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

Question

I need to develop an app who support iOS 4 - iOS 7, is it possible to do it on XCode 5.0.2 on Mavericks?

The lowest deployment target xcode provide me is iOS 6.

Answer 1

In Project > Target > Build Settings > Architectures:

change the Architectures from Standard Architectures (including 64-bit) to Standard Architectures ,or if you want to support older devices, type armv6 armv7 armv7s in Others

Then, you can change the deployment target to lower versions (just type 4.0 or 4.3 in the Deployment Target box)

But why do you want to target iOS 4 still?

Answer 2

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

Answer 3

Try this , You will get the deployement Target till from lowerEnd . Its mainly due to the >XCode 5.0 choses by default architecture as including 64-bit this only supports >IOS 6.0

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

Xcode can build your app with both 32-bit and 64-bit binaries included. This combined binary requires a minimum deployment target of iOS 7 or later.

Note: A future version of Xcode will let you create a single app that supports the 32-bit runtime on iOS 6 and later, and that supports the 64-bit runtime on iOS 7.