No, filter does not scan the whole stream. It's an intermediate operation, which returns a lazy stream (actually all intermediate operations return a lazy stream). To convince you, you can simply do the following test:

List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);
int a = list.stream()
            .peek(num -> System.out.println("will filter " + num))
            .filter(x -> x > 5)
            .findFirst()
            .get();
System.out.println(a);

Which outputs:

will filter 1
will filter 10
10

You see that only the two first elements of the stream are actually processed.

So you can go with your approach which is perfectly fine.

Answer from Alexis C. on Stack Overflow
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Medium
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February 8, 2020 - package com.mkyong.java8; import java.util.Arrays; import java.util.List; import java.util.Optional; public class Java8FindFirstExample1 { public static void main(String[] args) { List<Integer> list = Arrays.asList(1, 2, 3, 2, 1); Optional<Integer> first = list.stream().findFirst(); if (first.isPresent()) { Integer result = first.get(); System.out.println(result); // 1 } else { System.out.println("no value?"); } Optional<Integer> first2 = list .stream() .filter(x -> x > 1).findFirst(); if (first2.isPresent()) { System.out.println(first2.get()); // 2 } else { System.out.println("no value?"); } } }
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