checkedint call removal

[via Digitalmars-d]

On Monday, 28 July 2014 at 15:52:23 UTC, John Colvin wrote:
> On Monday, 28 July 2014 at 15:20:44 UTC, Ola Fosheim Grøstad 
> wrote:
>> If asserts were used as optimization constraints
>
> all available code is fair game as optimisation constraints. 
> What you are asking for is a special case for `assert` such 
> that the optimiser is blind to it.
>
> bool foo(int a)
> {
>     //let's handwrite a simple assert
>     if(a >= 0)
>     {
>         exit(EXIT_FAILURE);
>     }
>     //and then do something.
>     return a < 0;
> }
>
> Of course the compiler is free to rewrite that as
>
> bool foo(int a)
> {
>     if(a >= 0)
>     {
>         exit(EXIT_FAILURE);
>     }
>     return true;
> }
>
> Why should the situation be different if I use the builtin 
> `assert` instead?

The problem is that in release mode, the asserts are removed. It 
would then be a very big mistake to still take them into account 
for optimizations, as if they were actually there.

To take your code:

     assert(a >= 0);
     return a < 0;

is equivalent to

     assert(a >= 0);
     return true;

but only in non-release mode. In release mode, this effectively 
becomes

     return a < 0;

which is _not_ equivalent to

     return true;

I believe this is was Ola is protesting about, and I agree with 
him. Such optimizations must only happen if the check stays.