Is it possible to create an iOS 4 - iOS 7 application compatible with Xcode 5.0.2 on Mavericks?

Question

I need to develop an application that supports iOS 4 - iOS 7, can this be done on XCode 5.0.2 on Mavericks?

The youngest xcode of the deployment target is iOS 6.

Answer 1

In the project> Goal> Build Settings> Architectures:

change Architectures to Standard architectures (including 64-bit) to Standard architectures or if you want to support older devices, enter armv6 armv7 armv7s in other

You can then change the deployment target to lower versions (just enter 4.0 or 4.3 in the "Deployment Location" field)

But why do you want to target iOS 4?

Answer 2

Yes, support for iOS4 is possible, the problem is support for ARMv6 (iPhone 3G and iPod Touch 2G). To support ARMv6, you need the iOS SDK 5.x, which is available in Xcode 4.4. But you also need to support iPhone widescreen resolution, so you need to use two different versions of Xcode for it

Answer 3

Try this, you will get the deployment target from lowerEnd. This is mainly due to the fact that> Xcode 5.0 uses architecture by default , including 64-bit , it only supports iOS 6.0

iPhone 5S is powered by A7 64bit processor . From apple docs

Xcode can create your application with both 32-bit and 64-bit binaries enabled. This combo binary requires a minimum deployment target of iOS 7 or later.

Note. The future version of Xcode will allow you to create one application that supports the 32-bit runtime on iOS 6 and later and supports the 64-bit runtime on iOS 7.