Eric Roode wrote:
>
> Steve Fink wrote:
> >I am merely suggesting that the compiler detect, when it can, that
> >you're trying to use the value of a variable without ever having
> >assigned a value to that variable. And in THAT message, you had better
> >know the name of the variable, since it's the basis of the analysis. And
> >yes, it only handles simple named variables.
>
> Um, "when it can"? Isn't this considered a Hard Problem by the
> computer scientists?
The halting problem is unsolvable. That doesn't mean you can't figure
out whether the complete program
$x = 1
halts.
But that doesn't even matter that much here; I'm saying that if the
compiler can definitely determine that you are using an uninitialized
variable, it should warn. If you're using something that appears like it
might be uninitialized, but can't be proven either way without solving
that pesky halting problem, then it (if you ask it) says "you might have
screwed up." 90% of the time, it's right -- if the right set of freak
occurrences happen, you really do use the value of an uninitialized
variable. As an example, this may produce a spurious warning:
$y = 1 if $x;
$y = 2 if ! $x;
...because the compiler is unable to be sure that ($x) || (!$x) is
definitely true. (And in fact, it might not be in the presence of tying.
Which is why, when doing this sort of thing, compilers usually
completely ignore the actual conditions being tested, and assume that
they're independent and sometimes true, sometimes false.)
> >Example:
> >
> >1 my ($x, $y, $z);
> >2 $z = 1;
> >3 my $logfile = "/tmp/log";
> >4 $x = 1 if cond();
> >5 print $x+$y;
> >6 undef $z;
> >7 print $z;
> >
> >--> use of uninitialized variable $y in line 5 (compile time)
> >--> possible use of uninitialized variable $x in line 5 (compile time)
> >--> variable $logfile defined in line 3 but never used (compile time)
> >--> use of undefined value in line 7 (run time)
>
> How about:
>
> foo();
> $x = 1 unless defined($x);
> print $x;
>
> Generate a warning, or not?
$x is a global. The compiler cannot detect all possible assignments to
or accesses of globals, so it never warns about them.
If you inserted my $x at the top of that code, it would most likely
produce the "possible use" warning. Or not; this is a simple enough case
that it might be able to infer the right answer.
I am certainly not saying that the "possible use" warning should be
enabled by default. But please, argue over that one separately from the
others. It's the most likely to annoy.
> Or:
> foo();
> print $x;
>
> Generate a warning, or not? Which one? Remember, foo() may initialize $x.
Same thing. If $x is lexical, it gives a definite warning. If $x is a
global, it says nothing. You're right; I need to point this out in the
RFC.
