如何使用make和C99进行编译? [英] How to use make and compile as C99?

问题描述

我正在尝试使用Makefile编译linux内核模块:

I'm trying to compile a linux kernel module using a Makefile:

obj-m += main.o

all:
    make -C /lib/modules/$(shell uname -r)/build M=$(PWD) modules

clean:
    make -C /lib/modules/$(shell uname -r)/build M=$(PWD) clean

哪个给我:

main.c:54: warning: ISO C90 forbids mixed declarations and code

我需要切换到C99.阅读后，我注意到我需要添加一个标志-std = c99，不确定是否应该将其添加到何处.

I need to switch to C99. After reading I noticed I need to add a flag -std=c99, not sure where it suppose to be added.

如何更改Makefile，使其可以编译为C99?

How do I change the Makefile so it will compile as C99?

推荐答案

与makefile无关. ISO C90禁止在块或文件开头以外的任何地方声明变量-像这样

It's got nothing to do with the makefile. ISO C90 forbids declaring variables anywhere but in the beginning of a block or the file - like this

int main(int argc, char **argv) {
   int a; /* Ok */
   int b = 3; /* Ok */

   printf("Hello, the magic number is %d!\n", b);
   int c = 42; /* ERROR! Can only declare variables in the beginning of the block */
   printf("I also like %d.. but not as much as %d!\n", c, b);

   return 0;
}

因此必须对此进行修改...

Thus it has to be modified to this...

int main(int argc, char **argv) {
   int a; /* Ok */
   int b = 3; /* Ok */
   int c = 42; /* Ok! */

   printf("Hello, the magic number is %d!\n", b);
   printf("I also like %d.. but not as much as %d!\n", c, b);

   return 0;
}

您只能在源代码中修复"该问题，而不能在makefile中.

You can only "fix" that in the source code, not in the makefile.

此规则在C99中已经放宽，但是我认为将变量定义，声明和初始化与下面的代码分开是个好主意:)

This rule has been relaxed in C99, but in my opinion it's a good idea to separate variable definitions, declarations and initializations from the code below it :)

因此，要更改您的makefile使其可以用C99进行编译，您需要在makefile所引用的"build"目录中更改Makefile，并在"gcc"行中添加"-std = c99"源文件.

So to change your makefile to make it compile with C99, you need to change the Makefile in the "build" directory that your makefile is referencing, and add the "-std=c99" at the "gcc" line compiling the source file.

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