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Perhaps there's a sandbox escape hiding in the workaround they did to get swiftc to use their jemalloc build?
by m_eiman 5y ago
Perhaps there's a sandbox escape hiding in the workaround they did to get swiftc to use their jemalloc build?
- saagarjha 5y agoIt’s a build variable in Xcode telling it which compiler to use. At that point you have arbitrary code execution anyways as you control the build process.