CODEWITHSUNDEEP
javascriptfunctional-programmingarrow-functions
Isaac
Isaac

Reputation: 12884

Where is the parameter coming from?

function createMathOperation(operator) {
  console.log(operator); //(augend, addend) => augend + addend
  return (value, other) => {
    return operator(value, other)
  }
}

const add = createMathOperation((augend, addend) => augend + addend)

add(1,2)//3

I found the above function definition from lodash. I am trying to understand it but to no avail.

Right inside createMathOperation, I try to log operator and this is the value 

(augend, addend) => augend + addend

I guess value and other is 1 and 2 but how?

And how return operator(value, other) works when operator is (augend, addend) => augend + addend

Can anyone convert it to longer human readable form for a better understanding instead?

Upvotes: 5

Views: 139

Answers (3)

FINDarkside
FINDarkside

Reputation: 2445

createMathOperation returns function, which adds two numbers. Here's more readable version:

function createMathOperation(fn) {
  console.log(fn);
  return function(value, other){
    return fn(value, other);
  };
}

const add = createMathOperation(function (augend, addend) {
  return augend + addend;
});

add(1,2)//3

I renamed 'operator' to 'fn' to make it less confusing (syntax highlighting colored it blue for some reason).

Upvotes: 1

Titian Cernicova-Dragomir
Titian Cernicova-Dragomir

Reputation: 250106

This is the essence of functional programming you can pass in a function, return a function, and call the function you received as a parameter:

function createMathOperation(operator) {
  console.log(operator); // This is a the function that performs the computation 
  // We return a new function (using arrow syntax) that receives 2 arguments and will call the original operator we passed in to createMathOperation
  // The code inside this function is not executed here, the function is not invoked. 
  // The caller can take the returned function and executed 0-n times as they wish. 
  return (value, other) => { 
    // when we invoke add this is the code that gets called and the arguments we pass to add end up in value and other
    console.log("Getting ready to compute " + value + " operator " + other); 
    return operator(value, other) // since operator is a function we just invoke it as we would any other function with the two arguments we got from whoever called us.
  }
}

// add will contain the wrapped function that has our extra console.log 
const add = createMathOperation((augend, addend) => augend + addend)

// The 'Getting ready ...' text has not been printed yet, nobody invoked the function that was returned yet, the next line will do so.
console.log(add(1,2)) 
// will output:
// Getting ready to compute 1 operator 2
// 3

A note on => is just syntactic sugar over a function expression, it has extra semantics around this, but for this example, (augend, addend) => augend + addend is equivalent to function (augend, addend){ return augend + addend; }

Upvotes: 4

wiesion
wiesion

Reputation: 2455

Your code in good old JS would look like:

var createMathOperation = function(operator) {
  // operator is scope-locked within this operation wrapper
  console.log('operator:', operator);
  return function(value, other) {
    // The wrapper returns an anonymous function that acts as a call-wrapper
    // for your original function
    console.log('value:', value);
    console.log('other:', other);
    return operator(value, other)
  }
}

var add = createMathOperation(function(augend, addend) {
  // This is what is being called at line 9 - return operator(value, other)
  return augend + addend;
});

console.log('result 1+2:', add(1,2));

In general i don't see much use to all of this, you could just do const add = (a, v) => a + v; and have the same result.

Upvotes: 0

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