checkedint call removal

[Daniel Murphy via Digitalmars-d]

"Ary Borenszweig"  wrote in message news:lrbbtl$778$1 at digitalmars.com...

> If assume stays in release mode then that's the difference. The problem 
> then is that assert doesn't stay in release mode so the compiler can't 
> make any assumption about the code that follows the "assert" because 
> asserts are removed in release mode.

assume never creates a runtime check, assert does in non-release mode. 
assert's condition can't be 'assumed' in non-release mode, or the compiler 
could simply assume the check passed an non emit it.

> In my opinion, removing asserts in release-mode is the big problem here. 
> How expensive can an assert be? Do you really want to debug that? Wouldn't 
> it be better to get an exception?

That's the whole point of assert+release.  If you always want an exception, 
simply use enforce.

> If the compiler *never* removes asserts, then assert and assume have the 
> same meaning.

No, because assert puts in a runtime check in non-release.

Hopefully my table helps.  I'm not sure which assert I'm talking about half 
the time.