How to generate random array of ints using Stream API Java 8?

Question

If you want primitive int values, do not call IntStream::boxed as that produces Integer objects by boxing.

Simply use Random::ints which returns an IntStream:

int[] array = new Random().ints(size, lowBound, highBound).toArray();

There's no reason to boxed(). Just receive the Stream as an int[].

int[] array = intStream.limit(limit).toArray();

To generate random numbers from range 0 to 350, limiting the result to 10, and collect as a List. Later it could be typecasted.

However, There are no guarantees on the type, mutability, serializability, or thread-safety of the List returned.

List<Object> numbers =  new Random().ints(0,350).limit(10).boxed().collect(Collectors.toList());

and to get thearray of int use

int[] numbers =  new Random().ints(0,350).limit(10).toArray();

tl;dr

ThreadLocalRandom     // A random number generator isolated to the current thread.
.current()            // Returns the current thread's `ThreadLocalRandom` object.
.ints( low , high )   // Pass the "origin" (inclusive) and "bound" (exclusive).
.limit( 100 )         // How many elements (integers) do you want in your stream?
.toArray()            // Convert the stream of `int` values into an array `int[]`. 

ThreadLocalRandom

You can do it using ThreadLocalRandom.

The ints method generates an IntStream within your specified bounds. Note the the low is inclusive while the high is exclusive. If you want to include your high number, just add one while calling the ints method.

int[] randInts = ThreadLocalRandom.current().ints( low , high ).limit(100).toArray();

See this code run live at IdeOne.com.