functions with return

Hey, in a course i'm doing, there was a question asking that you create a secondary array with the original array items lengths. The solution i added is what's inside of the function. However, it wouldn't register as being correct unless i put return before strings.map. I'm not really sure why it's important that's there. Is this because that's how it returns the array it creates back to you? And with the second return returning string.length, it that how it pulls the value it's looking for?
function getStringSizes(strings) {
return strings.map(string => {
return string.length;
});
}

// Sample usage - do not modify
console.log(getStringSizes(["a", "abc"])); // [1, 3]
console.log(getStringSizes(["Sam", "Alex", "Charlie"])); // [3, 4, 7]
console.log(getStringSizes(["Hello", "Blue"])); // [5, 4]
function getStringSizes(strings) {
return strings.map(string => {
return string.length;
});
}

// Sample usage - do not modify
console.log(getStringSizes(["a", "abc"])); // [1, 3]
console.log(getStringSizes(["Sam", "Alex", "Charlie"])); // [3, 4, 7]
console.log(getStringSizes(["Hello", "Blue"])); // [5, 4]
I'd appreciate any explanation. Thank you in advance.
5 Replies
glutonium
glutonium2mo ago
ok so is it that u thought adding that return within the map() will also return from the function? in that case , let's refactor the code maybe that'll help u understand why returning from inside the map() won't return fron rhe function
function getStringSizes(strings) {
strings.map(mapCallback);
}

function mapCallback(string) {
return string.length;
}
function getStringSizes(strings) {
strings.map(mapCallback);
}

function mapCallback(string) {
return string.length;
}
can u now understand how returning from the call back can't return from the getStringSizes? return statements r scoped to their parent function. so when u say it return, it only returns from the parent function, not any of the grandparent function. which is why despite returning from within the map(), u still need to return from the getStringSizes()
missymae
missymae2mo ago
when writing a function, the instructions inside are done in order; and yes, the return statement is what the function will send back when it completes, the return value. It's optional. in JS you can write functions in multiple ways. Here you also use an arrow function, which automatically returns a value - unless it's an {object} in braces, in which case it needs the
return
return
. So you can remove the braces and return from the map and have the same result with
function getAgain(string){
return string.map(string => string.length)
}
function getAgain(string){
return string.map(string => string.length)
}
or you can rewrite the whole works as a const and use arrow functions to avoid writing any returns and have something like this
const getStringSizes= (strings) => strings.map(string => string.length);
const getStringSizes= (strings) => strings.map(string => string.length);
The best way to learn the patterns is just use them and mess around, see what works and what breaks.
b1mind
b1mind2mo ago
Keep in mind that its not avoiding returns its just making them implied I realize you said avoid writing now 🤦‍♂️ brain but still good to clarify (which you I also missed you do LOL mbmbmb, going back to my coffee)
snxxwyy
snxxwyyOP2mo ago
ah yeah i see why it doesn't return now, thank you. i'm used to adding return xyz as the last line of a function not adding it before anything, i think that's where i got confused. And thank you everyone else for the pointers!
ἔρως
ἔρως2mo ago
people usually forget this, but, you can return almost anywhere in a function the use of arrow functions with the implied return will be very useful for you
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