checkedint call removal

[Ary Borenszweig via Digitalmars-d]

On 7/30/14, 4:17 AM, "Ola Fosheim Grøstad" 
<ola.fosheim.grostad+dlang at gmail.com>" wrote:
> On Wednesday, 30 July 2014 at 03:32:50 UTC, Walter Bright wrote:
>> On 7/29/2014 7:08 PM, Timon Gehr wrote:
>>> Of course version(assert) is a language feature. Always double-check
>>> your claims.
>>>
>>> It's even documented: http://dlang.org/version.html
>>
>> You're right. My mistake. I'd forgotten about that.
>>
>>
>>>> Not a bit. The distinction utterly escapes me.
>>> This is unfortunate, but I don't know what else to say.
>>
>> I don't either. I still have no idea what the difference between
>> assume(i<6) and assert(i<6) is supposed to be.
>
> main(){
>    foobar();
>    assert(false==true)
> }
>
> With assert():
>
> main(){
>    foobar();
> }

Could you show an example that's a bit more complex and useful? I'm with 
Walter that I don't see any difference between assert and assume.

For example:

---
int x = ...;
int y = ...;

assert(x + y == 3);

writeln(x + y);
---

If the assert fails then the program crashes in non-release mode. So 
after the assert the compiler knows that "x + y == 3" so that it can 
replace "writeln(x + y)" with "writeln(3)" (assuming the compiler is the 
smartest on earth).

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.

With assume... what's the difference? You say the compiler should 
replace it anyway because that's what you assume? There's nothing left 
at release mode to verify that for you.

Could you elaborate this example using assert and assume?

Thanks!