Java 如何从 List<Integer> 获取 IntStream？
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How do I get an IntStream from a List<Integer>?
提问by fredoverflow
I can think of two ways:
我可以想到两种方法：
public static IntStream foo(List<Integer> list)
{
return list.stream().mapToInt(Integer::valueOf);
}
public static IntStream bar(List<Integer> list)
{
return list.stream().mapToInt(x -> x);
}
What is the idiomatic way? Maybe there is already a library function that does exactly what I want?
什么是惯用方式？也许已经有一个库函数可以完全满足我的要求？
采纳答案by gontard
I guess (or at least it is an alternative) this way is more performant:
我猜（或至少它是一种替代方法）这种方式性能更高：
public static IntStream baz(List<Integer> list)
{
return list.stream().mapToInt(Integer::intValue);
}
since the function Integer::intValueis fully compatible with ToIntFunctionsince it takes an Integerand it returns an int. No autoboxingis performed.
因为该函数Integer::intValue是完全兼容的，ToIntFunction因为它需要一个Integer并且它返回一个int。不执行自动装箱。
I was also looking for an equivalent of Function::identity, i hoped to write an equivalent of your barmethod :
我也在寻找等效的Function::identity，我希望编写等效的bar方法：
public static IntStream qux(List<Integer> list)
{
return list.stream().mapToInt(IntFunction::identity);
}
but they didn't provide this identitymethod. Don't know why.
但他们没有提供这种identity方法。不知道为什么。
回答by Naman
An alternate way to transform that would be using Stream.flatMapToIntand IntStream.ofas:
另一种转换方法是使用Stream.flatMapToInt和IntStream.of作为：
public static IntStream foobar(List<Integer> list) {
return list.stream().flatMapToInt(IntStream::of);
}
Note: Went through few linked questions before posting here and I just couldn't find this suggested in them either.
注意：在这里发帖之前，我已经解决了几个链接问题，但我也没有在其中找到建议。
