Amstrong number using lambda Expressions in Java 8

Question

I am fairly new to functional programming. Following is my code to check the sum of powered digits(Arsmstrong number check). Number is an input for example: 370 which is an amstrong number, the return statement has to return the sum as 370. I need to achieve this by functional programming only. How can I change the value 3 to be dynamic and equal to the number of digits.

return String.valueOf(number).chars().mapToObj(e -> Character.getNumericValue(e)).collect(Collectors.toList())
        .stream()
        .reduce(0, (sum,value) -> {
            sum += (int)Math.pow(value, 3);
            return sum;
        });

Answer 1

Here is an IntPredicate that can test for an Armstrong number.

• get the number of digits (in this case exponent) using Math.log10
• stream the values divided by ten to expose the least significant digit of each quotient.
• reduce the stream to a sum by raising last digit (via quotient % 10) to the number of digits
• returns a boolean on the result of comparing the original value to the sum.

IntPredicate isArmstrong = val -> {
            int exp = (int) Math.log10(val) + 1;
            return IntStream.iterate(val, v -> v > 0, v -> v / 10)
                    .reduce(0, (sum, quotient) -> sum + (int) Math
                            .pow(quotient % 10, exp)) == val;
        };

System.out.println(isArmstrong.test(153));  
System.out.println(isArmstrong.test(8208)); 
System.out.println(isArmstrong.test(9923)); 

prints

true
true
false

This can also be used to filter a stream of ints. ...filter(isArmstrong)... and return the values in an array, collection, or just printing as they pass thru the filter.