我正在尝试从其源代码编译Python2.6,并采取以下步骤,并遇到错误消息:
./configure
...
[root@KuGouIdc Python-2.6.6]# ls
config.log Demo Makefile.pre Objects Python
config.status Doc Makefile.pre.in Parser README
configure Grammar Makefile.pre.in.cflags PC RISCOS
configure.in Include Makefile.pre.in.fix-parallel-make PCbuild setup.py
configure.in.check-for-XML_SetHashSalt install-sh Makefile.pre.in.lib64 pyconfig.h setup.py.add-RPATH-to-pyexpat
configure.in.disable-pymalloc-on-valgrind Lib Makefile.pre.in.no-static-lib pyconfig.h.in setup.py.expat
configure.in.expat LICENSE Makefile.pre.in.systemtap pyconfig.h.in.disable-pymalloc-on-valgrind setup.py.lib64
configure.in.rpath Mac Misc pyconfig.h.in.systemtap systemtap-example.stp
configure.in.systemtap Makefile Modules pyfuntop.stp Tools
[root@KuGouIdc Python-2.6.6]# make -n all
gcc -pthread -c -fno-strict-aliasing -g -O2 -DNDEBUG -g -fwrapv -O3 -Wall -Wstrict-prototypes -I. -IInclude -I./Include -DPy_BUILD_CORE -o Modules/python.o ./Modules/python.c
make: *** No rule to make target `libpython2.6.a', needed by `python'. Stop.
我查看了 Makefile 详细信息,确实找不到任何规则来制作目标“libpythonXXX.a”。我google了一下,好像没有人遇到过和我一样的问题?
在 Makefile 中,它确实有一个规则来制作目标“libpythonXXX.so”。