Ale*_*ing 11 java lambda arraylist toarray java-8
我正在使用Codingbat网站,特别是AP-1中的这种方法
public String[] wordsWithout(String[] words, String target) {
ArrayList<String> al = new ArrayList<>(Arrays.asList(words));
al.removeIf(s -> s.equals(target));
return al.toArray(new String[al.size()]);
}
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这个实现工作,以及它当前提交的内容,但是当我将return语句更改为
return al.toArray(String[]::new);
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据我所知,这应该工作,给出以下错误:
no suitable method found for toArray((size)->ne[...]size])
method java.util.Collection.<T>toArray(T[]) is not applicable
(cannot infer type-variable(s) T
(argument mismatch; Array is not a functional interface))
method java.util.List.<T>toArray(T[]) is not applicable
(cannot infer type-variable(s) T
(argument mismatch; Array is not a functional interface))
method java.util.AbstractCollection.<T>toArray(T[]) is not applicable
(cannot infer type-variable(s) T
(argument mismatch; Array is not a functional interface))
method java.util.ArrayList.<T>toArray(T[]) is not applicable
(cannot infer type-variable(s) T
(argument mismatch; Array is not a functional interface)) line:4
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有人会这么好解释为什么这不起作用?
Eug*_*ene 12
这个:
al.toArray(String[]::new)
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仅在java-11之后支持:
default <T> T[] toArray(IntFunction<T[]> generator) {
return toArray(generator.apply(0));
}
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如果你想使用那个方法参考,你必须stream()首先,如:
al.stream().toArray(String[]::new)
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