Why is armv6 a valid architecture for iOS 5?

Question

Can someone tell me why creating a new project with the goal of deploying iOS for iOS 5.0 includes armv6 as a valid architecture in the project> Default build settings? In the info.plist file, by default, only armv7 elements are displayed in the capabilities of the required device.

As I understand it, if I create for iOS 5 or later that will only work on armv7 devices (i.e. iPhone 3GS or later, iPod 3G or later and iPad). Thus, I should be able to remove all armv6 links and install Build Active Architecture only YES.

Can anyone confirm or clarify? Thanks.

Xcode v4.3.2, Basic SDK 5.1, iOS Deployment for iOS 5.0, iPhone Device Family

PS - I understand that removing armv6 will reduce the size of the binary.

Answer 1

$ (VALID_ARCHS) defines the set of architectures supported by the Xcode target. $ (ARCHS) tells Xcode which architectures to build. In fact, for each target, an intersection of $ (VALID_ARCHS) and $ (ARCHS) is created.

$ (VALID_ARCHS) is usually used to avoid creating a target for some architecture that otherwise supports the project as a whole. If you do not need to do this, you can safely ignore it.