如何从shell脚本中检测到它正在M1苹果硬件上运行？
我希望能够运行命令行命令，这样我就可以编写一个if-语句，它的主体只有在使用M1处理器的mac上运行时才会执行(当然，至少是macOS Big Sur(。
uname -m
将返回arm64而不是x86_64
if [[ $(uname -m) == 'arm64' ]]; then
echo M1
fi
或者，正如@chepner建议的
uname -p
将返回arm而不是i386
if [[ $(uname -p) == 'arm' ]]; then
echo M1
fi
另一个工具是arch:
if [[ $(arch) == 'arm64' ]]; then
echo M1
fi
我发现sysctl -n machdep.cpu.brand_string报告了Apple M1，尽管该进程是在Rosetta下运行的。
更新：为Apple M1 Pro、Apple M2、Apple M2 Max等做好准备。！
当使用原生shell(比如/bin/bash -i或/bin/zsh -i(时，Klas Mellbourn的答案会按预期工作。
如果使用通过Intel/Rosetta Homebrew安装安装的shell，则uname -p返回i386，uname -m返回x86_64，如Datasun的注释所示。
为了获得跨环境工作的东西(Apple Silicon Native、Rosetta Shell、Linux、Raspberry Pi 4s(，我使用了dorothy dotfile生态系统中的以下内容：
is-mac && test "$(get-arch)" = 'a64'
如果你没有使用dorothy，dorothy的相关代码是：
https://github.com/bevry/dorothy/blob/1c747c0fa6bb3e6c18cdc9bae17ab66c0603d788/commands/is-mac
test "$(uname -s)" = "Darwin"
https://github.com/bevry/dorothy/blob/1c747c0fa6bb3e6c18cdc9bae17ab66c0603d788/commands/get-arch
arch="$(uname -m)" # -i is only linux, -m is linux and apple
if [[ "$arch" = x86_64* ]]; then
if [[ "$(uname -a)" = *ARM64* ]]; then
echo 'a64'
else
echo 'x64'
fi
elif [[ "$arch" = i*86 ]]; then
echo 'x32'
elif [[ "$arch" = arm* ]]; then
echo 'a32'
elif test "$arch" = aarch64; then
echo 'a64'
else
exit 1
fi
sysctl -n machdep.cpu.brand_string输出Apple M1，但在Raspberry Pi 4 Ubuntu服务器上输出以下内容：
> sysctl -n machdep.cpu.brand_string
Command 'sysctl' is available in the following places
* /sbin/sysctl
* /usr/sbin/sysctl
The command could not be located because '/sbin:/usr/sbin' is not included in the PATH environment variable.
This is most likely caused by the lack of administrative privileges associated with your user account.
sysctl: command not found
> sudo sysctl -n machdep.cpu.brand_string
sysctl: cannot stat /proc/sys/machdep/cpu/brand_string: No such file or directory
/usr/bin/arch就是你所需要的。既然已经有一个程序可以为你做这件事，为什么还要把事情复杂化呢。