Re: using undeclared function returning bool results in wrong return value

[Martin Sebor via Gcc]

On 2/17/21 2:05 PM, Thanos Makatos via Gcc wrote:

I run into a problem that I'm not sure whether it's a bug in my program (most 
likely) or something wrong with GCC (highly unlikely, I know, hence why I 
haven't sent this to gcc-bugs). The problem is using a function that returns a 
bool, defined in another source file without a declaration, and the program 
compiled with -O. In my test, function foo1 returns true while it should be 
returning false.

Here are the files and repro:

/* main.c */
#include <stdio.h>
#include <stdbool.h>
#include <assert.h>

bool foo2(int);

void main(void) {
         assert(!foo2(0));
         assert(!foo1(0));
}

/* foo.c */
#include <stdbool.h>

bool foo1(int n) {
         return n == 1 || n == 2 || n == 4;
}

bool foo2(int n) {
         return foo1(n);
}

# gcc --version
gcc (Debian 8.3.0-6) 8.3.0
# gcc main.c foo.c -O
In file included from main.c:3:
main.c: In function 'main':
main.c:9:10: warning: implicit declaration of function 'foo1'; did you mean 
'foo2'? [-Wimplicit-function-declaration]
   assert(!foo1(0));
           ^~~~
# ./a.out
a.out: main.c:9: main: Assertion `!foo1(0)' failed.
Aborted

I get the same behavior if I use char instead of bool.

The problem goes away if:
1. I declare foo1 in main.c (just like I do for foo2), or
2. compile w/o -O, or
3. use n == 3 instead of n == 4 in the return statement, or
4. use short or int instead of bool/char.

Am I causing undefined behavior because I'm not declaring foo1?

Yes.

Implicit function declarations are assumed to return int so defining
one that doesn't match is undefined.  If the function is defined to
return a smaller type the value may be passed in only a part of
a register (with the remaining bits left unchanged) and the caller
will be looking at the whole thing.

Martin

Please CC me in your response, I'm not subscribed to this list.