java Java泛型方法的返回类型
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Return Type of Java Generic Methods
提问by Jops
I wonder why generic methods which return nothing voidare (or can be) declared this way:
我想知道为什么不返回任何内容的泛型方法void（或可以）以这种方式声明：
public static <E> void printArray( E[] inputArray ) {
// Display array elements
for ( E element : inputArray ){
System.out.printf( "%s ", element );
}
System.out.println();
}
It seems like <E>is the type of the returned object, but the method returns nothing in fact. So what is the real meaning of <E>in this case specifically and in generic methods generally?
看起来像是<E>返回对象的类型，但实际上该方法没有返回任何内容。那么<E>在这种情况下具体和一般方法中的真正含义是什么？
回答by Jops
This question suits one of my old notes. I hope this illustration helps:
这个问题适合我的一个旧笔记。我希望这个插图有帮助：
回答by Jon Skeet
The <E>is the generic type parameter declaration. It means "this method has a single type parameter, called E, which can be any type".
该<E>是泛型类型参数声明。这意味着“此方法具有单个类型参数，称为E，可以是任何类型”。
It's not the return type - that comes afterthe type parameter declaration, just before the method name. So the return type of the printArraymethod in your question is still void.
这不是返回类型-随附后的类型参数声明，只是方法名之前。所以printArray你问题中方法的返回类型仍然是void.
See section 8.4 of the JLSfor more details about method declarations.
有关方法声明的更多详细信息，请参阅JLS 的第 8.4 节。
回答by JB Nizet
It's not the type of the returned object. It indicates that E, in the method signature, is a generic type and not a concrete type. Without it, the compiler would look for a class named Efor the argument of the method.
它不是返回对象的类型。它表明E，在方法签名中，是泛型类型而不是具体类型。没有它，编译器会寻找一个以E方法参数命名的类。
回答by Razvan
The < E > is called a formal type parameter. It is not the return type of the method. It basically says that the method can accept as parameters arrays of different types (E[] inputArray).
< E > 被称为形式类型参数。它不是方法的返回类型。它基本上是说该方法可以接受不同类型的参数数组（E [] inputArray）。
回答by Imran Tufail
Eis used as a placeholder for the actual type that will be passed to Gen function when this function will call.
E用作实际类型的占位符，该类型将在调用此函数时传递给 Gen 函数。
suppose Ecan be replaced by integer
假设E可以用整数代替
