[英]Java: compilation error with generic function reference
Could someone help me to fix this compilation error, please? 有人可以帮我解决此编译错误吗?
I 'd like to define a map method in the generic interface called Suite and use it like this: 我想在称为Suite的通用接口中定义一个map方法,并像这样使用它:
Suite < Integer > suite2 = Suite.from("34", "78", "23").map(Integer::parseInt);
assertEquals(3, suite.size());
assertEquals(34, (int)suite.get(0));
assertEquals(78, (int)suite.get(1));
assertEquals(23, (int)suite.get(2));
The the call to the same method with a function and parameter compile well: 对具有函数和参数的相同方法的调用编译良好:
Suite<Integer> suite1 = Suite.from(1, 2).map(x -> x * 2);
assertEquals(2, suite.size());
assertEquals(2, (int)suite.get(0));
assertEquals(4, (int)suite.get(1));
So I've defined the method in the interface like this 所以我在这样的界面中定义了方法
public interface Suite<E> {
public <E> Suite<E> map(int i, Function<? super E, ? extends E> f);
}
Note: this is almost the same protytype as the map method of Stream class (except the paramter i ) 注意:这几乎与Stream类的map方法具有相同的原型(参数i除外)。
My problem is in my test, this line does not compile: 我的问题是在测试中,此行无法编译:
map(Integer::parseInt)
because of these errors: 由于这些错误:
- The type Integer does not define
toString(Object)
that is applicable here.Integer类型未定义此处适用的
toString(Object)
。- Type mismatch: cannot convert
Suite<Object>
toSuite<String>
类型不匹配:无法将
Suite<Object>
转换为Suite<String>
I'm tried to redefine the function with a Supplier<E>
but it does not work. 我试图用
Supplier<E>
重新定义该功能,但是它不起作用。
Function<? super E, ? extends E> function
Integer is not a super class of String, hence this fails: 整数不是String的超类,因此失败:
Suite <String> suite1 = Suite.span(1, Integer::toString);
reference: Difference between <? 参考: 区别<? super T> and <?
超级T>和<? extends T> in Java
在Java中扩展T>
Usually functional mapping expects type changing, eg argument and result types are diverged: 通常,函数映射期望类型更改,例如,参数和结果类型是不同的:
https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html#map-java.util.function.Function- https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html#map-java.util.function.Function-
Suite < Integer > suite2 = Suite.from("34", "78", "23").map(Integer::parseInt);
Suite <Integer> suite2 = Suite.from(“ 34”,“ 78”,“ 23”)。map(Integer :: parseInt);
Here you change String
type to Integer
, thus you can't use common E
generic name like 在这里,您将
String
类型更改为Integer
,因此您不能使用通用的E
通用名称,例如
public <E> Suite<E> map(int i, Function<? super E, ? extends E> f);
but: 但:
public <T, R> Suite<R> map(int i, Function<? super T, ? extends R> f);
or: 要么:
public interface Suite<E> {
public <R> Suite<R> map(int i, Function<? super E, ? extends R> f);
}
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