I need to find the sum of integers in an array using recursion.
The following code throws an error :
var sum = function(array) {
if(array.length === 0) return 0;
while(array.length) {
sum = array[0] + sum(array.slice(1));
return sum;
}
}
while this works :
var sum = function(array) {
if(array.length === 0) return 0;
while(array.length) {
return array[0] + sum(array.slice(1));
}
}
The difference is the way the sum is returned. Can someone explain?
As mentioned in comments, the while loop never loops - it executes a return statement before testing the loop condition a second time.
Even so, why does the first version work the first time it is called but not the second?
sum = array[0] + sum(array.slice(1));
stores the right hand expression value in sum, after the function call to sum returns. That variable is in scope, so the assignment is valid.
But the assignment doesn't occur until after the call to sum (on the right hand side) returns. Each time sum returns, assignment overwrites the variable sum with the partial, and eventually the final result.
So the first call works successfully, but leaves sum set to the result of the first call. Since you can't call a number, it errors the second time you call it.
Related
I'm trying to wrap my head around the return statement but I can't see a reason why I should use one. My code works just fine without using one...
In both examples my console prints out 10 not matter if I use return or not.
Without return:
var sum = 5;
function myFunction() {
sum += 5;
}
myFunction();
console.log(sum);
With return:
var sum = 5;
function myFunction() {
return sum += 5;
}
myFunction();
console.log(sum);
By default, functions return the value undefined. If you want the function to return some other value, you need to have a return statement.
You may also use a return statement to halt execution of the function based on some logic, again returning a value that has some meaning or just undefined.
In the first example in the OP, the function is called and the return value is not used for anything, so it doesn't matter what the return value is and a return statement isn't necessary.
In another scenario, the return value might be important, e.g. a function that generates an integer random number between 0 and 10:
function getRandomInteger(){
return Math.floor(Math.random() * 11);
}
function showRandomNumber() {
document.getElementById('s0').textContent = getRandomInteger();
}
<button onclick="showRandomNumber()">Show random number</button>
<span id="s0"></span>
In the above, the getRandomInteger function needs to return a specific value, so it uses a return statement. The showRandomNumber function just displays the random number, so it doesn't need a return statement as the caller (the listener on the button) doesn't care what the return value is.
Here is what is happening with your example:
var sum = 5; //Sets the sum to =5
function myFunction() {
return sum += 5; // += reassigns the global (sum) to 10
}
myFunction();
console.log(sum);
A better example would be this:
sum = 5;
function myFunction() {
var sumOther = sum + 5;
return sumOther;
}
console.log(“sum:” + sum); // 5
console.log(“myFunction:” + myFunction()); // 10
This is how you would get the run of the function and not the global variable ‘sum’
This is because you use (global) variable declared outside the function (function modify it directly so there is no need to return value). However it is not good way of write functions because they are less reusable because they need proper global context to be used. Better is to use something like this:
function myFunction(num) {
return num += 5;
}
var sum = 5;
var result = myFunction(5) ;
console.log(result); // -> 10
this function can be easily used ind different context because it have parameter num and works only on it (function uses only values declared inside its declaration and body)
The trick in your case a is scope of a function. In both cases variable sum is defined in global scope. When you calling the function, it goes through the following steps:
Look whether var sum is defined in current step Since it not defined
inside the function, go one step over and take a look into outer
scope. Yes, it is defined here, start using it.
Perform calculation.
You are using incremental operator sum += 5 so actually it can be
written as sum = sum + 5. If you will look into second case,
you'll notice that variable have increased value now. And since that
variable is taken from global scope, your function just mutates it.
NOTE: At this point no matter whether you are returning something
from function or not. Your function just mutated outer variable.
The last step - exit from function. As I said earlier, return value
matters only if you want to use result of function call directly,
for instance: var result = myFunction()
Why does javascript's implementation of reduce skip execution for the first iteration?
[1,2,3].reduce((acc, val) => {
console.log('acc',acc);
console.log('val',val)
return acc + val;
});
// acc 1
// val 2
// acc 3
// val 3
// 6
I notice that the first statement execution never runs (in this case, I would have expected there to be 6 console logs, 2 for each element). This was very unexpected behavior when I was trying to execute a function with a side effect within each iteration with reduce.
In other languages that I've used, every iteration of the list passed executes. Are there examples otherwise?
Why does this happen and why is the implementation of javascript's native Array reduce like this?
========================= EDIT 1/Solution ========================
To make sure it goes through the first iteration, give it an initial value (the 2nd argument here/ 0 in this case)
[1,2,3].reduce((acc, val) => {
console.log('acc',acc);
console.log('val',val)
return acc + val;
}, 0);
That is because of the fact that on every iteration, the first value is treated as a return value (or the accumulator).
Straight from here, you can see
The accumulator accumulates the callback's return values; it is the
accumulated value previously returned in the last invocation of the
callback, or initialValue, if supplied (see below).
If we look at the source code here, we can see how it's implemented:
Array.prototype.myReduce = function(callback, initialVal) {
var accumulator = (initialVal === undefined) ? undefined : initialVal;
for (var i = 0; i < this.length; i++) {
if (accumulator !== undefined)
accumulator = callback.call(undefined, accumulator, this[i], i, this);
else
accumulator = this[i];
}
return accumulator;
};
In the else structure, we can see that if the value is undefined, we set it to the i-th subindex in the array; which, for the first iteration, is the first one. After that, it becomes the callback (return) value of the iterations which follow.
If you want, you can backtrack and check the output.
I posted a question not too long ago this morning regarding a kata that I was trying to solve. In that question, (found here if interested Kata Question) I needed to add a return statement to my function so that I would avoid the following error Value is not what was expected.
