SwiftUI can not return View on function - swift

Question

I'm trying return View based selected menu item on function. But it throws error:
Protocol 'View' can only be used as a generic constraint because it has Self or associated type requirements.
There my code:
enum MenuItem {
case Main
case Report
}
struct Menu: View {
#State var activeItem: MenuItem = .Main
private func getActiveView() -> View {
switch activeItem {
case .Main:
return DashboardView()
case .Report:
return ReportView()
}
}
var body: some View {
...
getActiveView()
...
}
}
struct DashboardView: View {
var body: some View {
Text("Contact")
}
}
struct ReportView: View {
var body: some View {
Text("Contact")
}
}
Im new on SwiftUI. Any ideas how to return View?

Answer 1

SwiftUI 2
Here is a solution tested with Xcode 12b / iOS 14
struct Menu: View {
#State var activeItem: MenuItem = .Main
// make function ViewBuilder
#ViewBuilder
private func getActiveView() -> some View {
switch activeItem {
case .Main:
DashboardView() // don't use 'return' as it disables ViewBuilder
case .Report:
ReportView()
}
}
var body: some View {
getActiveView()
}
}

Answer 2

SwiftUI gives us a type-erased wrapper called AnyView that we can return.
Tested Solution:
struct Menu: View {
#State var activeItem: MenuItem = .Main
func getActiveView() -> some View {
switch activeItem {
case .Main:
return AnyView(DashboardView())
case .Report:
return AnyView(ReportView())
}
}
var body: some View {
getActiveView()
}
}
Note: type-erased wrapper effectively forces Swift to forget about what specific
type is inside the AnyView, allowing them to look like they are the
same thing. This has a performance cost, though, so don’t use it
often.
For more information you can refer to the this cool article: https://www.hackingwithswift.com/quick-start/swiftui/how-to-return-different-view-types