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Re: Compiling Zsh for Mac PowerPC and Intel
- X-seq: zsh-users 14059
- From: Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx>
- To: Zsh Users List <zsh-users@xxxxxxxxxx>
- Subject: Re: Compiling Zsh for Mac PowerPC and Intel
- Date: Wed, 22 Apr 2009 08:09:05 -0700
- In-reply-to: <d2ecb10b0904220631n4e9e256ctf2250acab9f73ca3@xxxxxxxxxxxxxx>
- Mailing-list: contact zsh-users-help@xxxxxxxxxx; run by ezmlm
- References: <d2ecb10b0904220631n4e9e256ctf2250acab9f73ca3@xxxxxxxxxxxxxx>
On Apr 22, 9:31am, TjL wrote:
}
} My question is, how do I compile zsh so I end up with a "universal" binary?
I'm no expert, but "man cc" seems to indicate that you have to pass the
-arch option to the compiler multiple times.
This did it for me:
make clean
make CFLAGS='-arch i386 -arch ppc -arch ppc64' LDFLAGS='${CFLAGS}'
And now:
schaefer[398] file Src/zsh
Src/zsh: Mach-O universal binary with 3 architectures
Src/zsh (for architecture i386): Mach-O executable i386
Src/zsh (for architecture ppc7400): Mach-O executable ppc
Src/zsh (for architecture ppc64): Mach-O 64-bit executable ppc64
I have no idea whether it'll actually *run* on all three platforms.
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