Re: assert(false, "...") doesn't terminate program?!

[Walter Bright]

On 10/29/2012 7:51 AM, Don Clugston wrote:> On 27/10/12 20:39, H. S. Teoh wrote:
>> On Sat, Oct 27, 2012 at 08:26:21PM +0200, Andrej Mitrovic wrote:
>>> On 10/27/12, H. S. Teoh <hst...@quickfur.ath.cx> wrote:
>>>>          writeln("how did the assert not trigger??!!");    // how did we 
get
>>>> here?!
>>>
>>> Maybe related to -release?
>> [...]
>>
>> Haha, you're right, the assert is compiled out because of -release.
>>
>> But I disassembled the code, and didn't see the "auto x = 1/toInt()"
>> either. Is the compiler optimizing that away?
>
> Yes, and I don't know on what basis it thinks it's legal to do that.
Because x is a dead assignment, and so the 1/ is removed. Divide by 0 faults are 
not considered a side effect. I think the code would be better written as:

    if (toInt() == 0) throw new Error();

If you really must have a divide by zero fault,

    if (toInt() == 0) divideByZero();

where:

    void divideByZero()
    {
         static int x;
         *cast(int*)0 = x / 0;
    }