静态< T扩展数&可比< ;?超级号码>> [英] static <T extends Number & Comparable<? super Number>>
问题描述
我有一个静态方法的类:
public class Helper {
public static < T扩展数字&可比< ;?超级号码>>布尔inRange(T值,T minRange,T maxRange){
//等价(值> = minRange&& value< = maxRange)
if(value.compareTo(minRange)> = 0&& amp; value.compareTo(maxRange)< = 0)
return true;
else
返回false;
}
}
我尝试调用这个方法:
整数值= 2;
整数最小值= 3;
整数最大值= 8;
Helper.inRange(value,min,max);
Netbeans编译器向我显示以下错误消息:
< block类>
方法inRange类Helper不能应用于给定的类型;
required:T,T,T
found:java.lang.Integer,java.lang.Integer,java.lang.Integer
reason:推断的类型不符合声明的bound(s )
推断:java.lang.Integer
bound(s):java.lang.Number,java.lang.Comparable
有什么想法?
谢谢。
试试< T扩展Number&比较< T>>
。
Integer
implements Comparable< Integer>
,它与 Comparable <? super Number>
(整数不是Number的超类)。 可比< ;? extends Number>
不会工作,因为Java会认为?
可以是 Number $ c的任何子类$ c>,并且传递一个
,而不是 T
到 compareTo
不会被编译,因为它需要一个<$ c $的参数c> T
。
编辑:如newacct所述,< T extends Number&可比< ;?因为
会接受任何?,所以超级T>> / code>其中
T
是一个子类,像往常一样,子类的一个实例可以作为一个参数给予超类。
I have following class with one static method:
public class Helper {
public static <T extends Number & Comparable<? super Number>> Boolean inRange(T value, T minRange, T maxRange) {
// equivalent (value >= minRange && value <= maxRange)
if (value.compareTo(minRange) >= 0 && value.compareTo(maxRange) <= 0)
return true;
else
return false;
}
}
I try to call this method:
Integer value = 2;
Integer min = 3;
Integer max = 8;
Helper.inRange(value, min, max) ;
Netbeans compiler show me this error message:
method inRange in class Helper cannot be applied to given types; required: T,T,T found: java.lang.Integer,java.lang.Integer,java.lang.Integer reason: inferred type does not conform to declared bound(s) inferred: java.lang.Integer bound(s): java.lang.Number,java.lang.Comparable
Any ideas?
thanks.
Try <T extends Number & Comparable<T>>
.
E.g. Integer
implements Comparable<Integer>
, which is not compatible with Comparable<? super Number>
(Integer is not a superclass of Number). Comparable<? extends Number>
would not work either because Java would then think the ?
could be any subclass of Number
, and passing a T
to compareTo
would then not compile because it expects a parameter of ?
, not T
.
Edit: as newacct said, <T extends Number & Comparable<? super T>>
will work too (and be slightly more general) since then compareTo
will then accept any ?
of which T
is a subclass, and as usual, an instance of a subclass can be given as a parameter where a superclass is expected.
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