Java流中Groovy的int.times
我试图在Java8中创建有序整数流:Java流中Groovy的int.times,java,groovy,java-stream,Java,Groovy,Java Stream,我试图在Java8中创建有序整数流:[1,2,3,4,…]。在Groovy中,我将使用I.e.5.times{-lambda-}。在Java中,我发现只有一种方法可以做到这一点,但它不是很漂亮: final IntStream.Builder builder = IntStream.builder(); for (int i = 0; i < 5; i++) { builder.add(i); } final IntStream stream
[1,2,3,4,…]
。在Groovy中,我将使用I.e.5.times{-lambda-}
。在Java中,我发现只有一种方法可以做到这一点,但它不是很漂亮:
final IntStream.Builder builder = IntStream.builder();
for (int i = 0; i < 5; i++) {
builder.add(i);
}
final IntStream stream = builder.build();
final IntStream.Builder=IntStream.Builder();
对于(int i=0;i<5;i++){
增加(i);
}
final IntStream=builder.build();
你可以这样做
IntStream stream = IntStream.range(1, 5);