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Re: how to make gcc warn about arithmetic signed overflow
- From: Jonathan Wakely <jwakely dot gcc at gmail dot com>
- To: "James K. Lowden" <jklowden at schemamania dot org>
- Cc: gcc-help <gcc-help at gcc dot gnu dot org>
- Date: Mon, 23 Sep 2013 08:54:57 +0100
- Subject: Re: how to make gcc warn about arithmetic signed overflow
- Authentication-results: sourceware.org; auth=none
- References: <20130921164609 dot GC3086 at a dot lan> <CAH6eHdTToM+TMy55m5HYo39DC8nA0RrTma1Bp5OnhUtPErMfOA at mail dot gmail dot com> <20130921174229 dot GD3086 at a dot lan> <CAH6eHdQzJNQY4Meysi259RheSaGscKUF28OW43OvOD1rf6FkxQ at mail dot gmail dot com> <20130923000355 dot fa2a964c dot jklowden at schemamania dot org>
On 23 September 2013 05:03, James K. Lowden wrote:
> On Sat, 21 Sep 2013 19:30:02 +0100
> Jonathan Wakely <jwakely.gcc@gmail.com> wrote:
>
>> > its value can be changed using pointers
>>
>> No, that's not true. You can't change the value of a const object in a
>> valid program.
>
> I don't know if we're talking C or C++ at this point, but const_cast
> will surely let you change the value of a const object without treading
> into undefined behavior.
No, it surely won't!
If an object is defined as const in the first place then it is
undefined behaviour to change it.
1.9 [intro.execution]/4:
"Certain other operations are described in this International Standard
as undefined (for example, the effect of attempting to modify a const
object)."
5.2.11 [expr.const.cast]/7:
"[ Note: Depending on the type of the object, a write operation
through the pointer, lvalue or pointer
to data member resulting from a const_cast that casts away a
const-qualifier may produce undefined
behavior (7.1.6.1). — end note ]"
And the definitive reference, 7.1.6.1 [dcl.type.cv]/4:
"Except that any class member declared mutable (7.1.1) can be
modified, any attempt to modify a const object during its lifetime
(3.8) results in undefined behavior."
You can't even do it by destroying an a const objehct and recreating a
new object at the same address:
3.8 [basic.life]/9
"Creating a new object at the storage location that a const object
with static, thread, or automatic storage duration occupies or, at the
storage location that such a const object used to occupy before its
lifetime ended results in undefined behavior."
> Regarding the OP's query
>
>> > int r = ab * bc;
>
> although the provided example is simple enough, it's the compiler's
> job is to generate object code, not to do static analysis.
>
> Even if the values are const, in the general case they could be
> modified by another module or another thread. The compiler simply
> doesn't have enough information to warn of every runtime overflow.
Unless they're automatic variables and the compiler can determine
their addresses haven't been taken or haven't escaped the current
scope. Escape analysis should be able to do that, ideally.