如何编译使用-std = gnu99一个Linux内核模块？ [英] How to compile a Linux kernel module using -std=gnu99?
问题描述
我最近学会了如何编程简单的字符驱动程序和同时与code我注意到，我得到了很多抛出我的C99 code以下GCC警告玩耍：
I've recently learned how to program simple character drivers and while playing around with the code I noticed that I get a lot of the following GCC warnings thrown for my C99 code:
warning: ISO C90 forbids mixed declarations and code
我想这是因为主要的Linux内核Makefile文件设置为使用非C99标准进行编译。我搜索周围，我发现这个答案在这里计算器：<一href=\"http://stackoverflow.com/questions/2935047/how-to-use-make-and-compile-as-c99/5543589#5543589\">How使用化妆和编译为C99？
于是，我试着自然在我的Makefile如下：
So I naturally tried the following in my Makefile:
ccflags-y := -std=gnu99
不幸的是这并不能平息海湾合作委员会的警告。我检查的详细输出使并验证GCC确实用在了 -std = gnu99 上涨了执行结束;所以我有点糊涂了。
Unfortunately this didn't silence the GCC warnings. I checked the verbose output of make and verified that GCC is indeed executed with the -std=gnu99 tacked on at the end; so I'm a bit confused.
如何正确使用 -std = gnu99 选项编译Linux内核模块？
How do I properly compile a Linux kernel module using the -std=gnu99 option?
修改
我注意到GCC输出显示此选项： -Wdeclaration-后声明。这是为什么我收到的警告甚至与 -std = gnu99 选项？
I noticed the GCC output shows this option: -Wdeclaration-after-statement. Is this why I am getting the warnings even with the -std=gnu99 option?
推荐答案
原来， -std = gnu99 的确实能使工作的;我开始卸下编译器标志后看到关于C99的功能错误。因此，这意味着别的东西是导致警告打印出除了 -std = 选项。
It turns out that -std=gnu99 does in fact work; I began seeing errors regarding C99 features after removing the compiler flag. So that meant something else was causing the warnings to print out besides the -std= option.
通过通过详细的输出解析后做出V = 1 ，我发现了 -Wdeclaration-后声明选项作为海湾合作委员会执行的一部分。这是我看见的ISO C90混合声明警告的原因。
After parsing through the verbose output via make V=1, I discovered the -Wdeclaration-after-statement option as part of the GCC execution. This was the cause of the ISO C90 mixed declaration warnings I saw.
要禁用ISO C90警告，它传递给GCC： -Wno声明-后声明
To disable the ISO C90 warnings, pass this to GCC: -Wno-declaration-after-statement.
例如：
ccflags-y := -std=gnu99 -Wno-declaration-after-statement
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