cc1plus：警告：命令行选项" -Wstrict-prototypes& quot;对Ada / C / ObjC有效，但对C ++无效 [英] cc1plus: warning: command line option "-Wstrict-prototypes" is valid for Ada/C/ObjC but not for C++

问题描述

我正在构建用于Python的C ++扩展。我在编译过程中看到此警告-类型为

I am building a C++ extension for use in Python. I am seeing this warning being generated during the compilation process - when a type:

python setup.py build_ext -i

是什么原因引起的，如何解决？

What is causing it, and how do I fix it?

顺便说一句，这是我的安装文件的副本：

BTW, here is a copy of my setup file:

#!/usr/bin/env python

    """
    setup.py file for SWIG example
    """

    from distutils.core import setup, Extension


    example_module = Extension('_foolib',
                               sources=['example_wrap.cxx', 
                                        '../wrapper++/src/Foo.cpp'
                                       ],
                               libraries=["foopp"]
                               )

    setup (name = 'foolib',
           version = '0.1',
           author      = "Me, Myself and I",
           description = """Example""",
           ext_modules = [example_module],
           py_modules = ["example"],
           )

I在Ubuntu上使用gcc 4.4.3

I am using gcc 4.4.3 on Ubuntu

推荐答案

我可以回答部分问题，为什么会收到消息。

I can answer part of the question, why you're getting the message.

构建过程中的某些事情是使用选项 -Wstrict-prototypes 在C ++源文件上调用gcc。对于C和Objective-C，这会导致编译器警告未声明参数类型的老式函数声明。

Something in your build process is invoking gcc on a C++ source file with the option -Wstrict-prototypes. For C and Objective-C, this causes the compiler to warn about old-style function declarations that don't declare the types of arguments.

对于C ++，此选项不会没有道理；

For C++, this option doesn't make sense; such declarations aren't even allowed by the language (prototypes are mandatory).

（我不知道为什么消息中提到Ada； -严格的原型对Ada的意义远不及对C ++的意义。虽然这没什么大不了的，但是我已经提交了此错误报告，自2015年12月6日起标记为已解决/已修复。）

(I don't know why the message mentions Ada; -Wstrict-prototypes makes even less sense for Ada than for C++. It's not a huge deal, but I've submitted this bug report, marked as RESOLVED/FIXED as of 2015-12-06.)

解决方法是删除gcc调用的 -Wstrict-prototypes 选项。但是由于您不是直接调用gcc，所以很难知道该怎么做。

The solution should be to remove the -Wstrict-prototypes option from the invocation of gcc. But since you're not invoking gcc directly, it's difficult to know how to do that.

我能够使用您的设置重现警告.py ，在手动创建虚拟 example_wrap.cxx 文件后：

I was able to reproduce the warning using your setup.py, after manually creating a dummy example_wrap.cxx file:

% python setup.py build_ext -i
running build_ext
building '_foolib' extension
gcc -pthread -fno-strict-aliasing -DNDEBUG -g -fwrapv -O2 -Wall -Wstrict-prototypes -fPIC -I/usr/include/python2.7 -c example_wrap.cxx -o build/temp.linux-i686-2.7/example_wrap.o
cc1plus: warning: command line option "-Wstrict-prototypes" is valid for Ada/C/ObjC but not for C++
...

所以这可能是Python build_ext 中的一个小错误。

So it's probably a minor bug in Python's build_ext.

但是由于这只是一个警告，而不是致命错误，我想您可以放心地忽略它。 gcc会警告无意义的选项，但它只会忽略它。

But since it's only a warning, not a fatal error, I'd say you can safely ignore it. gcc warns about the meaningless option, but then it just ignores it.

编辑：

通过Python-2.7.2的源代码， configure.in 的这一部分可能是罪魁祸首：

Looking through the Python-2.7.2 sources, this section of configure.in might be the culprit:

case $GCC in
yes)
    if test "$CC" != 'g++' ; then
        STRICT_PROTO="-Wstrict-prototypes"
    fi

（I' m假定是在使用 build_ext 时调用的。）

(I'm assuming that's invoked when using build_ext.)

打开 -Wstrict-仅当不是作为 g ++ 调用编译器时，才选择prototypes 选项，但在您的情况下，它使用的是 gcc 命令来编译C ++源代码。在 Lib / distutils / command / build_ext.py 中， build_extension（）并不关注源文件调用 self.compiler.compile（）时的语言，仅当调用 self.compiler.link_shared_object（）时。 （这似乎很奇怪；对于gcc以外的编译器，您不一定能够使用相同的命令来编译C和C ++，并且使用 g ++ 命令，即使您未链接也是如此。）

It turns on the -Wstrict-prototypes option only if the compiler is not being invoked as g++ -- but in your case it's using the gcc command to compile C++ source code. And in Lib/distutils/command/build_ext.py, build_extension() doesn't pay attention to the source file language when invoking self.compiler.compile(), only when invoking self.compiler.link_shared_object(). (Which seems odd; for compilers other than gcc, you wouldn't necessarily be able to use the same command to compile C and C++ -- and it makes more sense to use the g++ command anyway, even if you're not linking.)

更新：提交了Python错误报告： https://bugs.python.org/issue9031 ，并作为此副本的副本关闭： https ：//bugs.python.org/issue1222585 ，在我撰写本文时仍会打开。

UPDATE: A Python bug report was submitted: https://bugs.python.org/issue9031, and closed as a duplicate of this one: https://bugs.python.org/issue1222585, which is still open as I write this.

但是正如我所说，这只是一个警告，你可以放心地忽略它。也许Python维护人员可以使用以上信息在将来的版本中解决该问题。

But as I said, it's only a warning and you can probably safely ignore it. Perhaps the Python maintainers can use the above information to fix the problem in a future release.

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