What is the most elegant way to check if all values in a boolean array are true?

I have a boolean array in java:

boolean[] myArray = new boolean[10];

What's the most elegant way to check if all the values are true?

public static boolean areAllTrue(boolean[] array)
{
    for(boolean b : array) if(!b) return false;
    return true;
}

Arrays.asList(myArray).contains(false)

In Java 8, you could do:

boolean isAllTrue = Arrays.asList(myArray).stream().allMatch(val -> val == true);

Or even shorter:

boolean isAllTrue = Arrays.stream(myArray).allMatch(Boolean::valueOf);

Note: You need Boolean[] for this solution to work. Because you can't have a primitives List.

It depends how many times you're going to want to find this information, if more than once:

Set<Boolean> flags = new HashSet<Boolean>(myArray);
flags.contains(false);

Otherwise a short circuited loop:

for (i = 0; i < myArray.length; i++) {
  if (!myArray[i]) return false;
}
return true;

I can't believe there's no BitSet solution.

A BitSet is an abstraction over a set of bits so we don't have to use boolean[] for more advanced interactions anymore, because it already contains most of the needed methods. It's also pretty fast in batch operations since it internally uses long values to store the bits and doesn't therefore check every bit separately like we do with boolean[].

BitSet myBitSet = new BitSet(10);
// fills the bitset with ten true values
myBitSet.set(0, 10);

For your particular case, I'd use cardinality():

if (myBitSet.cardinality() == myBitSet.size()) {
    // do something, there are no false bits in the bitset
}

---

Another alternative is using Guava:

return Booleans.contains(myArray, true);

That line should be sufficient:

BooleanUtils.and(boolean... array)

but to calm the link-only purists:

> Performs an and on a set of booleans.

In Java 8+, you can create an IntStream in the range of 0 to myArray.length and check that all values are true in the corresponding (primitive) array with something like,

return IntStream.range(0, myArray.length).allMatch(i -> myArray[i]);

This is probably not faster, and definitely not very readable. So, for the sake of colorful solutions...

int i = array.length()-1;
for(; i > -1 && array[i]; i--);
return i==-1

boolean alltrue = true;
for(int i = 0; alltrue && i<booleanArray.length(); i++)
   alltrue &= booleanArray[i];

I think this looks ok and behaves well...

Simply convert the array to a String using Arrays.toString and test if it contains false.

Demo:

import java.util.Arrays;

public class Main {
	public static void main(String args[]) {
		// Test
		System.out.println(areAllValuesTrue(new boolean[] { true, true, true, true }));
		System.out.println(areAllValuesTrue(new boolean[] { true, false, false, true }));
		System.out.println(areAllValuesTrue(new boolean[] { false, false, false, false }));
	}

	static boolean areAllValuesTrue(boolean[] arr) {
		return !Arrays.toString(arr).contains("false");
	}
}

Output:

true
false
false

You can check all value items are true or false by compare your array with the other boolean array via Arrays.equal method like below example :

private boolean isCheckedAnswer(List<Answer> array) {
    boolean[] isSelectedChecks = new boolean[array.size()];
    for (int i = 0; i < array.size(); i++) {
        isSelectedChecks[i] = array.get(i).isChecked();
    }

    boolean[] isAllFalse = new boolean[array.size()];
    for (int i = 0; i < array.size(); i++) {
        isAllFalse[i] = false;
    }

    return !Arrays.equals(isSelectedChecks, isAllFalse);
}

Kotlin: if one elemnt is false then not all are selected

return list.filter { isGranted -> isGranted.not() }.isNotEmpty()

OK. This is the "most elegant" solution I could come up with on the fly:

boolean allTrue = !Arrays.toString(myArray).contains("f");

Hope that helps!