Dog*_*Dog 27 java generics type-inference java-7 java-8
此代码在Java 8中编译,但无法在Java 7中编译:
class Map<K,V> {
static <K,V> Map<K,V> empty() {return null;}
Map<K,V> put(K k, V v) {return null;}
V get(K k) {return null;}
}
class A {
static void f(Map<Integer,String> m){}
public static void main(String[] args) {
f(Map.empty());
}
}
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它不会推断出Map返回的具体类型Map.empty():
$ javac7 A.java
A.java:10: error: method f in class A cannot be applied to given types;
f(Map.empty());
^
required: Map<Integer,String>
found: Map<Object,Object>
reason: actual argument Map<Object,Object> cannot be converted to Map<Integer,String> by method invocation conversion
1 error
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如果您将f呼叫更改为,则编译f(Map.<Integer,String>empty());.在Java 8中,它无需借助于此工作.
但是如果你将f调用更改为f(Map.empty().put(1,"A").put(2,"B"));,它将无法再次编译,在Java 7和8上.为什么?
$ $javac7 A.java
A.java:10: error: method f in class A cannot be applied to given types;
f(Map.empty().put(1,"A").put(2,"B"));
^
required: Map<Integer,String>
found: Map<Object,Object>
reason: actual argument Map<Object,Object> cannot be converted to Map<Integer,String> by method invocation conversion
1 error
$ $javac8 A.java
A.java:10: error: incompatible types: Map<Object,Object> cannot be converted to Map<Integer,String>
f(Map.empty().put(1,"A").put(2,"B"));
^
Note: Some messages have been simplified; recompile with -Xdiags:verbose to get full output
1 error
$ $javac8 -Xdiags:verbose A.java
A.java:10: error: method f in class A cannot be applied to given types;
f(Map.empty().put(1,"A").put(2,"B"));
^
required: Map<Integer,String>
found: Map<Object,Object>
reason: argument mismatch; Map<Object,Object> cannot be converted to Map<Integer,String>
1 error
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gon*_*ard 32
为什么?
因为泛型类型的类型推断尚未扩展为链式调用.
什么是目标类型的概念已经扩展到包括方法参数.
这就是为什么这段代码:
f(Map.empty());
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编译.
但是这段代码没有,因为这是一个链式调用:
f(Map.empty().put(1,"A").put(2,"B"));
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您还可以在JSR-000335 Lambda表达式中找到JavaTM编程语言评估最终版本的小段落(特别是D部分):
允许推理"链"有一些兴趣:在a().b()中,将类型信息从b的调用传递给a的调用.这为推理算法的复杂性增加了另一个维度,因为部分信息必须在两个方向上传递; 它只适用于a()的返回类型的擦除对于所有实例化(例如List)是固定的.此特征不适合多聚表达模型,因为目标类型不能轻易导出; 但也许还有其他增强功能,可以在将来添加.
也许在Java 9中.