How to use make and compile as C99?

Question

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

Which gives me:

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

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.

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

[How to compile a Linux kernel module using -std=gnu99?](https://stackoverflow.com/questions/15910064/how-to-compile-a-linux-kernel-module-using-std-gnu99) — Shay, Oct 09 '19 at 13:19

score 23 · Answer 1 · edited Dec 13 '11 at 14:08

23

The correct way to add compiler flags when compiling modules is by setting the ccflags-y variable. Like this:

ccflags-y := -std=gnu99

See Documentation/kbuild/makefiles.txt in the kernel tree for more information.

Note that I'm using the gnu99 standard instead of c99 since the Linux kernel heavily relies on GNU extensions.

edited Dec 13 '11 at 14:08

Maus

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answered Apr 04 '11 at 19:53

mtvec

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Hi, I add ccflags-y to the Makefile for my own module, but the compiler still warns that "ISO C90 forbids mixed declarations and code". Why? – basicthinker Apr 22 '12 at 03:47
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@basicthinker: maybe [this](http://stackoverflow.com/questions/15910064/how-to-compile-a-linux-kernel-module-using-std-gnu99) helps – Ciro Santilli OurBigBook.com Jun 23 '13 at 14:47

score 20 · Answer 2 · answered Jan 16 '11 at 02:30

20

You could just add

CFLAGS=-std=c99

To the top of your makefile, or you can make the code compliant with C90 (as LukeN suggests.)

answered Jan 16 '11 at 02:30

ocodo

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LukeN · Accepted Answer · 2010-05-29T12:55:08.170

-8

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;
}

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;
}

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

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 :)

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.

edited May 29 '10 at 12:55

answered May 29 '10 at 12:42

LukeN

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CFLAGS is much more common, preferred, and less fragile than editing each invocation of the compiler. – May 29 '10 at 13:00
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After living in the OO (Java) world for a long time and recently getting back into using C on a more day to day basis I have to disagree with the point about separating variable information out. Keeping things in as small a scope as possible seems more important. Some variables will only be needed inside of a while or for loop for example. – powerj1984 May 11 '11 at 03:01
@powerj1984 You may use brace {} scoping in this case. LukeN is correct as in *what would be accepted by the kernel main line*. Whether that is a good thing all around, I won't say and I guess it is not what the OP asked. – artless noise Mar 03 '14 at 23:08
The kernel allows c11 now, then no need to stay with this old pedantic style. – Alexis Feb 09 '22 at 08:10

How to use make and compile as C99?

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