How to assign lambda expression to a variable?

Question

I have a kind of calculator:

sum = value1 + value2;
    multiplication = value1 * value2;
    division = (value1 / value2);
    average = (sum / 2);

    if (operation.compareToIgnoreCase("add") == 0)
        result = sum;
    else if (operation.compareToIgnoreCase("mult") == 0)
        result = multiplication;
    else if (operation.compareToIgnoreCase("div") == 0)
        result = division;
    else if (operation.compareToIgnoreCase("avg") == 0)
        result = average;

And i have to turn this to lambda expressions:

result = (request.getValue1(), request.getValue2()) -> {
        if (request.getOperation().compareToIgnoreCase("add") == 0) return request.getValue1() + request.getValue2();
        else if (request.getOperation().compareToIgnoreCase("mult") == 0) return request.getValue1() * request.getValue2();
        else if (request.getOperation().compareToIgnoreCase("div") == 0) return request.getValue1() / request.getValue2();
        else if (request.getOperation().compareToIgnoreCase("avg") == 0) return (request.getValue1() + request.getValue2())/2;
    }

But it doesnt work. Can you guys help? Thanks

In addition to everything else, the point of using polymorphism is to use a Map<String,something> instead of those if statements. — chrylis -cautiouslyoptimistic-
– chrylis -cautiouslyoptimistic-, Commented Sep 14, 2018 at 13:03

Lorelorelore · Accepted Answer · 2019-06-03 20:53:54Z

8

For example:

BiFunction<Number,Number,Number> sum = (value1, value2) -> value1 + value2;

and you can use it like this:

sum.apply(value1,value2)

more examples here.

Also you can store your set of BiFunction in a Map:

Map<String,BiFunction<Number, Number, Number>> operations = new HashMap<>();
operations.put("sum", sum);

If you do the same for all operations, you call them as it follows:

operations.get("sum").apply(value1, value2);

So, the result can be computed as is follows:

result = (value1, value2, operation) -> operations.get(operation).apply(value1, value2);

edited Jun 3, 2019 at 20:53

answered Sep 14, 2018 at 12:56

Lorelorelore

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4 Comments

Eugene Over a year ago

much cleaner would be to define an enum for this

Lorelorelore Over a year ago

Yes, you can improve the solution using enums. This is just a basic example, but enumeration fits the use case pretty well and the get operation should handle unknown operations.

fps Over a year ago

Why the lambda at the end? what is this 3-arg lambda?

Lorelorelore Over a year ago

To replace the last lambda reported in question.

Youcef LAIDANI · Accepted Answer · 2018-09-14 13:06:03Z

3

You can create a map with a String and DoubleBinaryOperator or an IntBinaryOperator depends on the type you want like so :

Map<String, DoubleBinaryOperator> map = new HashMap<>();
map.put("add", (v1, v2) -> v1 + v2);
map.put("mult", (v1, v2) -> v1 * v2);
map.put("div", (v1, v2) -> v1 / v2);
map.put("avg", (v1, v2) -> (v1 * v2) / 2);

Then you can call that map with the operation you want for example :

// an example
String operation = "add";
Double reslt = map.get(operation).applyAsDouble(3, 5);

edited Sep 14, 2018 at 13:06

answered Sep 14, 2018 at 13:00

Youcef LAIDANI

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Comments

Ondřej Fischer · Accepted Answer · 2018-09-17 20:53:39Z

0

from the example it's not very clear, what result should be. So let me rewrite it including types:

Function<Request, Double> calculator = (request) -> {
    if (request.getOperation().compareToIgnoreCase("add") == 0) return request.getValue1() + request.getValue2();
    else if (request.getOperation().compareToIgnoreCase("mult") == 0) return request.getValue1() * request.getValue2();
    else if (request.getOperation().compareToIgnoreCase("div") == 0) return request.getValue1() / request.getValue2();
    else if (request.getOperation().compareToIgnoreCase("avg") == 0) return (request.getValue1() + request.getValue2())/2;
    else throw new IllegalArgumentException();
};

double result = calculator.apply(new Request("add", 2, 4));

edited Sep 17, 2018 at 20:53

answered Sep 14, 2018 at 13:02

Ondřej Fischer

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