How to create a sequence of numbers

Short answer

IntStream.range(a, b) gives the numbers from a up to b exclusive, rangeClosed() inclusive:

void main() {
    IO.println(IntStream.range(1, 5).boxed().toList());
    IO.println(IntStream.rangeClosed(1, 5).boxed().toList());
}

Output:

[1, 2, 3, 4]
[1, 2, 3, 4, 5]

With a step

iterate() with a condition works like a for loop:

void main() {
    IO.println(IntStream.iterate(0, n -> n <= 20, n -> n + 5).boxed().toList());
    IO.println(IntStream.iterate(10, n -> n > 0, n -> n - 3).boxed().toList());
}

Output:

[0, 5, 10, 15, 20]
[10, 7, 4, 1]

Instead of a loop

A range often replaces a counting loop:

void main() {
    int sumOfSquares = IntStream.rangeClosed(1, 10).map(n -> n * n).sum();
    IO.println(sumOfSquares);
}

Output:

An infinite stream

iterate() without a condition and generate() are infinite — limit them with limit():

void main() {
    IO.println(Stream.iterate(1, n -> n * 2).limit(8).toList());
    IO.println(Stream.generate(() -> "ab").limit(3).collect(Collectors.joining()));
}

Output:

[1, 2, 4, 8, 16, 32, 64, 128]
ababab

Common mistake

Forgetting boxed() before toList(). IntStream has no toList() method, because a list cannot hold int values:

void main() {
    var numbers = IntStream.range(0, 3).toList();
}

Output:

Main.java:2: error: cannot find symbol

Thanks for reading!

Short answer

With a step

Instead of a loop

An infinite stream

Common mistake

Functions used in this recipe

Thanks for reading!