Xcode 14.3 CLI command to trigger Rosetta build/run

Question

Created Apr ’23

Replies 7

Boosts 11

Views 3.4k

Participants 7

Hi, as Xcode 14.3 dropped Rosetta 2 Support and added a run option directly in Xcode 14.3 (Rosetta Simulators), I wondered how I can trigger a build for a i.e. iPhone 14 Pro (Rosetta) via CLI. I already checked the man page of xcbuild and xcrun.

xcrun simctl list also outputs only the simulators and the device IDs which brings me to the assumption that the building and running for Rosetta is a parameter/flag set by Xcode. But which and where?

Happy for any sort of help, resource, or hint.

Boost

Answer 1

liorho OP

Apr ’23

Having the same issue, wondering what will be the response for this

1

Answer 2

Valeriy OP

May ’23

I'm in the same boat too...

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Answer 3

Jride88 OP

May ’23

Having the exact same issue. Has anyone had any luck with this?

0

Answer 4

Lemoni OP

Jun ’23

same issue, any update ?

0

Answer 5

DevinWu OP

Jun ’23

same issue, please help~

0

Answer 6

Noitham OP

Jun ’23

@Lemoni @Jride88 @Valeriy @liorho

Give this a go:

Append arch=x86_64 to build with a Rosetta simulator.

For example: xcodebuild -workspace App.xcworkspace -scheme App -destination "platform=iOS Simulator,name=iPhone 13,arch=x86_64" build

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Answer 7

Lemoni OP

Jun ’23

@Noitham not work..

xcodebuild test -workspace OmniSDK.xcworkspace -scheme OmniAPITests -destination 'platform=iOS Simulator,name=iPhone 14,arch=x86_64'

--- xcodebuild: WARNING: Using the first of multiple matching destinations: { platform:macOS, arch:x86_64, variant:Mac Catalyst, id:00008103-001D69941A79001E }

Error: building for Mac Catalyst, but linking in object file built for iOS Simulator...

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