4|5|6|1L|7|@]|8|@]|9|$]]|$1|O|3|P|4|Q|6|1M|7|@]|8|@]|9|$]]|$1|R|3|S|4|Q|6|1N|7|@$J|1O|K|1P|L|T]|$J|1Q|K|1R|L|T]|$J|1S|K|1T|L|T]|$J|1U|K|1V|L|T]]|8|@]|9|$]]|$1|U|3|-4|4|5|6|1W|7|@]|8|@]|9|$]]|$1|V|3|W|4|5|6|1X|7|@$J|1Y|K|1Z|L|M]]|8|@]|9|$]]|$1|X|3|Y|4|5|6|20|7|@]|8|@]|9|$]]|$1|Z|3|10|4|5|6|21|7|@]|8|@]|9|$]]|$1|11|3|-4|4|5|6|22|7|@]|8|@]|9|$]]|$1|12|3|13|4|Q|6|23|7|@]|8|@]|9|$]]|$1|14|3|15|4|Q|6|24|7|@]|8|@]|9|$]]|$1|16|3|17|4|Q|6|25|7|@]|8|@]|9|$]]|$1|18|3|-4|4|5|6|26|7|@]|8|@]|9|$]]|$1|19|3|1A|4|5|6|27|7|@]|8|@]|9|$]]|$1|1B|3|-4|4|5|6|28|7|@]|8|@]|9|$]]]|1C|$]]

<blockquote>
 Huh? How do two different generic types have the same signature? There completely separate!
</blockquote>

The root of your problem appears to be that you do not have a good grasp of Java terminology.

The compiler is not telling you that two generic types have the same signature:

<ul>
<li>Types (generic or otherwise) don't have signatures in the formal sense. Methods have signatures.</li>
<li><code>A</code> and <code>B</code> are not generic types. They are type parameters.</li>
<li>The things that the compiler tells you have the same signature are <code>put(A)</code> and <code>put(B)</code>. These are methods, not types.</li>
</ul>

EDIT

OK ... so you do understand the terminology.

If you use technical terminology incorrectly, don't be surprised if:

<ul>
<li>some people misinterpret your questions (and answers), </li>
<li>some people call you out for using terminology incorrectly, and </li>
<li>some people discount what you ask / say as being uninformed. </li>
</ul>

But the case I'm worried most about (and the real reason I took the time to answer your question!!) is when some poor confused Java newbie does a Google search, finds your SO question full of sloppy terminology, and gets even more confused.

I stumbled upon an interesting error that I've never seen before, and can't explain why

Consider the following class

<pre><code>public class Sandbox&lt;A,B&gt; {
 public void put(B b) {
 }
 public void put(A a) {
 }
}
</code></pre>

Looks okay to my eyes. So I compile it and then get this

<pre><code>name clash: put(B) and put(A) have the same erasure
</code></pre>

Huh? How do two different generic types have the same signature? There completely separate!

I'm probably missing something completly basic, but I've just not run into this issue before. I've band-aid fixed the problem by calling the methods <code>putA</code> and <code>putB</code>, but I'm really curious to why this error happened in the first place. 

Would someone mind explaining?

Two separate generic types have the same erasure?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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社区新版编辑器体验调研

我偶然发现了一个我以前从未见过的有趣的错误，并且无法解释为什么考虑下面的类public class Sandbox<A,B> { public void put(B b) { } public void put(A a) { }}在我看来没问题。所以我编译了它，然后得到了这个name clash: put(B) and put(A) have the same erasure哈?两个不同的泛型如何

问两个独立的泛型类型有相同的擦除吗？
EN

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问两个独立的泛型类型有相同的擦除吗？EN

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问两个独立的泛型类型有相同的擦除吗？
EN