强制类型而不创建临时变量
Coerce type without creating temporary variable
我有以下功能:
<K, P, Q> Map<K, List<Q>> convert(Map<K, ? extends List<? extends P>> input) {
return HashMap.empty();
}
我是这样调用的:
Map<Integer, List<? extends String>> stringMap = HashMap.empty();
Map<Integer, List<? extends Double>> doubleMap = HashMap.empty();
Map<Integer, List<Boolean>> stringMapConverted = convert(stringMap);
Map<Integer, List<Boolean>> merged = stringMapConverted.merge(convert(doubleMap));
但是,我想执行此操作而不必先创建临时变量。
像这样:
Map<Integer, List<? extends String>> stringMap = HashMap.empty();
Map<Integer, List<? extends Double>> doubleMap = HashMap.empty();
Map<Integer, List<Boolean>> merged = convert(stringMap).merge(convert(doubleMap));
但是,我在这里遇到编译错误:
'merge(io.vavr.collection.Map<? extends java.lang.Integer,? extends io.vavr.collection.List<java.lang.Object>>)' in 'io.vavr.collection.Map' cannot be applied to '(io.vavr.collection.Map<java.lang.Integer,io.vavr.collection.List<? extends java.lang.Double>>)'
Required type: Map<? extends Integer,? extends List<Object>>
Provided: Map<Integer,List<? extends Double>>
我也试过显式转换:
Map<Integer, List<Boolean>> merged2 = ((Map<Integer,List<Boolean>>)convert(stringMap)).merge(doubleMap);
但它抛出一个编译错误:
Inconvertible types; cannot cast 'io.vavr.collection.Map<java.lang.Integer,io.vavr.collection.List<java.lang.Object>>' to 'io.vavr.collection.Map<java.lang.Integer,io.vavr.collection.List<java.lang.Boolean>>'
Map.merge()
签名如下:
Map<K, V> merge(Map<? extends K, ? extends V> that);
地图和列表 类 来自 vavr。
调用convert
时使用type witness, or so called TypeArguments
mentioned in 15.12 in SE11 JLS指定参数类型。
MethodInvocation:
MethodName ( [ArgumentList] )
TypeName . [TypeArguments] Identifier ( [ArgumentList] )
ExpressionName . [TypeArguments] Identifier ( [ArgumentList] )
Primary . [TypeArguments] Identifier ( [ArgumentList] )
super . [TypeArguments] Identifier ( [ArgumentList] )
TypeName . super . [TypeArguments] Identifier ( [ArgumentList] )
ArgumentList: Expression {, Expression}
注意在使用TypeArguments
时不能省略实例名->(this).
Map<Integer, List<Boolean>> merged = this.<Integer, String, Boolean>convert(stringMap).merge(convert(doubleMap));
我有以下功能:
<K, P, Q> Map<K, List<Q>> convert(Map<K, ? extends List<? extends P>> input) {
return HashMap.empty();
}
我是这样调用的:
Map<Integer, List<? extends String>> stringMap = HashMap.empty();
Map<Integer, List<? extends Double>> doubleMap = HashMap.empty();
Map<Integer, List<Boolean>> stringMapConverted = convert(stringMap);
Map<Integer, List<Boolean>> merged = stringMapConverted.merge(convert(doubleMap));
但是,我想执行此操作而不必先创建临时变量。
像这样:
Map<Integer, List<? extends String>> stringMap = HashMap.empty();
Map<Integer, List<? extends Double>> doubleMap = HashMap.empty();
Map<Integer, List<Boolean>> merged = convert(stringMap).merge(convert(doubleMap));
但是,我在这里遇到编译错误:
'merge(io.vavr.collection.Map<? extends java.lang.Integer,? extends io.vavr.collection.List<java.lang.Object>>)' in 'io.vavr.collection.Map' cannot be applied to '(io.vavr.collection.Map<java.lang.Integer,io.vavr.collection.List<? extends java.lang.Double>>)'
Required type: Map<? extends Integer,? extends List<Object>>
Provided: Map<Integer,List<? extends Double>>
我也试过显式转换:
Map<Integer, List<Boolean>> merged2 = ((Map<Integer,List<Boolean>>)convert(stringMap)).merge(doubleMap);
但它抛出一个编译错误:
Inconvertible types; cannot cast 'io.vavr.collection.Map<java.lang.Integer,io.vavr.collection.List<java.lang.Object>>' to 'io.vavr.collection.Map<java.lang.Integer,io.vavr.collection.List<java.lang.Boolean>>'
Map.merge()
签名如下:
Map<K, V> merge(Map<? extends K, ? extends V> that);
地图和列表 类 来自 vavr。
调用convert
时使用type witness, or so called TypeArguments
mentioned in 15.12 in SE11 JLS指定参数类型。
MethodInvocation:
MethodName ( [ArgumentList] )
TypeName . [TypeArguments] Identifier ( [ArgumentList] )
ExpressionName . [TypeArguments] Identifier ( [ArgumentList] )
Primary . [TypeArguments] Identifier ( [ArgumentList] )
super . [TypeArguments] Identifier ( [ArgumentList] )
TypeName . super . [TypeArguments] Identifier ( [ArgumentList] )
ArgumentList: Expression {, Expression}
注意在使用TypeArguments
时不能省略实例名->(this).
Map<Integer, List<Boolean>> merged = this.<Integer, String, Boolean>convert(stringMap).merge(convert(doubleMap));