Now I have my second iteration of my kata solution to try out and here it is:
function isMerge(s, part1, part2) {
var pointer = 0
splitString = s.split('');
splitString.forEach(function(character) {
if (part1.includes(character) || part2.includes(character)) {
pointer++;
return true;
} else {
return false;
}
});
}
isMerge('codewars','cdw','oears')
I am still getting Value is not what was expected errors when I try to execute the code and this time I'm confused as to why in particular this happens.
For starters, taken from the MDN guide
The return statement ends function execution and specifies a value to be returned to the function caller.
expression
The expression to return. If omitted, undefined is returned instead.
Look at my if/else logic I am specifying a return true and return false condition in my forEach loop to see if all the chars from part1 and part2 are in the string. I am returning something so why is it that I have a Value is not what was expected?.
Second of all, by definition of the return statement, the function is supposed to stop when it reaches that keyword. However, when I place a console.log(character) in the logic, I can see on my console that all of the characters are being outputted so the function is not breaking at all when return true is executed. Why is that?
Third, I am confused as to when to use the return keyword in general. Consider these examples from the MDN docs for ForEach.
Example 1:
function logArrayElements(element, index, array) {
console.log('a[' + index + '] = ' + element);
}
// Notice that index 2 is skipped since there is no item at
// that position in the array.
[2, 5, , 9].forEach(logArrayElements);
// logs:
// a[0] = 2
// a[1] = 5
// a[3] = 9
Example 2:
function Counter() {
this.sum = 0;
this.count = 0;
}
Counter.prototype.add = function(array) {
array.forEach(function(entry) {
this.sum += entry;
++this.count;
}, this);
// ^---- Note
};
var obj = new Counter();
obj.add([2, 5, 9]);
obj.count
// 3
obj.sum
// 16
Not a single return statement to in these examples.
Now look at this .every example.
function isBigEnough(element, index, array) {
return element >= 10;
}
[12, 5, 8, 130, 44].every(isBigEnough);
And finally, from my previous question, I need to add a second return statement like this to avoid the value error.
function isBigEnough(element, index, array) {
return element >= 10;
}
function whenToUseReturn(array) {
return array.every(isBigEnough);
}
whenToUseReturn([12, 5, 8, 130, 44]);
So....... in conclusion, for my original function that started this how am I supposed to exit the loop when I reach false and return it and likewise when all the characters are in the string, how do I return a 'cumulative' true and avoid a Value error. I hope this makes sense and I can clarify with edits to better illustrate my point.
I am returning something so why is it that I have a Value is not what was expected?.
The return statement returns from the callback you pass to forEach, not from isMerge. return statements don't cross function boundaries. isMerge doesn't contain a return statement, hence it returns undefined. If we rewrite the function slightly it might become clearer:
function doSomething(part1, part2) {
return function(character) {
if (part1.includes(character) || part2.includes(character)) {
return true;
} else {
return false;
}
}
}
function isMerge(s, part1, part2) {
splitString = s.split('');
splitString.forEach(doSomething(part1, part2));
}
isMerge('codewars','cdw','oears')
This is equivalent to your code. As you can see, there is no return statement in isMerge.
Not a single return statement to in these examples.
There are no return statements in the forEach examples because forEach doesn't do anything with the return value of the callback, so there is no point in returning anything.
forEach is just a different way to iterate over an array, but it doesn't produce a value like reduce or every.
how am I supposed to exit the loop when I reach false and return it and likewise when all the characters are in the string, how do I return a 'cumulative' true and avoid a Value error.
You cannot exit a forEach "loop". If you have to stop the iteration early, you need to use a normal for (for/in, for/of) loop.
To return and produce a value, you can use your original solution that uses every.
My friend, since you decided to go the "callback way" using .each and the like, you should consider using callbacks, since you cannot return anything in this case. If you do not wish to go the callback way, just use standard javascript, such as:
splitString.forEach(function(character) {
Replace with
for(var i = 0 ; i < splitString.length; i++){
And now you can return. Using "each" to loop an array is just plain unnecessary and prevents you to return.
I implemented the following recursive JS function to get the the sum of elements in a array. This function works fine, when input [1,2,3] it returns 6, which is OK.
function sumOfNumbers(array) {
if (array.length == 1) {
return array[0];
} else {
last = array.length - 1;
return array[last] + sumOfNumbers(array.slice(0, last));
}
}
However, when changing the order of the sum to:
return sumOfNumbers(array.slice(0,last)) + array[last];
It returns 5 for [1,2,3]. Does anybody knows why?
Because the variable last is global, and you're changing it with your call to sumOfNumbers(array.slice(0,last)), before this part: array[last] sees it.
I am trying some loops and in this specific I do not understand why is sumSoFar = 0.
This is a task from Learn Street.com
function sum(n) {
var sumSoFar = 0;
var currentNumber = 1;
while (currentNumber <= n) {
sumSoFar += currentNumber;
currentNumber += 1;
}
return sumSoFar;
}
sumSoFar is initialized to zero at the start of the function because the rest of the function uses the += operator which takes the current value of sumSoFar and adds a value to that and then assigns the new value back to sumSoFar. If it wasn't initialized to zero, then it would not have an initial value and the first += operation on it would not have the desired result.
Initializing it to zero gives it the desired starting point so that the first time through the while loop, it adds currentNumber to that initial value of zero and assigns that to sumSoFar which is the desired behavior.
FYI, you can see your function work here: http://jsfiddle.net/jfriend00/kXrV8/ and in the latest version, you can try different input arguments to your sum(n) function.
If you don't initialize sumSoFar to zero you can't add values to it because it wouldn't be a number.