var arr = [1,2,3,4,5];
arr2 = arr;
arr = arr2.concat([6,7,8,9,10]);
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
arr2;
[1, 2, 3, 4, 5] as in the code above, the array is a reference type object, and arr2 holds a reference to arr.
but why after operating on arr, the output arr2 is still the old value before arr.
concat function returns a new reference because concat does not change the original array
var arr = [1,2,3,4,5];
arr2 = arr; // arr2 = [1,2,3,4,5]
arr = arr2.concat([6,7,8,9,10]);
//concatarr
// arr2 [1,2,3,4,5],arr[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]as described upstairs;
Array.concat () returns a new array, so arr is no longer the same as the reference to arr2.
the details are
var arr = [1,2,3,4,5]; //heap_1[1,2,3,4,5] arr
arr2 = arr; //arrarr2arr arr2 heap_1
arr = arr2.concat([6,7,8,9,10]);
//Array.concat() heap_2 arr heap_2;
//Array.concat()arr2heap_1