我正在尝试为我们的 Xcode 交叉编译设置 CI。交叉编译测试了 ARMv7 和 ARMv8。事情看起来不错,除非需要链接 ARMv8:
clang++ -DNDEBUG -g2 -O3 -fPIC -pipe -Wall -miphoneos-version-min=7 -arch arm64 \
-isysroot /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS10.2.sdk \
-stdlib=libc++ -c cryptlib.cpp
clang++ -DNDEBUG -g2 -O3 -fPIC -pipe -Wall -miphoneos-version-min=7 -arch arm64 \
-isysroot /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS10.2.sdk \
-stdlib=libc++ -c cpu.cpp
...
clang++ -o cryptest.exe -DNDEBUG -g2 -O3 -fPIC -pipe -Wall -miphoneos-version-min=7 -arch arm64 \
-isysroot /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS10.2.sdk \
-stdlib=libc++ test.o bench1.o bench2.o ... ./libcryptopp.a
Undefined symbols for architecture arm64:
"CryptoPP::CRC32_Update_ARMV8(unsigned char const*, unsigned long, unsigned int&)", referenced from:
CryptoPP::CRC32::Update(unsigned char const*, unsigned long) in libcryptopp.a(crc.o)
"CryptoPP::CRC32C_Update_ARMV8(unsigned char const*, unsigned long, unsigned int&)", referenced from:
CryptoPP::CRC32C::Update(unsigned char const*, unsigned long) in libcryptopp.a(crc.o)
ld: symbol(s) not found for architecture arm64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
make: *** [cryptest.exe] Error 1
我们显然不运行输出工件cryptest.exe
。我们只是编译和链接来测试东西。
该代码在 LLVM Clang 下测试良好。
所有 ARMv8/Aarch64 机器都有 CRC-32 和 CRC-32C;但加密扩展是可选的。该错误没有多大意义。
Clang 是否缺少 ARMv8/Aarch64 的 CRC32?
以下是导致错误的代码。
#if defined(__ARM_FEATURE_CRC32)
void CRC32_Update_ARMV8(const uint8_t *s, size_t n, uint32_t& c)
{
for(; !IsAligned<uint32_t>(s) && n > 0; s++, n--)
c = __crc32b(c, *s);
for(; n > 4; s+=4, n-=4)
c = __crc32w(c, *s);
for(; n > 0; s++, n--)
c = __crc32b(c, *s);
}
#endif