why setState does not change my variable Flutter? - flutter

Question

I have a variable to which i assign a value inside gridView.Builder and there is a button, when clicked on which my variable should change, I use setState for this, but it does not change, what could be the reason?
class _CatalogItemsState extends State<CatalogItems> {
Set<int> _isFavLoading = {};
bool isFavorite = false;
#override
Widget build(BuildContext context) {
return Scaffold(
appBar: AppBar(
title: Text(name!),
),
body: Padding(
padding: const EdgeInsets.only(left: 10, right: 10),
child: Column(
children: [
Expanded(
child: FutureBuilder<List<Product>>(
future: productFuture,
builder: (context, snapshot) {
if (snapshot.connectionState == ConnectionState.waiting) {
return buildGridShimmer();
} else if (snapshot.hasData) {
final catalog = snapshot.data;
if (catalog!.isEmpty) {
return const Center(
child: Text(
'Нет товаров',
style: TextStyle(
fontSize: 25, fontWeight: FontWeight.bold),
),
);
}
return buildCatalog(catalog);
} else {
print(snapshot.error);
return const Text("No widget to build");
}
}),
),
],
),
),
);
}
SliverGrid(
gridDelegate: SliverGridDelegateWithFixedCrossAxisCount(),
delegate: SliverChildBuilderDelegate(childCount: product.length,
(BuildContext context, int index) {
final media =
product[index].media?.map((e) => e.toJson()).toList();
final photo = media?[0]['links']['local']['thumbnails']['350'];
final productItem = product[index];
isFavorite = productItem.is_favorite; // this is the new value of the variable, work good
IconButton(
icon: Icon(_isFavLoading.contains(index) || isFavorite ? Icons.favorite : Icons.favorite_border, color: Colors.red,),
onPressed: () {
setState(() {
isFavorite = !isFavorite;
});
print('t: ${isFavorite}'); // the value of the variable does not change
},
)

Answer 1

This is because you're not really updating your product object. You must change its value when icon is pressed
onPressed: () {
setState(() {
productItem.is_favorite = !isFavorite;
});
print('t: ${productItem.is_favorite}'); // the value of the variable does not change
}