Why Use a Spread Operator Followed by an Iterable Instead of Just the Iterable Itself When Instantiating a Collection in Dart?

D'errah 0
Viewed 45

For example with the following List:

var originalList = ['follow', 'spread', 'operators.'];

When referencing it to make another List, why use the spread operator:

var spreadList = ['See? I', ...originalList]

Instead of without the spread operator:

var lonelyList = ['Actually I dont', originalList]

When the results for spreadList and lonelyList respectively are:

[See? I, follow, spread, operators.]
[Actually I dont, follow, spread, operators.]
iterable dart flutter
2 Answers

In this statement:

var spreadList = ['See? I', ...originalList]

With the spread operator you're inserting each individual String element from originalList as it's own element into spreadList. The end result being a:

<String>['See? I', 'follow', 'spread', 'operators']

But in this statement:

var lonelyList = ['Actually I dont', originalList]

You're inserting the entire List as the second element. The end result being a:

<dynamic>['Actually I dont', ['follow', 'spread', 'operators']]

You probably didn't notice the second set of square brackets on the output

Rob Castillo0

Below is my code:

var originalList = ['follow', 'spread', 'operators.'];
var spreadList = ['See? I', ...originalList];
var lonelyList = ['Actually I dont', originalList];
print(spreadList);
print(lonelyList);

Here is my result:

I/flutter (26526): [See? I, follow, spread, operators.]
I/flutter (26526): [Actually I dont, [follow, spread, operators.]]

You can see that when using ...originalList it removes the square brackets. This means that when using ...originalList it adds multiple strings, but using originalList adds a list.

My Car0
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