My code looks like this
var res = [];
var temp = [];
function Permutations(target, size) {
if (size === 0) {
res.push(temp);
console.log(res);
return;
}
for (let i = 0; i < target.length; i++) {
if (target[i] !== null) {
temp.push(target[i]);
target[i] = null;
Permutations(target, size - 1);
target[i] = temp.pop();
}
}
}
Permutations([1, 2, 3], 2);
console.log(res);When I run my code, I can see my res stores each permutation as it is is being executed. However, when I log it outside the function, all the stored value disappeared.
[ [ 1, 2 ] ]
[ [ 1, 3 ], [ 1, 3 ] ]
[ [ 2, 1 ], [ 2, 1 ], [ 2, 1 ] ]
[ [ 2, 3 ], [ 2, 3 ], [ 2, 3 ], [ 2, 3 ] ]
[ [ 3, 1 ], [ 3, 1 ], [ 3, 1 ], [ 3, 1 ], [ 3, 1 ] ]
[ [ 3, 2 ], [ 3, 2 ], [ 3, 2 ], [ 3, 2 ], [ 3, 2 ], [ 3, 2 ] ]
[ [], [], [], [], [], [] ] // This is my console.log outside the function
The array temp holds is the same array throughout the complete execution of your code. And res.push(temp); adds this same array (not a copy of it) to your res array.
Here a related question about how Objects are handled in JavaScript: Is JavaScript a pass-by-reference or pass-by-value language?
So your code results in res having N times the same array.
You could copy the element stored in temp to a new array using [...temp], and push that to your res array.
var res = [];
var temp = [];
function Permutations(target, size) {
if (size === 0) {
res.push([...temp]);
return;
}
for (let i = 0; i < target.length; i++) {
if (target[i] !== null) {
temp.push(target[i]);
target[i] = null;
Permutations(target, size - 1);
target[i] = temp.pop();
}
}
}
Permutations([1, 2, 3], 2);
console.log(res);