I was anticipating that the code would generate an undefined result, but it actually returned a function instead

Question

I was anticipating that the code would generate an undefined result, but it actually returned a function instead

Is anyone able to explain why this code is printing function when I expect it to print undefined? I am just starting out with JavaScript.

function createGreeter(greeting){
    function greet(){
        console.log(greeting,name)
    }
    return greet
}
    
let g1=createGreeter('Good Morning')
console.log(typeof g1)
let g2=createGreeter('Good Evening')

javascript

Answer 1

Answer №1

It seems like you're interested in creating a function that takes a greeting and returns another function that takes a name (while keeping a reference to the greeting variable in its outer lexical scope when returned) and combines those strings when the function is called.

// `createGreeter` function accepts a greeting string
// and returns a new function that takes a name input
// then, it concatenates both strings when called
function createGreeter(greeting) {
  return function (name) {
    return `${greeting}, ${name}.`;
  }
}

// Both of these functions return a greet function for a specific occasion
const goodevening = createGreeter('Good evening');
const expectingyou = createGreeter('I\'ve been expecting you');

// Now we can call these functions with a name argument
console.log(goodevening('Blofeld'));
console.log(expectingyou('Mr. Bond'));

Answer 2

It seems like you're interested in creating a function that takes a greeting and returns another function that takes a name (while keeping a reference to the greeting variable in its outer lexical scope when returned) and combines those strings when the function is called.

// `createGreeter` function accepts a greeting string
// and returns a new function that takes a name input
// then, it concatenates both strings when called
function createGreeter(greeting) {
  return function (name) {
    return `${greeting}, ${name}.`;
  }
}

// Both of these functions return a greet function for a specific occasion
const goodevening = createGreeter('Good evening');
const expectingyou = createGreeter('I\'ve been expecting you');

// Now we can call these functions with a name argument
console.log(goodevening('Blofeld'));
console.log(expectingyou('Mr. Bond'));

Answer 3

Answer №2

Within the function greet, you are returning the function itself.

If you wish to store the result of the greet function in a variable, you must invoke it:

Instead of using return greet, consider using return greet()

Answer 4

Within the function greet, you are returning the function itself.

If you wish to store the result of the greet function in a variable, you must invoke it:

Instead of using return greet, consider using return greet()

Answer 5

Answer №3

Within the code, it is indicated on line 5 to return greet. The variable greet references the function greet itself. It might be advisable to adjust line 5 to return greet(), thereby triggering the execution of the greet() function and subsequently returning its result, which in this instance happens to be undefined.

Answer 6

Within the code, it is indicated on line 5 to return greet. The variable greet references the function greet itself. It might be advisable to adjust line 5 to return greet(), thereby triggering the execution of the greet() function and subsequently returning its result, which in this instance happens to be undefined.

I was anticipating that the code would generate an undefined result, but it actually returned a function instead

Answer №1

Answer №2

Answer №3

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