checkedint call removal

[John Colvin via Digitalmars-d]

On Wednesday, 30 July 2014 at 15:12:58 UTC, Ary Borenszweig wrote:
> On 7/30/14, 11:44 AM, Daniel Murphy wrote:
>> "Ary Borenszweig"  wrote in message 
>> news:lravtd$2siq$1 at digitalmars.com...
>>
>>> Now, if you compile in release mode, according to Walter, all 
>>> the
>>> "asserts" are gone (which, as a side note, is something I 
>>> don't like:
>>> in every case it should throw an AssertError). So the 
>>> question is: can
>>> the compiler still replace that writeln call? It should, but 
>>> since
>>> there's nothing there preventing x + y to be different than 3 
>>> (the
>>> assertion is gone), the compiler can't replace it anymore.
>>
>> That's the whole point - the compiler theoretically can 
>> optimize as if
>> the assert is checked.
>>
>> (This example uses assert(0) because this behaviour is 
>> actually in the
>> spec)
>>
>> if (x != 3) assert(0);
>> if (x == 3) deleteAllMyFiles();
>>
>> The compiler is allowed to treat assert(0) as unreachable - 
>> and if it's
>> unreachable then it must be impossible for x to be != 3.
>>
>> So it becomes:
>>
>> deleteAllMyFiles();
>>
>> He's asking for assert to mean 'check this condition' and 
>> assume to mean
>> 'optimize as if this is a mathematical identity'.
>
> And how is that different if instead of:
>
> if (x != 3) assert(0);
>
> you write:
>
> assume(x != 3);
>
> ?

assert(0) is treated as a special case, different from 
assert(someRuntimeExpression)