Are the effects of delete array[i] and array[i] = undefined identical? I'm aware that using the former can result in a sparse array - but does the latter have the same outcome?
Are the effects of delete array[i] and array[i] = undefined identical? I'm aware that using the former can result in a sparse array - but does the latter have the same outcome?
The action of the latter option does not result in the same outcome. It only assigns the value undefined to that specific index without deleting it.
Using the delete keyword will actually remove the index from the array, although it does not adjust the length accordingly. This can lead to a situation where you have undefined indexes, creating what is known as a "sparse" array with more length than actual items.
You can easily observe this behavior by applying various array methods to the array. They will skip over elements with the value undefined, but will still acknowledge indexes that are not defined altogether (there is indeed a distinction).
var arr = [0, 1, 2, 3];
delete arr[1]; // delete the second index
arr.forEach(function(item) {
console.log(item); // 0, 2, 3
});
console.log('--------')
var arr2 = [0, 1, 2, 3];
arr2[1] = undefined; // set the second index to "undefined"
arr2.forEach(function(item) {
console.log(item); // 0, undefined, 2, 3
});They actually do not function in the same way. According to MDN's article on the delete operator:
Removing elements from an array
Deleting an element from an array does not change the length of the array, even if the last element is deleted.
When you use the delete operator to remove an element from an array, that element is no longer part of the array. In the example provided,
trees[3]is removed usingdelete.var trees = ['redwood', 'bay', 'cedar', 'oak', 'maple']; delete trees[3]; if (3 in trees) { // this code block will not be executed }If you want to keep an array element but assign it an undefined value, it's better to directly set it as undefined instead of using the delete operator. In the following example,
trees[3]is given an undefined value, while still maintaining the existence of the array element:var trees = ['redwood', 'bay', 'cedar', 'oak', 'maple']; trees[3] = undefined; if (3 in trees) { // this code block will be executed }
Utilizing both methods, one has the ability to transform a compact array into a sparse one or maintain its density. The way in which they are employed varies.
Consider this...
let numbers = [1, 2, 3];
delete numbers[10]; // This will return true but 'numbers' remains non-sparse
delete numbers[1]; // This will return true and 'numbers' is now sparse
let values = [1, 2, 3];
values[1] = undefined; // 'values' remains non-sparse
values[10] = undefined; // 'values' is now sparse
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