I'm trying to write a recursive function with Javascript, but apparently I miss something and I cannot make it return at the point that I want. In the following example, I would expect to see that baz=18 in the logs, however baz is undefined.
https://jsfiddle.net/n8fg1h3v/1/
Any ideas?
function foo(arr) {
if (arr[0]==7) {
return 18
} else {
arr = arr.slice(1,arr.length);
foo(arr);
}
}
var arr1 = [9,1,2,3,4,7,6];
var baz = foo(arr1);
console.log(baz)
You need a return of calling foo inside of the function.
return foo(arr);
function foo(arr) {
if (arr[0] == 7) {
return 18;
} else {
arr = arr.slice(1, arr.length);
return foo(arr);
}
}
var arr1 = [9, 1, 2, 3, 4, 7, 6];
var baz = foo(arr1);
console.log(baz)
7 is in the array?
if then else statement with a return inside if did not work as I would have expected. With adding the return foo(arr) as you suggested it works fine in the simple example and also in my real problem. Thanks! (I cannot accept the answer for another ~10 min, I will do it just after that.)I would simply do this:
function foo(arr) {
var el = arr.find(el => el === 7);
return el ? 18 : '7 was not found';
}
var bar = foo([9, 1, 2, 3, 4, 7, 6]);
console.log(bar);
var baz = foo([9, 1, 2, 3, 4, 8, 6]);
console.log(baz);some, because it breaks the iteration, but it has nothing to do with recursion.
Array.prototype.find() breaks the iteration.. I understand your point about recursion, but here I do not think that it's necessary
arr = ...) in Javascript wherever you want, it is quite uncommon within recursive algorithms. Pass the expression as an argument instead (foo(arr.slice(...)))