这是一个迷你系列：*，《Java: The Good Parts》 http://www.amazon.com/Java-Good-Parts-Jim-Waldo/dp/059680373... http://www.amazon.com/PHP-Good-Parts-Delivering-Best/dp/0596...
it's a miniseries: *, the Good Parts http://www.amazon.com/Java-Good-Parts-Jim-Waldo/dp/059680373... http://www.amazon.com/PHP-Good-Parts-Delivering-Best/dp/0596...
所以很容易验证你不能把 int 加到一个强类型的字符串列表里，但仅限于源码层面。（我想是这样，我对 javac / JVM 内部机制没那么熟）你说得对。以下是 Jim Waldo 在《Java: The Good Parts》中的解释：“无法更改 VM 或字节码［由于向后兼容性的要求］意味着参数化……
So it was easy to check that you couldn't add a int to a strongly typed list of strings, but only in source. (I think, I'm not as well read as javac / JVM internals) You're right. Here's Jim Waldo's explanation from Java: The Good Parts: "the inability to change the VM or bytecodes [due to requirements of backwards compatibility] meant that parameterized…
这是一个迷你系列：*，《Java: The Good Parts》 http://www.amazon.com/Java-Good-Parts-Jim-Waldo/dp/059680373... http://www.amazon.com/PHP-Good-Parts-Delivering-Best/dp/0596...
it's a miniseries: *, the Good Parts http://www.amazon.com/Java-Good-Parts-Jim-Waldo/dp/059680373... http://www.amazon.com/PHP-Good-Parts-Delivering-Best/dp/0596...
所以很容易验证你不能把 int 加到一个强类型的字符串列表里，但仅限于源码层面。（我想是这样，我对 javac / JVM 内部机制没那么熟）你说得对。以下是 Jim Waldo 在《Java: The Good Parts》中的解释：“无法更改 VM 或字节码［由于向后兼容性的要求］意味着参数化……
So it was easy to check that you couldn't add a int to a strongly typed list of strings, but only in source. (I think, I'm not as well read as javac / JVM internals) You're right. Here's Jim Waldo's explanation from Java: The Good Parts: "the inability to change the VM or bytecodes [due to requirements of backwards compatibility] meant that parameterized…