如何从 List<Integer> 中获取 IntStream?
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);
}
什么是惯用的方式?也许已经有一个库函数可以完全满足我的要求?
What is the idiomatic way? Maybe there is already a library function that does exactly what I want?
推荐答案
我猜(或者至少是另一种选择)这种方式性能更高:
I guess (or at least it is an alternative) this way is more performant:
public static IntStream baz(List<Integer> list)
{
return list.stream().mapToInt(Integer::intValue);
}
由于函数 Integer::intValue
完全兼容 ToIntFunction
因为它需要一个 Integer
并返回一个 int
.不执行自动装箱.
since the function Integer::intValue
is fully compatible with ToIntFunction
since it takes an Integer
and it returns an int
. No autoboxing is performed.
我也在寻找 Function::identity
,我希望写一个与你的 bar
方法等效的方法:
I was also looking for an equivalent of Function::identity
, i hoped to write an equivalent of your bar
method :
public static IntStream qux(List<Integer> list)
{
return list.stream().mapToInt(IntFunction::identity);
}
但他们没有提供这个 identity
方法.不知道为什么.
but they didn't provide this identity
method. Don't know why.
相关文章