Re: GCC not detecting use of uninitialized variable?

[Willy Tarreau]

On Wed, Oct 27, 2021 at 09:47:31PM -0400, Alan Stern wrote:
> On Wed, Oct 27, 2021 at 10:48:31PM +0200, Willy Tarreau wrote:
> > On Wed, Oct 27, 2021 at 04:12:49PM -0400, Alan Stern wrote:
> > > The following code does not generate a warning when compiled with GCC 
> > > 11.2.1:
> > > 
> > > 
> > > int foo;
> > > 
> > > void cc_test(void)
> > > {
> > > 	int	a, b;
> > > 
> > > 	a = 0;
> > > 	a = READ_ONCE(foo);	// Should be: b = READ_ONCE(foo)
> > > 	do {
> > > 		a += b;
> > > 		b = READ_ONCE(foo);
> > > 	} while (a > 0);
> > > 	WRITE_ONCE(foo, a);
> > > }
> > > 
> > > 
> > > But if the loop is changed to execute only once -- replace the while 
> > > test with "while (0)" -- then gcc does warn about the uninitialized use 
> > > of b.
> > > 
> > > Is this a known problem with gcc?  Is it being too conservative about 
> > > detecting uses of uninitialized variables?
> > 
> > I already had similar issues not being detected in loops.  I guess the
> > reason is simple: it might not be trivial for the compiler to prove
> > that the value was not set on any path leading to the first use,
> > because one of these paths is the loop itself after the instruction was
> > assigned. I've been so much used to it that I think it has always been
> > there and I can live with it.
> 
> Well, in this case there's only one path leading to the first use, since 
> the path that is the loop itself will never be the first use.  It seems 
> like a rather surprising oversight.

For the first iteration yes but not the next ones. And each time I met
a similar bug not being detected it was exactly in this situation. For
example the warning about "variable X is set but not used" tends to
disappear in such loops:

      extern int blah();
      int ret()
      {
          int a;
          do { a = 1; } while (blah());
          return 0;
      }

says "variable 'a' is set but not used". Just change "a=1" to "a++" and
it disappears:

      extern int blah();
      int ret()
      {
          int a;
          do { a++; } while (blah());
          return 0;
      }

And the asm code shows that the a++ code is optimized away, explaining
why there is no "may be used uninitialized" while it appears if you
return a instead of 0.

With that said, it could also depend on the gcc version and/or some
kernel options, as gcc-7, 8 and 9 do emit the warning for me on your
code when I build it by hand. You may want to double-check this
aspect before asking GCC people.

Willy