在Python 3.x中，为什么磁盘上没有itertools共享对象? [英] In Python 3.x, why is there not an itertools shared-object on disk?

问题描述

itertools C模块是否包含在主程序中3.x中的Python二进制文件?

Is the itertools C module included somehow in the main Python binary in 3.x?

假定已构建并包含C模块，它看起来像是:

Assuming that the C module is built and included, which it appears to be:

>>> import inspect
>>> import itertools
>>>
>>> inspect.getsourcefile(itertools)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/python3.4/inspect.py", line 571, in getsourcefile
    filename = getfile(object)
  File "/usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/python3.4/inspect.py", line 518, in getfile
    raise TypeError('{!r} is a built-in module'.format(object))
TypeError: <module 'itertools' (built-in)> is a built-in module

我在系统上找不到用于Python 3.x的itertools.so，但是有一个用于2.7的.

I can't find an itertools.so for Python 3.x on my system, but there's one for 2.7.

我注意到其他一些C模块作为磁盘上的共享对象(locate '.so' | grep -E '^/usr/local/' | grep '.so'例如mmap.so)存在，那么itertools有什么用呢?如果没有共享库，怎么用?

I noted that some other C modules exist as shared objects (locate '.so' | grep -E '^/usr/local/' | grep '.so' e.g. mmap.so) on disk, so what's the deal with itertools? How can I use it if there's not a shared library?

推荐答案

makefile中的提示靠近inspect.py的Python包装器:

There are hints in the makefile that's near the Python wrapper of inspect.py:

/usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/python3.4/config-3.4m/Makefile

我们可以看到itertools.c源代码的构建规则:

We can see the build rules for the itertools.c source:

1668 Modules/itertoolsmodule.o: $(srcdir)/Modules/itertoolsmodule.c; $(CC) $(PY_CORE_CFLAGS)  -c $(srcdir)/Modules/itertoolsmodule.c -o      Modules/itertoolsmodule.o

然后对其进行一点跟踪以查看其是否已捆绑在其中:

And then trace it a little to see that it's being bundled in:

 24 MODOBJS= ..  Modules/itertoolsmodule.o  ... Modules/xxsubtype.o

 462 # objects that get linked into the Python library
 463 LIBRARY_OBJS_OMIT_FROZEN=   \
...
 470         $(MODOBJS)
 471
 472 LIBRARY_OBJS=   \
 473         $(LIBRARY_OBJS_OMIT_FROZEN) \
 474         Python/frozen.o
 ...
 553 # Build the interpreter
 554 $(BUILDPYTHON): Modules/python.o $(LIBRARY) $(LDLIBRARY) $(PY3LIBRARY)
 555     $(LINKCC) $(PY_LDFLAGS) $(LINKFORSHARED) -o $@ Modules/python.o $(BLDLIBRARY) $(LIBS) $(MODLIBS) $(SYSLIBS) $(LDLAST)
 556
 557 platform: $(BUILDPYTHON) pybuilddir.txt
 558     $(RUNSHARED) $(PYTHON_FOR_BUILD) -c 'import sys ; from sysconfig import get_platform ; print(get_platform()+"-"+sys.version[0:     3])' >platform
 589 # Build static library
 ...
 598     $(AR) $(ARFLAGS) $@ $(MODOBJS)
 599     $(RANLIB) $@

 944 $(LIBRARY_OBJS) $(MODOBJS) Modules/python.o: $(PYTHON_HEADERS)

或者如果通过distutils创建，则路径将类似于:/usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/python3.4/_sysconfigdata.py

Or if made via distutils, the path will be something like: /usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/python3.4/_sysconfigdata.py

并假定将其内置到动态库中

And assuming that this gets built into a dynamic library:

Ξ ~ → strings /usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/libpython3.4.dylib | grep itertools
itertools
itertools._tee_dataobject
itertools._tee
itertools._grouper
itertools.groupby
itertools.repeat
itertools.product
...

这意味着在构建时，itertools.c模块将包含在libpython动态库中，这意味着它现在已成为标准库的一部分.

Which means that at build time, the itertools.c module gets included in the libpython dynamic library, meaning that it's now part of the standard library.

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