当有多个 arguments 时，方法引用如何在后台工作

Question

How does compiler ensure that equivalent lambda for below statement

BinaryOperator<String> concatOperator = String::concat; 

is

BinaryOperator<String> concatOperator = (resultString, inputString) -> resultString.concat(inputString);

and not

BinaryOperator<String> concatOperator = (resultString, inputString) -> inputString.concat(resultString);

Answer 1

This behaviour is well-documented in the JLS

15.13.3. Run-Time Evaluation of Method References

If the compile-time declaration is an instance method, then the target reference is the first formal parameter of the invocation method . Otherwise, there is no target reference.

If the compile-time declaration is an instance method, then the arguments to the method invocation expression (if any) are the second and subsequent formal parameters of the invocation method . Otherwise, the arguments to the method invocation expression are the formal parameters of the invocation method.

and it seems reasonable and intuitive. If you take a method with arity n ( n > 2 ), it becomes obvious that the target reference should be the first parameter, not the last, not the one in the middle.

Answer 2

The line of code using method references is categorized under the types of method reference as -

Reference to an Instance Method of an Arbitrary Object of a Particular Type

where the first lambda argument is inferred as an object of type String on which the method named concat is invoked with the parameter value equivalent to the second lambda argument. In the above case as:

BinaryOperator<String> concatOperator = (result, input) -> result.concat(input);