BiFunction 的区别<X, X>和 BinaryOperator<X>

Question

I am not able to understand that how come BinaryOperator<Integer> could be placed at the place of A in the code below, but not BiFunction<Integer, Integer> ?

A foo = (a, b) -> { return a * a + b * b; };
int bar = foo.apply(2, 3);
System.out.println(bar);

Could someone please help me understand it.

Answer 1

BinaryOperator is a special BiFunction . So you can assign the same expression to both of them. Check this out.

BinaryOperator<Integer> foo = (a, b) -> {
    return a * a + b * b;
};
BiFunction<Integer, Integer, Integer> barFn = (a, b) -> {
    return a * a + b * b;
};

If you look at the source code, it would be

public interface BinaryOperator<T> extends BiFunction<T,T,T> {
   // Remainder omitted.
}

Answer 2

The Bifunction and the BinaryOperator are same but the only difference here is the argument type and the return type of interfaces.

Consider a case where you want to concatenate two strings and return the result. In this case, you can choose either one of them but BinaryOperator is a good choice to go with because if you focus on the arguments and the return type, they all are the same.

BinaryOperator<String> c=(str,str1)->str+str1;

you can do the same with Bifunction but now see the difference here:

BiFunction<String,String,String> c=(str,str1)->str+str1;

Now consider a case in which we want to add two integers and return a string. Here we can only choose the BiFunction and not the BinaryOperator:

BiFunction<Integer,Integer,String> c=(a,b)->"Answer="+(a+b);