Adding up BigDecimals using Streams

Original answer

Yes, this is possible:

List<BigDecimal> bdList = new ArrayList<>();
//populate list
BigDecimal result = bdList.stream()
        .reduce(BigDecimal.ZERO, BigDecimal::add);

What it does is:

1. Obtain a List<BigDecimal>.
2. Turn it into a Stream<BigDecimal>

3. Call the reduce method.

   3.1. We supply an identity value for addition, namely BigDecimal.ZERO.

   3.2. We specify the BinaryOperator<BigDecimal>, which adds two BigDecimal‘s, via a method reference BigDecimal::add.

Updated answer, after edit

I see that you have added new data, therefore the new answer will become:

List<Invoice> invoiceList = new ArrayList<>();
//populate
Function<Invoice, BigDecimal> totalMapper = invoice -> invoice.getUnit_price().multiply(invoice.getQuantity());
BigDecimal result = invoiceList.stream()
        .map(totalMapper)
        .reduce(BigDecimal.ZERO, BigDecimal::add);

It is mostly the same, except that I have added a totalMapper variable, that has a function from Invoice to BigDecimal and returns the total price of that invoice.

Then I obtain a Stream<Invoice>, map it to a Stream<BigDecimal> and then reduce it to a BigDecimal.

Now, from an OOP design point I would advice you to also actually use the total() method, which you have already defined, then it even becomes easier:

List<Invoice> invoiceList = new ArrayList<>();
//populate
BigDecimal result = invoiceList.stream()
        .map(Invoice::total)
        .reduce(BigDecimal.ZERO, BigDecimal::add);

Here we directly use the method reference in the map method.