gcc的-Wc90-c99-compat标志如何工作？

Question

I've a code like

#include <stdio.h>
#include <stdbool.h>
main() {
    bool ok = true;
    printf("%u\n", ok);
}

and I'm compiling it like

gcc -std=c90 -Wc90-c99-compat a.c

but I don't get any warnings.

I expected some kind of warning since bool is not defined in C90 . The gcc documentation states that the -Wc90-c99-compat flag is used to

[w]arn about features not present in ISO C90, but present in ISO C99.

Here you can try it yourself . My version of gcc is 7.3.0, TIO's version is 8.1.1.

Am I doing something wrong or misunderstanding something?

Update. Something like

#include <stdio.h>
main() {
    _Bool ok = 1;
    printf("%u\n", ok);
}

does give off a warning: warning: ISO C90 does not support boolean types . Try it here .

I've also tried preprocessing the first version with

gcc -std=c90 -Wc90-c99-compat -E a.c

and the relevant part becomes

main() {
    _Bool ok = 1;
     printf("%u\n", ok);
}

but no errors raised about that.

Update.

As others suggested I've filed a bug report .

Answer 1

[this is more a comment than an aswer; AFAICT this looks like a gcc bug, but someone may come with a cogent explanation of it]

Your example could be reduced to:

# 2 "a.c" 3
main() {
    _Bool ok = 1;
}

With the bug (feature?) being that cc -std=c90 -Wc90-c99-compat -Wsystem-headers doesn't warn about the c99 _Bool type, despite the -Wsystem-headers option.

The 3 at the end of the preprocessor line directive means that the code following it is from a system header (see the c preprocessor docs ) -- and removing the 3 brings back the warning about _Bool .

If you check the cc -E output you will see that the expansions of the bool and true macros are wrapped in such lines, because they are defined in a system header ( stdbool.h ):

$ cat a.c
#include <stdbool.h>
main() {
    bool ok = true;
}
$ gcc -E a.c
...
main() {

# 3 "a.c" 3 4
   _Bool
# 3 "a.c"
        ok =
# 3 "a.c" 3 4
             1
# 3 "a.c"
                 ;
}