直接类型的Java vararg与通过扩展的通配符泛型之间有什么区别？

Question

以下两个Java方法声明有何不同：

public <S extends Item> void withExtra1(S... extra) {
    Collections.addAll(pool, extra);
}

和：

public void withExtra2(Item... extra) {
    Collections.addAll(pool, extra);
}

Answer 1

它们不会，因为S将被擦除为Item，所以两个签名最终都以Item...结尾。

Answer 2

Java规范说

Java编译器必须输出任何类，接口，构造函数或成员的通用签名信息，这些类，接口，构造函数或成员的Java编程语言中的通用签名将包括对类型变量或参数化类型的引用。

如果检查字节码，则可以看到具有泛型的方法具有不同的签名：

  public <S extends Item> void withExtra1(S...);
    descriptor: ([LItem;)V
    flags: (0x0081) ACC_PUBLIC, ACC_VARARGS
    Code:
      stack=2, locals=2, args_size=2
         0: aload_0
         1: getfield      #2                  // Field pool:Ljava/util/List;
         4: aload_1
         5: invokestatic  #3                  // Method java/util/Collections.addAll:(Ljava/util/Collection;[Ljava/lang/Object;)Z
         8: pop
         9: return
      LineNumberTable:
        line 9: 0
        line 10: 9
      LocalVariableTable:
        Start  Length  Slot  Name   Signature
            0      10     0  this   LItem;
            0      10     1 extra   [LItem;
      LocalVariableTypeTable:
        Start  Length  Slot  Name   Signature
            0      10     1 extra   [TS;
    Signature: #23                          // <S:LItem;>([TS;)V


  public void withExtra2(Item...);
    descriptor: ([LItem;)V
    flags: (0x0081) ACC_PUBLIC, ACC_VARARGS
    Code:
      stack=2, locals=2, args_size=2
         0: aload_0
         1: getfield      #2                  // Field pool:Ljava/util/List;
         4: aload_1
         5: invokestatic  #3                  // Method java/util/Collections.addAll:(Ljava/util/Collection;[Ljava/lang/Object;)Z
         8: pop
         9: return
      LineNumberTable:
        line 13: 0
        line 14: 9
      LocalVariableTable:
        Start  Length  Slot  Name   Signature
            0      10     0  this   LItem;
            0      10     1 extra   [LItem;

常量池中的[#23为#23 = Utf8 <S:LItem;>([TS;)V ] >>