nothrow and std.exception.ifThrown

[Meta]

On Friday, 30 April 2021 at 13:05:00 UTC, Steven Schveighoffer 
wrote:
> On 4/29/21 1:50 PM, Meta wrote:
>
>> 
>> The reason for this, apparently, is in the definition of 
>> `ifThrown`:
>> ```
>> CommonType!(T1, T2) ifThrown(E : Throwable = Exception, T1, 
>> T2)(lazy scope T1 expression, lazy scope T2 errorHandler) 
>> nothrow
>> ```
>> 
>> It's not marked as `nothrow` in the function's definition, so 
>> even if the delegate passed to ifThrown _is_ nothrow, the 
>> compiler can't tell. There's no easy way around this that I 
>> can think of OTOH that doesn't involve some effort on your 
>> part.
>
> Wait, I don't get what you are saying. You mean it should be 
> marked nothrow? It's a template, so it *should* be inferred 
> nothrow if it were actually nothrow.
>
> The current definition is not marked nothrow as you alluded, 
> and when I do mark it nothrow, it complains that the lazy 
> parameter used for the exception handler is not nothrow.
>
> It seems there's no way to infer the throwing of the lazy 
> parameter, lazy parameters are never nothrow.
>
> The higher order function DIP would I think help with this.
>
> -Steve

Change it to a delegate and it's the same thing. ifThrown being a 
template is irrelevant in this case because it is accepting the 
handler as a function argument, not a template argument. You:

1. Need to make it a delegate instead of a lazy argument.

2. Need to mark the delegate as nothrow.

For the function to be inferred as nothrow.