CODEWITHSUNDEEP
javascriptcallbackclosures
TheRealFakeNews
TheRealFakeNews

Reputation: 8193

Trying to understand closures (JavaScript)

function myFunc(inputFunc){
  var called = false;
  return function() {
    if (!called) {
      called = true;
      var storedResult = inputFunc();
      return storedResult;
    }
    else
      return storedResult;
    };
}

In the above code, I don't understand what purpose it serves to have the if-else statement returned in a function. Wouldn't it be the same effect if I just had the following instead?

function myFunc(inputFunc){
  var called = false;
  if (!called) {
    called = true;
    var storedResult = inputFunc();
    return storedResult;
  }
  else
    return storedResult;
  }

Upvotes: 1

Views: 28

Answers (1)

adeneo
adeneo

Reputation: 318362

Wouldn't it be the same...

Not really, the outer function returns a function, enclosing the called variable in it's scope so it doesn't change in later calls
Here's how the first code snippet would work

var instance = inputFunc();

var storedResult = instance(); // returns the result

var runItAgain = instance(); // probably returns `undefined`

Your second version wouldn't do any of that, it would just be

var storedResult = inputFunc(); // result

var runItAgain = inputFunc(); // result again, the "called" variable is always false

In other words, the first version returns the result once, and only once, here's a snippet

function myFunc(inputFunc) {
    var called = false;
    return function() {
        if (!called) {
            called = true;
            var storedResult = inputFunc();
            return storedResult;
        } else
            return storedResult;
    };
}

var instance = myFunc(function() {
	return 'result';
});

var log = [];

log.push( instance() ); // result
log.push( instance() ); // undefined
log.push( instance() ); // undefined
log.push( instance() ); // undefined

document.body.innerHTML = '<pre>' + JSON.stringify(log, null, 4) + '</pre>';

Upvotes: 4

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