Why does declaring a variable as type int require casting a stream? long type does not require a cast

Question

Why does one primitive type require casting and and the other does not?

/* This method uses stream operations to count how many numbers in a given array
* of integers are negative
*/    

public static void countNegatives(int[] nums) {
    long howMany = stream(nums) // or: int howMany = (int) stream(nums)
            .filter(n -> n < 0)
            .count();
    System.out.print(howMany);
}

Answer 1

The count() method of the Java IntStream class has been defined to return a long value. From the documentation :

long count() Returns the count of elements in this stream. This is a special case of a reduction and is equivalent to:

 return mapToLong(e -> 1L).sum();

In other words, this is how the Java is designed, ie it has nothing todo with Lambda.

Answer 2

count() returns long and not every long can fit into an int hence an explicit cast is required to store the result into an int . This is nothing to do with java-10. it's always been there in previous JDK's.

If you don't want to cast then the alternative would be:

...
.filter(n -> n < 0)
.map(e -> 1)
.sum();

but as one can see this is not as readable as your example because the code essentially says "give me a summation of the elements that pass the filter operation" instead of "give me a count of the elements that pass the filter operation".

So, ultimately if you need the result as an int then go for the cast.